Mathproblems

ISSN: 2217-446X, url: http://www.mathproblems-ks.org

Volume 6, Issue 1 (2016), Pages 516-558
Editors: Valmir Krasniqi, Sava Grozdev, Armend Sh. Shabani, Paolo
Perfetti, Mohammed Aassila, Mihály Bencze, Valmir Bucaj, Emanuele Callegari,
Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Omran Kouba, Cristinel Mor-
tici, Jozsef Sándor, Ercole Suppa, David R. Stone, Roberto Tauraso, Francisco
Javier Garcı́a Capitán.

PROBLEMS AND SOLUTIONS

Proposals and solutions must be legible and should appear on separate sheets, each
indicating the name of the sender. Drawings must be suitable for reproduction.
Proposals should be accompanied by solutions. An asterisk (*) indicates that nei-
ther the proposer nor the editors have supplied a solution. The editors encourage
undergraduate and pre-college students to submit solutions. Teachers can help by
assisting their students in submitting solutions. Student solutions should include
the class and school name. Solutions will be evaluated for publication by a com-
mittee of professors according to a combination of criteria. Questions concerning
proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail.com

Solutions to the problems stated in this issue should arrive before

June 19, 2017

Problems
145. Proposed by Paolo Perfetti, Department of Mathematics, University Tor Ver-
gata, Rome, Italiy. Let 0 ≤ x ≤ 1. Prove that xx ≤ x2 − x + 1 − x2 (1 − x)4 .

146. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca,

Romania. Let n ≥ 1 be an integer. Solve in M2 (Z) the equation X 2n+1 − X = I2 .

147. Proposed by Anastasios Kotronis, Athens, Greece. Let an be the sequence

defined by the relations
 
b−p−1 b − 2a 2a
an+3 − 1 + an+2 + an+1 + an = 0
n+3 n+3 n+3
and
(b − p)2 − p
a0 = 1, a1 = b − p ∧ a2 = a + ,
2
where a, b ∈ R and R 3 p 6∈ {−2, −1, 0, 1, . . .}.
a+b
e
(1) Show that limn→+∞ np+1 an = Γ(−p) and
 a+b

p+1 e
(2) Find limn→+∞ n n an − Γ(−p) if it exists.

516
 517

148. Proposed by D.M. Bătineţu-Giurgiu, “Matei Basarab” National College, Bucharest,

Romania, and Neculai Stanciu, “George Emil Palade” School, Buzǎu, Romania.
Find   
− sinh2 (t) − sinh2 (t)
cosh2 (t)
lim x (Γ(x + 1)) x
− (Γ(x + 2)) x+1
,
x→∞
where t ∈ R and Γ is the Gamma function.

149. Proposed by Arkady Alt, San Jose, California, USA. Let D be set of strictly
decreasing sequences of positive real numbers with first term equal to 1. For given
P∞ xp+q
real positive p, r and any xN = (x1 , x2 , ..., xn , ...) ∈ D. Let S (xN ) = n=1 pn if
xn+1
P∞ xp+r
series n=1 pn converges and S (xN ) = ∞ if it diverges. Find inf {S (xN ) | xN ∈ D} .
xn+1
150. Proposed by Cornel Ioan Vălean, Timiş, Rumania. Find
∞ X ∞
X H3
(−1)k+n k+n ,
n=1
k+n
k=1
1 1
where Hn = 1 + + · · · + denotes the nth harmonic number.
2 n
151. Proposed by Albert Stadler, Herrliberg, Switzerland. Prove that
∞ ∞  X∞ ∞
X 1X 1 3 π 1 X 1
2
= + coth π 2)
− .
n=1
n 1 + k 2 2 k(1 + k k(1 + k 2 )2
k=n k=1 k=1
518

Solutions
No problem is ever permanently closed. We will be very pleased considering for
publication new solutions or comments on the past problems.

138. Proposed by Leonard Giugiuc, National College Traian, Drobeta Turnu Sev-
erin, Romania. Let a, b, c, x, y and z be real numbers such that a+b+c+x+y+z = 3
and a2 + b2 + c2 + x2 + y 2 + z 2 = 9. Prove that abcxyz ≥ −2.
Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and
Technology Damascus, Syria. The variables a, b ,c, x, y, z will be denoted by
x1 , x2 , x3 , x4 , x5 , x6 for simplicity!. The set
( 6 6
)
X X
6 2
K = (x1 , . . . , x6 ) ∈ R , xi = 3, xi = 9
i=1 i=1
6
Q6
is a compact subset of R , so, the continuous function (x1 , . . . , x6 ) 7→ i=1 xi
attains its minimum µ on K, So, let (a1 , . . . , a6 ) ∈ K such that a1 ≤ a2 ≤ . . . ≤ a6
with ( 6 )
Y
6
µ = a1 a2 · · · a6 = min xi : (x1 , x2 , x3 , x4 , x5 , x6 ) ∈ R
i=1
Since (−2, 1, 1, 1, 1, 1) ∈ K we conclude that µ ≤ −2. This proves that none of the
ai ’s is zero, and that the number of negative ai ’s is odd. Therefore, we have three
cases: P5
(a) a5 < 0 < a6 . In this case we have a6 = 3 − k=1 ai > 3 and consequently
6
9 < a26 < i=1 a2i = 9 which is absurd.
P
(b) a3 < 0 < a4 . In this case we have
3
!2 3
X X
9= (ai + ai+3 ) ≤3 (ai + ai+3 )2
i=1 i=1
so,
6
X
3≤ a2i + 2(a1 a4 + a2 a5 + a3 a6 ) = 9 − 2(|a1 | a4 + |a2 | a5 + |a3 | a6 )
i=1
or
1
(|a1 | a4 + |a2 | a5 + |a3 | a6 ) ≤ 1.
3
The arithmetic mean- geometric mean inequality proves then that |µ| = |a1 | |a2 | |a3 | a4 a5 a6 ≤
1 which is also absurd.
(c) a1 < 0 < a2 . Here we have
6
!2 6
X X
(3 − a1 )2 = ai ≤5 a2i = 5(9 − a21 )
i=2 i=2

this is equivalent to a21 − a1 − 6 ≤ 0 and consequently −2 ≤ a1 < 0. It follows that

6  5  5
Y a2 + a3 + a4 + a5 + a6 3 − a1
0< ai ≤ = ≤1
i=2
5 5
 519

Q6
hence µ = i=1 ai ≥ −2. Consequently µ = −2 which is the desired conclusion.
Solution 2 by S.C. Locke, Department of Mathematical Sciences, Florida
Atlantic University and B. Reinhart (student), Oxbridge Academy.
We may assume that none of the variables a, b, c, x, y or z is zero, since then
abcxyz = 0 > −2.

We use the method of Lagrange multipliers. Let

F (a, b, c, x, y, z, λ, µ) = abcxyz+λ (a + b + c + x + y + z − 3)+µ a2 + b2 + c2 + x2 + y 2 + z 2 − 9 .

Then,

∂F
 
 ∂a 

 ∂F 

 ∂b  
bcxyz + λ + 2µa

∂F
 
 
   acxyz + λ + 2µb 
 ∂c   

 ∂F  
  abxyz + λ + 2µc 

 ∂x   abcyz + λ + 2µx 
∇F =  ∂F = .


 
  abcxz + λ + 2µy 

 ∂y   abcxy + λ + 2µy 

 ∂F  
  a + b + c + x+y+z−3


∂z
  2 2 2

 ∂F

 a + b + c + x2 + y 2 + z 2 − 9
 
 ∂λ 
 ∂F 
∂µ
→
−
We want ∇F = 0 . Thus, for any w ∈ {a, b, c, x, y, z} ,
∂F
w = 0 =⇒ 0 = abcxyz + λw + 2µw2 .
∂w
Hence,

λa + 2µa2 = λb + 2µb2 , λ (a − b) + 2µ a2 − b2 = 0, (a − b) (λ + 2µ (a + b)) = 0.
Either a = b, or λ = −2µ (a + b). Suppose that µ = 0. We have already stated
that we may assume a 6= 0. Thus, abcxyz + λa = 0 and bcxyz = λ. Similarly,
bcxyz = λ = acxyz = abxyz = abcyz = abcxz = abcxy, and a = b = c = x = y =
1 3
z = , which is impossible, since then a2 + b2 + c2 + x2 + y 2 + z 2 = 6= 9. Hence,
2 2
−λ
µ 6= 0. Now, we have a = b or a + b = . Thus, |{a, b, c, x, y, z}| ∈ {1, 2}, and
2µ
we’ve already ruled out |{a, b, c, x, y, z}| = 1. The remaining cases are, without loss
of generality,
2
(i) a = b = c = x = y, and then 5a + z = 3, 5a2 + z 2 = 9, 5a2 + (3 − 5a) = 9,
a = 1, z = −2, and abcxyz = −2.
(ii) a = b = c = x , y = z, and then abcxyz = a4 z 2 > 0.
(ii) a = b = c , x = y = z,3a + 3z = 3, 3a2 + 3z 2 = 9, a + z = 1, a2 + z 2 = 3,
2 3
az = 21 (a + z) − a2 + z 2 = −1, and abcxyz = (az) = −1 > −2.

Therefore, abcxyz > −2, with equality if five of the variables have value one and
one of the variables has value negative two.
520

Solution 3 by Moti Levy, Rehovot, Israel. The following notation and results
will be used in this solution:
1) The elementary symmetric polynomials in n variables are defined as:
X
ek := xj1 · · · xjk , k = 1, 2, . . . , n.
1≤j1 <j2 <···<jk ≤n

2) The elementary symmetric means in n variables are defined as:

ek
Ek := n
, E0 = 1.
k
3) The power sums are defined as:
n
X
pk := xki .
i=1

4) From Newton’s identities,

p1 = e1 , (1)
p2 = e1 p1 − 2e2 . (2)
5) The Newton’s inequalities are:
Ek−1 Ek+1 ≤ Ek2 . (3)
Let us denote, for convenience, x1 = a, x2 = b, x3 = c, x4 = x , x5 = y and x6 = z.
X5
The first constraint can be re-written as x6 = 3 − xi = 3 − e1 .
i=1
We will use the first constraint to eliminate x6 , so we are dealing with elementary
symmetric polynomials in 5 variables.
5
!2 5
X X
The second constraint becomes 3 − xi + x2i = 9,
i=1 i=1
or
2
(3 − p1 ) + p2 = 9,
which is equivalent to p2 − 6p1 + p21 = 0.
We reformulate the problem, in terms of elementary symmetric polynomials and
power sums as:
Show that
e5 (e1 − 3) ≤ 2, (4)
subjected to the constraint,
p2 − 6p1 + p21 = 0. (5)
Substitution of (2) in (5) gives
e2 = p1 (p1 − 3) , (6)
and substitution of (2) in (6) gives
p2 = p1 (6 − p1 ) . (7)
By Cauchy-Schwarz inequality,
p21 ≤ 5p2 . (8)
By (7) and (8)
p21 ≤ 5p1 (6 − p1 ) ,
 521

hence
e1 = p1 ≤ 5, (9)
or
E1 ≤ 1. (10)
Since by definition, p2 ≥ 0, it follows from (7) and (9) that
e1 = p1 ≥ 0.
It follows from (4) and (9) that
e5 (e1 − 3) ≤ e5 (5 − 3) ≤ 2,
or
E5 ≤ 1.
Thus, showing that E5 ≤ 1 subjected to the constraint (5) is equivalent to solving
our original inequality.
To show that E5 ≤ 1 we will use the Newton’s inequalities, which are:
E0 E2 = E2 ≤ E12 , (11)
E 1 E3 ≤ E22 , (12)
E 2 E4 ≤ E32 , (13)
E 3 E5 ≤ E42 . (14)
Clearly E5 = x1 x2 x3 x4 x5 ≤ |x1 | |x2 | |x3 | |x4 | |x5 | , therefore, if we show that E5 ≤ 1
for positive values of xi , we are done.
For positive values of xi , we have: E1 , E2 , E3 > 0.
Using the Newton’s inequalities (11) to (14), we obtain:
 2 2  2 3
E3 E2
E42 E2 E33 E1 E24 E18
E5 ≤ ≤ = 2 ≤ = ≤ = E15 .
E3 E3 E2 E22 E13 E13
But we have shown in (10), that for variables xi , i = 1, . . . , 5, which meet the
constraint, we have E1 ≤ 1, hence E5 ≤ 1.
Also solved by Richdad Phuc, Vietnam; Albert Stadler, Herrliberg,
Switzerland and the proposer.
139. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca,
Romania. Let a, b ∈ R, a < b, and let f : [a, b] → R be a Riemann integrable
function. Calculate
Z b
f (x)
lim dx.
n→∞ a 1 + sin2 x sin2 (x + 1) · · · sin2 (x + n)

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and

Technology Damascus, Syria. Note that,for real x
n n
X n+1 1 X
sin2 (x + k) = − cos(2x + 2k)
2 2
k=0 k=0
n
!
n+1 1 2ix
X
2ik
= − < e e
2 2
k=0
n + 1 cos(2x + n) sin(n + 1) n+3
= − <
2 2 sin 1 2
522

In particular, for all n ≥ 3 and x ∈ R we have

n
1 X 2 3
sin (x + k) ≤
n+1 4
k=0
and by the arithmetic mean-geometric mean inequality we conclude that for all
n ≥ 3 and x ∈ R we have
n  n+1
Y
2 3
sin (x + k) ≤
4
k=0
It follows that, for n ≥ 3 we have
Z b Z b Z b n
f (x) Y
f (x)dx − Qn 2 dx ≤ |f (x)| sin2 (x + k)dx
a a 1 + k=0 sin (x + k) a k=0
 n+1 Z b
3
≤ |f (x)| dx
4 a
Thus,
Z b Z b
f (x)
lim dx = f (x)dx.
n→∞ a 1 + sin2 x sin2 (x + 1) · · · sin2 (x + n) a

Solution 2 by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-

golia. ∀x ∈ R we have
1
≤1 (1)
1 + sin2 x sin2 (x + 1) · . . . · sin2 (x + n)
Using AM-GM inequality we have:
n+1
sin2 x + sin2 (x + 1) + . . . + sin2 (x + n)

2 2 2
sin x·sin (x+1)·. . .·sin (x+n) ≤ .
n+1
Other hand, we get:
n + 1 sin(n + 1) cos(n + 2x)
sin2 x + sin2 (x + 1) + . . . + sin2 (x + n) = − .
2 2 sin 1
Hence we have:
1 1
n+1 ≤ (2)
1 + sin2 x sin2 (x + 1) . . . sin2 (x + n)

1 sin(n+1) cos(n+2x)
1+ 2 − 2(n+1) sin 1

Let
1
gn (x) =  n+1 .
− sin(n+1) cos(n+2x)
1+
2(n+1) sin 1
1
2
 n+1  n+1
1 sin(n + 1) cos(n + 2x) 1 1
0< − ≤ + →0
2 2(n + 1) sin 1 2 2(n + 1) sin 1
hence
∀x ∈ [a, b] : lim gn (x) = 1.
n→∞
From (1) and (2), we have ∀x ∈ [a, b] :
1
lim =1
n→∞ 1 + sin2 x sin2 (x + 1) . . . sin2 (x + n)
 523

Using the Bounded Convergence Theorem, we get that

Z b
f (x)
lim dx
n→∞ a 1 + sin2 x sin2 (x + 1) · . . . · sin2 (x + n)
Z b 
1
= lim f (x)dx
a n→∞ 1 + sin2 x sin2 (x + 1) · . . . · sin2 (x + n)
Z b
= f (x)dx.
a

Solution 3 by Perfetti Paolo, Dipartimento di Matematica, Università

degli studi di
 Tor Vergata
S  5 7 Roma. Let consider

on the 7unit11circle

C the three
subsets: I1 = − π6 , π6 π 5

6 π, 6 π , I2 = ,
6 6 π , I3 = 6 π, 6 π .
Clearly C = I1 ∪ I2 ∪ I3 and let’s observe that the length of I1 is 2π/3.
Now consider the following set
sin2 (x + n + 1),
sin2 (x + n + 2), . . . , sin2 (x + n + 7))
2π
Out of the seven terms written, at least = 3 of them belong to I1 and this
2π/3
means that
1
sin2 (x + n + 1) · sin2 (x + n + 2) · · · sin2 (x + n + 7)) ≤
43
regardless the value of x. Now let’s divide the first N integers (0 included)
in blocks of length 7 getting N7 blocks plus the rest which is a set made of some
integers up 6. It follows
1
sin2 x sin2 (x + 1) · · · sin2 (x + N ) ≤ 3N/7
4
and this implies that
lim sin2 x sin2 (x + 1) · · · sin2 (x + n) = 0
n→∞

uniformly on [a, b]. It follows that

Z b Z b
f (x)
lim dx = f (x)dx
n→∞ a 1 + sin2 x sin2 (x + 1) · · · sin2 (x + n) a

Solution 4 by Michel Bataille, Rouen, France.

1
Let Kn (x) = 1+sin2 x sin2 (x+1)··· sin2 (x+n)
. We show that
Z b Z b
lim Kn (x)f (x) dx = f (x) dx.
n→∞ a a
Rb
From a known result, we have lim Kn (x) dx = b − a (see O. Furdui, Limits,
n→∞ a
Series, and Fractional Part Integrals, Springer, 2013, Problem 1.36, pp. 55-6).
Thus, if (α, β) ⊂ [a, b] and ξ(α,β) is the characteristic function of (α, β), we have
Rb Rβ Rb
lim Kn · ξ(α,β) = lim α Kn = β − α = a ξ(α,β) .
n→∞ a n→∞
Rb
Now, consider the functional L : f 7→ L(f ) = limn→∞ a Kn (x)f (x) dx. Clearly, L
is a linear functional on the linear space of Riemann integrable functions on [a, b].
Rb Rb
Since L(ξ(α,β) ) = a ξ(α,β) , by linearity we also have L(φ) = a φ whenever φ is a
step function from [a, b] to R.
524

Now, let f be any Riemann integrable function on [a, b] and let ε be any positive
Rb
real number. There exists a step function φ such that a |f − φ| ≤ ε. Then, from
Kn f − f = Kn (f − φ) + Kn φ − φ + (φ − f ) and 0 ≤ Kn (x) ≤ 1 for all x ∈ [a, b] we
deduce
Z b Z b Z b
Kn f − f = (Kn f − f )
a a a
Z b Z b Z b Z b
= Kn (f − φ) + Kn φ − φ+ (φ − f )
a a a a
Z b Z b Z b Z b
≤ Kn (f − φ) + Kn φ − φ + (φ − f )
a a a a
Z b Z b Z b Z b
≤ Kn |f − φ| + Kn φ − φ + |φ − f |
a a a a
Z b Z b Z b
≤ 2 |f − φ| + Kn φ − φ .
a a a
Rb Rb
Since L(φ) = limn→∞ a
Kn φ = a
φ, we deduce
Z b Z b Z b
lim sup Kn f − f ≤2 |f − φ| + 0 ≤ 2ε
n→∞ a a a

Rb Rb
and since this holds for any ε > 0, we must have lim supn→∞ a
Kn f − a
f = 0.
Rb Rb
Thus, limn→∞ a Kn f − a f = 0 so that
Z b Z b
L(f ) = lim Kn f = f and we are done.
n→∞ a a

Also solved by Moubinool Omarjee, Lycée Henri IV, Paris, France; Moti
Levy, Rehovot, Israel; Albert Stadler, Herrliberg, Switzerland; Ramya
Dutta, Chennai Mathematical Institute (student), India and the pro-
poser.
140. Proposed by Cornel Ioan Vălean, Timiş, Romania. Find
∞
X Hn3
n=1
(n + 1)2n

Where Hn = 1 + 12 + · · · + n1 denotes the nth harmonic number.

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and
Technology Damascus, Syria.
We will write H0 = 0 for convenience. Since Hn = O(log n), the following power
series
∞ ∞ ∞ ∞
X X X X Hn3 n+1
F (x) = Hn xn , G(x) = Hn2 xn , H(x) = Hn3 xn , I(x) = x ,
n=1 n=1 n=1 n=1
n+1

converge for x ∈ (−1, 1).

 525

(1) For all x ∈ (−1, 1) we have

∞ ∞
X X xn
(1 − x)F (x) = (Hn − Hn−1 )xn = = − log(1 − x).
n=1 n=1
n

thus
log(1 − x)
F (x) = − (1)
1−x
(2) For all x ∈ (−1, 1) we have
∞ ∞  2 !
X X 1
(1 − x)G(x) = (Hn2 − 2
Hn−1 )xn = Hn−1 + − 2
Hn−1 xn
n=1 n=1
n
∞ ∞ ∞
X 2Hn−1 n X xn X Hn n+1
= x + 2
= 2 x + Li2 (x)
n=1
n n=1
n n=1
n+1
∞
Z x X ! Z x
n log(1 − t)
=2 Hn t dt + Li2 (x) = −2 dt + Li2 (x)
0 n=1 0 1−t
= log2 (1 − x) + Li2 (x).
P∞
where Li2 (x) = n=1 xn /n2 is the Dilogarithm. Thus

log2 (1 − x) Li2 (x)

G(x) = + (2)
1−x 1−x
(3) For all x ∈ (−1, 1) we have
∞ ∞  3 !
X X 1
(1 − x)H(x) = (Hn3 − Hn−1
3
)xn = Hn−1 + 3
− Hn−1 xn
n=1 n=1
n
∞ 2 ∞ ∞
X 3Hn−1 X 3Hn−1 n X xn
= xn + x +
n=1
n n=1
n2 n=1
n3
∞ ∞
X Hn2 n+1 X Hn
=3 x +3 2
xn+1 + Li3 (x)
n=1
n + 1 n=1
(n + 1)
∞ ∞
Z x X ! Z x !
1 X H n
=3 Hn2 tn dt + 3 tn+1 dt + Li3 (x)
0 n=1 0 t n=1
n + 1
P∞
where Li3 (x) = n=1 xn /n3 is the Trilogarithm. Using the results of (2) we see
that
x
log2 (1 − t)
x
Z Z
3
(1 − x)H(x) = 3 G(t)dt + dt + Li3 (x)
0 0 2 t
3 x log2 (1 − t)
Z x Z
Li2 (t)
= − log3 (1 − x) + 3 dt + dt + Li3 (x)
0 1−t 2 0 t
Now, recall that for 0 < x < 1 we have
π2
Li2 (x) + Li2 (1 − x) = − log(x) log(1 − x)
6
526

So
x x x x
π2 log(t) log(1 − t) − t)
Z Z Z Z
Li2 (t) dt 2 (1
dt = − dt − dt
0 1−t 6 0 1−t 0 1−t 0 1−t
x 1
π2 log(t) log(1 − t)
Z Z
Li2 (t)
= − log(1 − x) − dt − dt
6 0 1−t 1−x t
x
π2 log(t) log(1 − t)
Z
= − log(1 − x) − dt + Li3 (1 − x) − Li3 (1)
6 0 1−t
But
0
log(t) log2 (1 − t) log(t) log(1 − t) 1 log2 (1 − t)

=− + ·
2 1−t 2 t
Hence
log2 (1 − t)
Z x Z x
Li2 (t) 1
dt + dt
0 1−t 2 0 t
π2 1
= − log(1 − x) + log(x) log2 (1 − x) + Li3 (1 − x) − Li3 (1)
6 2
Thus
log3 (1 − x) π 2 log(1 − x) Li3 (1 − x) 3 Li3 (1)
H(x) = − − +3 −
1−x 2 1−x 1−x 1−x
3 log(x) log2 (1 − x) Li3 (x)
+ + (3)
2 1−x 1−x
(4) Integrating (3) we obtain
1 π2
I(x) = log4 (1 − x) + log2 (1 − x) + 3ζ(3) log(1 − x) − 3 Li4 (1 − x) + 3ζ(4)
4 4
3 x log(t) log2 (1 − t)
Z Z x
Li3 (t)
+ dt +K(x) + dt
2 0 1−t 0 1−t
| {z } | {z }
J(x) K(x)

Now,
log(t) log2 (1 − t)
1−x Z 1
log(1 − t)
Z
J(1 − x) = dt = log2 (t)dt
0 1 − t x t
i1 Z 1
h
2 Li2 (t)
= − Li2 (t) log (t) + 2 log(t)dt
x x t
i1 Z 1
2
h Li3 (t)
= Li2 (x) log (x) + 2 Li3 (t) log(t) − 2 dt
x x t
= Li2 (x) log2 (x) − 2 Li3 (x) log(x) + 2 Li4 (x) − 2ζ(4)

h ix Z x − log(1 − t)
K(x) = − Li3 (t) log(1 − t) − Li2 (t)dt
0 t
Z x 0
= − Li3 (x) log(1 − x) − Li02 (t) Li2 (t)dt
0
1
= − Li3 (x) log(1 − x) − Li22 (x)
2
 527

It follows that,

1 π2 3
I(x) = log4 (1 − x) + log2 (1 − x) + 3ζ(3) log(1 − x) + Li2 (1 − x) log2 (1 − x)
4 4 2
1 2
− 3 Li3 (1 − x) log(1 − x)) − Li3 (x) log(1 − x) − Li2 (x) (4)
2

(5) Fortunately the values of Li2 (1/2) and Li3 (1/2) are known:

π2
 
1 1
Li2 = − log2 2
2 12 2
π2
 
1 7 1
Li3 = ζ(3) + log3 (2) − log(2).
2 8 6 12

We conclude immediately that

π4 log4 (2) π 2
 
1 1
I = ζ(3) log(2) − + + log2 (2)
2 2 288 24 12

1


and the announced answer follows since the desired sum is 2I . 2
P∞
Solution 2 by Moti Levy, Rehovot,
Israel. Let f (z) := n=1 Hn3 z n be the
generating function of the sequence Hn3 n≥1 .
An expression for f (z) can be found in a nice article by Professor István Mezõ,
”Nonlinear Euler Sums”, in the Pacific Journal of Mathematics, Vol.272, No. 1,
2014:

π2
 
1 3
f (z) = ln (1 − z) − ln3 (1 − z) + ln2 (1 − z) ln z + 3Li3 (1 − z) + Li3 (z) − 3ζ (3) ,
−
1−z
2 2
(15)
P∞ n
where Lik (z) := n=1 nz k is the polylogarithmic function.

z ∞
zX ∞ ∞
1X 3 z n Hn3 n
Z Z Z
1 1 X
f (t) dt = Hn3 tn dt = Hn t dt = z (16)
z 0 z 0 n=1 z n=1 0 n=1
n+1

1
Putting z = 2 in (16),

∞ 1
Hn3
X Z 2
=2 f (t) dt. (17)
n=1
(n + 1) 2n 0

1 1 Z 21 3 Z 1
π2 ln (1 − t) ln (1 − t) 3 2 ln2 (1 − t) ln t
Z 2
Z 2
f (t) dt = − dt − dt + dt
0 2 0 1−t 0 1−t 2 0 1−t
Z 21 Z 21 Z 21
Li3 (1 − t) Li3 (t) 1
+3 dt + dt − 3ζ (3) dt.
0 1−t 0 1 − t 0 1 − t
528

1
ln (1 − t)
Z 2 1
dt = − ln2 2, (18)
0 1−t 2
1
ln3 (1 − t)
Z 2 1
dt = − ln4 2, (19)
0 1−t 4
1
ln2 (1 − t) ln t π4 π2 2
Z  
2 1 4 1 7
dt = − − ln 2 − ln 2 + 2Li4 + ζ (3) ln 2, (20)
0 1−t 45 12 6 2 4
Z 21
π4
 
Li3 (1 − t) 1
dt = − Li4 , (21)
0 1−t 90 2
Z 21
Li3 (t) 1 4 1 1 4 7
dt = − π − π 2 ln2 2 + ln 2 + ζ (3) ln 2, (22)
0 1 − t 288 24 24 8
Z 21
1
dt = ln 2. (23)
0 1 − t
∞ 1
Hn3
Z 2 1 2 2 1 4 1 4
π + ζ (3) ln 2 ∼
X
=2 f (t) dt = π ln 2 + ln 2 − = 0.966 3.
n=1
(n + 1) 2n 0 6 12 144

Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-

golia; Refik Zeraoulia, Algeria and the proposer.
141. Proposed by Valmir Krasniqi, University of Prishtina, Republic of Kosova.
(p+1)!px−1 dn
Let p ∈ N, and let Ap (x) = x(x+1)···(x+p) . Prove that (−1)n dx n (ln φ(x)) > 0, for
√
x
Ap (x+1)
all n = 1, 2, 3 . . . and x > 0, where φ(x) = x .
Solution by Moti Levy, Rehovot, Israel. A positive function ϕ is said to be
logarithmically completely monotonic on an interval I if its logarithm ln f satisfies
n dn
(−1) (ln ϕ (x)) ≥ 0
dxn
for n = 1, 2, 3, . . . on I. √
x
Ap (x+1)
So we are asked here to show that ϕ (x) = x is strictly logarithmically
completely monotonic on the interval (0, ∞).

p!px
Γp (x) := . (24)
x (x + 1) · · · (x + p)

(p + 1)!px−1 p+1
Ap (x) = = Γp (x) .
x (x + 1) · · · (x + p) p
√x
Γ(x+1)
Feng Qi and Chao -Ping Chen actually showed in [1], that x is strictly
logarithmically completely monotonic.√
x
Ap (x+1)
I will follow here their footsteps , for x :
Let
p !  
x
Ap (x + 1) 1 p+1
f (x) := ln = − ln x + ln Γp (x + 1) .
x x p
 529

dn n dk n−k
Pn
d
Using Leibnitz’ rule dxn (u (x) v (x)) = k=0 k dxk (u (x)) dx n−k (v (x)) ,

dn
(f (x))
dxn
n−1 n  
(n − 1)! X n dk
   n−k  
(−1) p+1 d 1
=− n
+ k
ln Γp (x + 1) n−k
x k dx p dx x
k=0
    
n   k−1 p+1
n
(−1) n!

p+1
 n   X
d 1 n d d ln p Γ p (x + 1) dn−k
 
1
= + ln Γp (x + 1) +  
nxn p dxn x k dxk−1 dx dxn−k x
k=1
n  n   X n   k−1
dn−k
  
(−1) n! p+1 d 1 n d 1
= + ln Γp (x + 1) + (ψ p (x + 1))
nxn p dxn x k dxk−1 dxn−k x
k=1
n n n   k−1
dn−k
  X  
(−1) n! (−1) n! p+1 n d 1
= + ln Γp (x + 1) + (ψp (x + 1)) n−k
nxn xn+1 p k dx k−1 dx x
k=1
n n n n−k
(n − k)! dk−1
  X
(−1) n! (−1) n! p+1 n! (−1)
= + ln Γp (x + 1) + (ψp (x + 1))
nxn xn+1 p k! (n − k)! xn−k+1 dxk−1
k=1
n n n n−k
dk−1
  X
(−1) n! (−1) n! p+1 n! (−1)
= + ln Γ p (x + 1) + (ψp (x + 1))
nxn xn+1 p k! xn−k+1 dxk−1
k=1
n
!
n   X k k−1
(−1) n! x p+1 k x d
= + ln Γp (x + 1) + (−1) (ψp (x + 1))
xn+1 n p k! dxk−1
k=1

Let
  Xn k k−1
x p+1 k x d
g (x) := + ln Γp (x + 1) + (−1) (ψp (x + 1)) ,
n p k! dxk−1
k=1

so that
n
dn (−1) n!
(f (x)) = g (x) . (25)
dxn xn+1
One can check that
n
d 1 (−1) xn dn
(g (x)) = + (ψp (x + 1)) .
dx n n! dxn
n
d 1
Now we need the Laplace transforms of dx n (ψp (x)) and of xn , for x > 0 (can be

found in [2]):
Z ∞
1 1
= tn−1 e−xt dt
xn (n − 1)! 0
Z ∞
dn n+1 1 − e−(p+1)t n −xt
(ψ p (x)) = (−1) t e dt
dxn 0 1 − e−t
n
1 d 1 (−1) dn
n
(g (x)) = n
+ (ψp (x + 1))
x dx nx n! dxn
Z ∞
1 ∞ e−t 
Z
1 n−1 −xt −(p+1)t

= t e dt − 1 − e tn e−xt dt
n! 0 n! 0 1 − e−t
1 ∞
Z  
t  −(p+1)t
= 1− t 1−e tn−1 e−xt dt.
n! 0 e −1
530

< 1 and 0 < 1−e−(p+1)t < 1 for t > 0, then 1− et −1

t t
1 − e−(p+1)t >


Since 0 < et −1
R∞  
t
1 − e−(p+1)t tn−1 e−xt dt > 0.


0 for t > 0, and 0 1 − et −1
1 d d
Thus xn dx (g (x)) > 0 for x > 0, and dx (g (x)) > 0 for x > 0, which implies that
g (x) > g (0) = 0 on (0, ∞) . (26)
n dn n!
It follows from (25) and (26) that (−1) dx (f (x)) = xn+1 g (x) > 0 for x > 0
n
√
x
Ap (x+1)
and for all n = 1, 2, 3..., which implies that is strictly logarithmicallyx
completely monotonic.
References:
[1] Feng Qi and Chao -Ping Chen, ”A complete monotonicity property of the gamma
function”, J. Math. Anal. Appl. 296 (2004), pages 603-607.
[2] Valmir Krasniqi and Feng Qi, ”Complete monotonicity of a function involving
the p-psi function and alternative proofs”, G. Jour. Math. Anal. 2 (2014), no. 3,
204–208.
Also solved by the proposer.
142. Proposed by D.M. Bătineţu-Giurgiu, “Matei Basarab” National College,
Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School, Buzǎu,
Romania. Let (an )n≥1 such that a1 = 1 and an+1 = (n + 1)!an for all n ∈ N∗ . Let
(bn )n>1 be a positive sequence such that lim bn!n = b > 0. Compute
n→∞
√
2n
bn
lim n√ 2 .
n→∞ an
Solution 1 by Omran Kouba, Higher Institute for Applied √ Sciences and
√
Technology Damascus, Syria. The answer is e1/4 . Let un = 2n n!/ n2 an . Since
Yn n
Y
an = k! = k n+1−k we conclude that
k=1 k=1
n n
1 X 1 X
log un = log k − (n + 1 − k) log k
2n n2
k=1 k=1
n n n
1 k
X 1 1 X 1 X k n−1
= log + log n − 2 log k − 2 (n − k) log − log n
2n n 2 n n n 2n
k=1 k=1 k=1
n   n
1X k 1 k log n 1 X
= − log − − 2 log k
n n 2 n 2n n
k=1 k=1

So we have proved that

n    
1X k 1 k log n
log un = − log + O
n n 2 n n
k=1

But
n   Z 1  
1X k 1 k 1
lim − log = x− log x dx
n→∞ n n 2 n 0 2
k=1
1 Z 1
x2 − x

1−x 1
= log x + dx =
2 0 0 2 4
 531

√
2n
bn
Thus lim un = e1/4 . Finally, because lim bn!n = b > 0 we see that lim √
2n
n!
= 1,
n→∞ n→∞ n→∞
hence
√
2n
√
2n
bn b
lim n√ 2 = lim √ n un = e1/4
n→∞ an n→∞ 2n n!

as announced.
Solution 2 by Angel Plaza, University √ of Las Palmas de Gran Canaria,
√
Spain. The answer is e1/4 . Let un = 2n n!/ n2 an
Let L be the proposed limit. Then
s
n2 /2n bn/2
n2 bn ln ann
ln L = lim ln = lim
n→∞ an n→∞ n2
(n+1)/2
bn+1 bn/2
ln an+1 − ln n
an
= lim
n→∞ 2n + 1
(n+1)/2
bn+1 an
ln n/2 · an+1
bn
= lim
n→∞ 2n + 1
(n+1)/2
bn+1 1
ln n/2 · (n+1)!
bn
= lim
n→∞ 2n + 1
(n+2)/2 (n+1)/2
bn+2 bn+1
ln (n+1)/2 − ln n/2
bn+1 (n+2)! bn (n+1)!
= lim
n→∞ (2n + 3) − (2n + 1)
(n+2)/2
!
n/2
1 bn+2 bn (n + 1)!
= lim ln (n+1)/2
· (n+1)/2
2 n→∞ bn+1 (n + 2)! bn+1
 (n+2)/2  n/2
bn+2 bn
1 bn+1 · bn+1
= lim ln
2 n→∞ n+2
 n/2
n+2
1 (n + 2)
n+1
= lim ln
2 n→∞ n+2
1 1/2 1
= ln e = .
2 4

Also solved by Arkady Alt, San Jose, California, USA; Leonard Giugiuc,
National College Traian, Drobeta Turnu Severin, Romania; Moti Levy,
Rehovot, Israel; George-Florin Serban, Pedagogical High School, Braila,
Romania; Michel Bataille, Rouen, France and the proposers.
143. Proposed by Florin Stanescu, Serban Cioculescu school, city Gaesti, jud.
Dambovita, Romania. We consider two matrices A, B ∈ M2 (R), at least one of
them is not invertible. If A2 + AB + B 2 = 2BA, prove that AB = BA = O2 .

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and

Technology Damascus, Syria.
(a) A + B is singular. indeed by assumption det(BA) = det B det A = 0, so from
(A + B)2 = 3BA we conclude that det(A + B) = 0.
532

(b) Both A and B are singular. Indeed, from the fact that A + B is singular and
the two equalities:
(A − 2B)(A + B) = A2 + AB − 2BA − 2B 2 = −3B 2
(B + A)(B − 2A) = B 2 + AB − 2BA − 2A2 = −3A2
(c) Let α = A, β = B. Since det A = det B = det(A + B) = 0 we conclude that
A2 = αA, B 2 = βB and (A + B)2 = (α + β)(A + B). Taking traces of both sides of
the equality A2 +AB +BA+B 2 = (α+β)(A+B) we conclude that (AB) = (BA) =
αβ. Now, taking traces of both sides of the equality A2 + AB + B 2 = 2BA yields
α2 −αβ +β 2 = 0, and consequently α = β = 0. So, A2 = B 2 = (A+B)2 = O2 . This
implies that AB + BA = O2 and the A2 + AB + B 2 = 2BA becomes AB = 2BA.
These two equalities imply that AB = BA = O2 , which is the desired conclusion.
Solution 2 by Michel Bataille, Rouen, France. We shall use the following
known result: if M ∈ Mn (R) is not invertible, then M 2 = tr(M )M (this results
from the Hamilton-Cayley Theorem since the characteristic polynomial of M is
x2 − tr(M )x + det(M ) = x2 − tr(M )x).
We have (A + B)2 = A2 + AB + BA + B 2 = 3BA, hence det((A + B)2 ) =
9(det A)(det B) = 0 (the latter equality because A or B is not invertible). Thus
A + B is not invertible and so
3BA = (A + B)2 = tr(A + B)(A + B) (1).
Let m = tr(A + B) = tr(A) + tr(B). We first show that m = 0.
If neither A nor B is invertible, then A2 = tA and B 2 = (m − t)B where t = tr(A)
so that tr(A2 ) = t2 and tr(B 2 ) = (m − t)2 . Since tr(AB) = tr(BA), the hypothesis
A2 + AB + B 2 = 2BA gives t2 + (m − t)2 = tr(BA). But, from (1), we have
2
m2 = (tr(A + B))2 = tr((A + B)2 ) = 3tr(BA) and so t2 + (m − t)2 = m3 . This
2 2
rewrites as t − m +m


2 12 = 0, which implies m = 0.
Now, suppose that only one of A and B is invertible and assume that m 6= 0. Since
A + B is not invertible, we have (A + B)X = 0 for some nonzero column vector
X. Then, from (1), we also have BAX = 0 and so B 2 X = B(A + B)X = 0.
This cannot occur if B is invertible (since X 6= 0). If B is not invertible, then
BY = 0 for some column vector Y 6= 0 and so m 3 is an eigenvalue of B (because
BAY = m 3 (A + B)Y = m
3 AY and AY 6
= 0 since A is invertible). But X and
Y are independent vectors (X = αY implies BX = αBY = 0, a contradiction
since (A + B)X = 0 and AX 6= 0) and B 2 X = B 2 Y = 0, hence B 2 = O2 . This
contradicts the fact that B (hence also B 2 ) has a nonzero eigenvalue (namely m 3 ).
We conclude that m = 0.
Now, since m = 0, we already have BA = O2 and we deduce that A = O2 if B is
invertible and that B = O2 if A is invertible, in which cases AB = BA = O2 is
obvious. If neither A nor B is invertible, then A2 = tr(A)A and B 2 = tr(B)B and
so tr(A2 ) = (tr(A))2 and tr(B 2 ) = (tr(B))2 . Since tr(AB) = tr(BA) = tr(O2 ) = 0,
the hypothesis A2 + AB + B 2 = 2BA gives tr(A))2 + tr(B))2 = 0 so that tr(A) =
tr(B) = 0 (tr(A) and tr(B) being real numbers). We deduce A2 = B 2 = O2 and so
AB = 2BA = O2 , as desired.
Also solved by Leonard Giugiuc, National College Traian, Drobeta Turnu
Severin, Romania; George-Florin Serban, Pedagogical High School, Braila,
Romania and the proposer.
 533

144. Proposed by Anastasios Kotronis, Athens, Greece. Show that, as x → π − ,

Z ∞
y cosh(xy) 1 Li2 (e−2π ) ln(1 − e−2π ) 1
dy = + − − + O(π − x),
1 sinh(πy) (π − x)2 2π 2 π 2
2
where Li2 (x) = k≥1 xk2 denotes the Dilogarithm.
P

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and

Technology Damascus, Syria.
Let ε = π − x > 0. Clearly
cosh(xy) = cosh((π − ε)y) = sinh(πy)e−εy + e−πy cosh(εy)
It follows that
y cosh(xy) 2y cosh(εy)
= ye−εy + 2πy
sinh(πy) e −1
Hence
∞ ∞
(1 + ε)e−ε
Z Z
y cosh(xy) 2y
dy = + dy + g(ε)
1 sinh(πy) ε2 1 e2πy−1
where
∞
2y(cosh(εy − 1)
Z
g(ε) = dy
1 e2πy − 1
2
Noting that 0 ≤ cosh u − 1 ≤ u2 eu for u ≥ 0 and that e2πy − 1 ≥ eπy for y ≥ 1 we
conclude that
Z ∞ 3 εy Z ∞
2 y e 2 3 −(π−ε)y 6ε2
0 < g(ε) ≤ ε dy ≤ ε y e dy =
1 eπy 0 (π − ε)4
(1 + ε)e−ε 1 1 ε
In particular, g(ε) = O(ε2 ). Also, clearly we have = 2 − + +O(ε2 ).
ε2 ε 2 3
It follows that
Z ∞ Z ∞
y cosh(xy) 1 1 2y ε
dy = 2 − + 2πy − 1
dy + + O(ε2 )
1 sinh(πy) ε 2 1 e 3
Now it remains to evaluate the integral on the right:
Z ∞ Z ∞ Z e−2π
2y 2ye−2πy 1 ln x
2πy
dy = −2πy
dy = − 2 dx : x ← e−2πy
1 e −1 1 1−e 2π 0 1−x
 e−2π Z e−2π
1 1 ln(1 − x)
= ln(1 − x) ln x − 2 dx
2π 2 0 2π 0 x
1 Li2 (e−2π )
=− ln(1 − e−2π ) +
π 2π 2
−
So, we have shown that as x → π we have
Z ∞
y cosh(xy) 1 Li2 (e−2π ) 1 ln(1 − e−2π ) π − x
dy = 2
+ 2
− − + +O((π −x)2 )
1 sinh(πy) (π − x) 2π 2 π 3
which is stronger than the desired result.
Solution 2 by Ramya Dutta, Chennai Mathematical Institute (student),
India. Z ∞
sinh(xy) 1 x
We start with the integral dy = tan for x ∈ (0, π) · · · (1)
0 sinh(πy) 2 2
534

∞ ∞
e−πy (exy − e−xy )
Z Z
sinh(xy)
dy = dy
0 sinh(πy) 0 1 − e−2πny
∞ Z ∞
X
= e−(2n+1)πy (exy − e−xy ) dy
n=0 0
∞  
X 1 1
= −
n=0
(2n + 1)π − x (2n + 1)π + x
∞  
1 X 1 1
= −
x
2π n=0 n + 12 − 2π x
n + 12 + 2π
    
1 1 x 1 x 1 x
= ψ + −ψ − = tan
2π 2 2π 2 2π 2 2

where, we used the reflection formula for Digamma function,

ψ(1 − z) − ψ(z) = π cot(πz) for z ∈ (0, 1)

Differentiating under the integration sign in (1) we get,

Z ∞
y cosh(xy) 1 x
dy = sec2
0 sinh(πy) 4 2

Now, denoting x = (π − t), as x → π − we have t → 0+ , hence by the Laurent

2 

2
 
x t 2 t 4 1
series expansion, sec2 = csc + O t3 = 2 + + O t2 =


= +
2 2 t 12 t 3
4 1 2


+ + O (π − x) .
(π − x)2 3
Now, from Mean Value theorem, cosh(πy) − cosh(xy) = (π − x) sinh(θx y) for some
sinh(θx y) θx sinh(θx y)
θx ∈ (x, π) and as y → 0+ we have → , i.e., is bounded on
sinh(πy) π sinh(πy)
y ∈ [0, 1].
Hence,

Z 1 Z 1 Z 1
y cosh(xy) y cosh(πy) y sinh(θx y)
dy = dy − (π − x) dy
0 sinh(πy) 0 sinh(πy) 0 sinh(πy)
Z 1
= y coth(πy) dy + O(π − x)
0

Thus, combining the results,

Z ∞ Z 1
y cosh(xy) 1 1
dy = + − y coth(πy) dy + O ((π − x))
1 sinh(πy) (π − x)2 12 0
 535

and since,
1 1
2ye−2πy
Z Z
y coth(πy) dy = y+ dy
0 0 1 − e−2πy
∞ Z 1
1 X
= +2 ye−2πny dy
2 n=1 0
∞ 
e−2πn e−2πn

1 X 1
= +2 − −
2 n=1
4π 2 n2 2πn 4π 2 n2
1 1 1 1
= + + log(1 − e2π ) − 2 Li2 (e−2π )
2 12 π 2π
we have the desired result.
Also solved by Albert Stadler, Herrliberg, Switzerland; Refik Zeraoulia,
Algeria; Moti Levy, Rehovot, Israel and the proposer.
536

——————————————————————————————————-
MATHCONTEST SECTION
——————————————————————————————————-
This section of the Journal offers readers an opportunity to solve interesting and ele-
gant mathematical problems mainly appeared in Math Contest around the world
and most appropriate for training Math Olympiads. Proposals are always wel-
comed. The source of the proposals will appear when the solutions be published.

Proposals
100. Let (an )n∈N be a sequence of real numbers such that lim n (an − 1) = l ∈
n→∞
n  
1
Q
(−∞, ∞) and let p ≥ 1 be a natural number. Calculate lim an + √
p
kn
.
n→∞ k=1

101. Let f, g : [a, b] → R be two nonnegative continuous functions. Assume that f

attains its maximum at a unique point on [a, b] and g attains its maximum at the
same point as f and possibly at other points.
Rb
f n+1 (x)g(x)dx
a
1) Prove that lim Rb
= kf k∞ kgk∞ .
n→∞
f n (x)dx
a
2) Does the result hold under no assumption on f and g?

102. Let f ∈ C 3 (Rn , R) with f (0) = f 0 (0) = 0. Prove that there exist h ∈
C 3 (Rm , Sn (R)) , such that f (x) = xt h (x) x, when Sn (R) , is the set of symmetric
matrix, and xt is the transpose of x.
ei ln(pn )
P
103. Find the nature of the series pn when (pn )n≥1 is the prime number
n≥1
increasing order, and i imaginary complex number.

104. Let a, b, and c be positive real numbers. Prove that

 2  2  2
(6n + 1)a − b (6n + 1)b − c (6n + 1)c − a
+ + ≥ 27
n(b + c) n(c + a) n(a + b)
for any positive integer n ≥ 1.
 537

Solutions
1 1 1 2On n


95. Let n ∈ N and let On = 1 + 3 + ··· + 2n−1 . Calculate lim 1+ .
n→∞ n n

(Jozsef Wildt IMC 2016)

Solution 1 by Omran Kouba, Higher Institute Pn for Applied Sciences and
Technology, Damascus, Syria. Let Hn = k=1 1/k be the nth harmonic num-
ber. It is well-known that Hn = ln n + γ + O n1 where γ is the Euler-Mascheroni

constant. Clearly,
 
1
2On = 2H2n − Hn = ln n + 2 ln 2 + γ + O
n
It follows that
  n     2 
1 2On 2On ln n
ln 1+ = n ln 1 + − ln n = 2On − ln n + O
n n n n
 2 
ln n
= 2 ln 2 + γ + O
n
 n

Hence lim ln n1 1 + 2O


n
n
= 2 ln 2 + γ, and consequently
n→∞
 n
1 2On
lim 1+ = 4eγ .
n→∞ n n
Solution 2 by Angel Plaza, University of Las Palmas de Gran Canaria,
Spain. Let H − n denote the nth harmonic number Hn = 1 + 12 + · · · + n1 , and also
γn = Hn − ln n. It is well-know that lim γn = γ, where γ is the Euler-Marcheroni
n→∞
constant. Then On = H2n − 21 Hn and therefore, the proposed limit may be written
as  n
1 2H2n − Hn
lim 1+ .
n→∞ n n
It follows that  n
1 2H2n − Hn 1 n
xn = 1+ = (1 + an )
n n n
   
2H2n − Hn 2 ln n − ln n ln(4n)
where an = . Also, we note that an = O =O .
n n n
Thus,
a2
 
ln xn = n an − n + . . . − ln n
2
 2 !
ln(4n)
= nan − ln n − n O
n
 2 !
ln(4n)
= 2γ2n + 2 ln(2n) − γn − ln n − ln n − n O
n
 2 !
ln(4n)
= 2γ2n − γn + ln 4 − n O
n
538

and hence, lim ln xn = γ + ln 4 and therefore the proposed limit equals 4eγ .
n→∞
Also solved by Paolo Perfetti, Department of Mathematics, Univer-
sity Tor Vergata, Rome, Italiy; Albert Stadler, Herrliberg, Switzerland;
Michel Bataille, Rouen, France; Arkady Alt, San Jose, California, USA.
96. Let p be a positive real number and let (an )n≥1 be a sequence defined by
an
a1 = 1, an+1 = 1+a p .
n
Find those real values q 6= 0 such that following series
∞ q
(pn)−1/p − an .
P
converges
n=1

(Jozsef Wildt IMC 2016)

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and
Technology, Damascus, Syria. First, the case p = 1 is easy since we have
an = 1/n for every n ≥ 1 in this case, and the considered series converges for every
q > 0 (all the terms are 0 in this case). In what follows we suppose that p 6= 1.
A simple proof by induction shows that (an )n≥1 is positive decreasing. So it must
converge and its limit ` must satisfy ` + `1+p = ` that is ` = 0. Let bn = 1/an . The
sequence (bn )n≥1 is monotone increasing and satisfy lim bn = +∞. Moreover,
n→∞
 
bn+1
∀ n ≥ 1, − 1 bpn = (bn+1 − bn )bnp−1 = 1 (∗)
bn
Now,
Z bn+1 Z 1
bpn+1 − bpn = ptp−1 dt = p(bn+1 − bn ) (bn + s(bn+1 − bn ))p−1 ds
bn 0
Z 1 Z 1
p
p−1 p−1
= bn + sb1−p
n ds = p (1 + sapn ) ds
bp−1
n 0 0

The dominated convergence theorem shows that lim (bpn+1 − bpn ) = p. Using
n→∞
Cezáro’s lemma we conclude that lim (bpn /n) = p. Again, from the above for-
n→∞
mula we have
Z 1
p p p (1 + sapn )p−1 − 1
bn (bn+1 − bn − p) = p ds
0 apn
and the dominated convergence theorem shows that
Z 1
p p p p(p − 1)
lim bn (bn+1 − bn − p) = p(p − 1) s ds =
n→∞ 0 2
Therefore, using lim (bpn /n) = p, we get
n→∞

p−1
lim n(bpn+1 − bpn − p) =
n→∞ 2
or,
 
p−1 1
bpn+1 − bpn =p+ +o
2n n
1 1
Recalling that 1 + 2 + ··· +
= log n + O(1), it follows:
n
  
p p−1 p − 1 log n log n
bn = pn + log n + o(log n) = pn 1 + +o
2 2p n n
 539

So
  −1/p
−1/p p − 1 log n log n
an = (pn) 1+ +o
2p n n
  
−1/p p − 1 log n log n
= (pn) 1− +o
2p2 n n
 
1 p − 1 log n log n
= 1/p 1/p − 2+1/p 1+1/p + o
p n 2p n n1+1/p
Thus
q logq n
an − (np)−1/p ∼
nq(1+1/p)
∞ q
(pn)−1/p − an
P
and the series converges if and only if q(1 + 1/p) > 1 or equiv-
n=1
alently q > p/(p + 1).
Solution 2 by Michel Bataille, Rouen, France. By an easy induction, we
obtain an > 0 for all n ∈ N. It follows that an+1 < an for all n ∈ N so that
{an }n≥1 is decreasing and bounded below, hence convergent. Its limit ` satisfies
` 1
` = 1+` p , hence ` = 0. If p = 1, then an = n for all n ∈ N (by induction) and
1 q
so (pn)− p − an is defined only for q > 0 and vanishes for all n ≥ 1 and all
∞ 1 q
(pn)− p − an is convergent if and only if
P
q > 0. Thus, for p = 1 the series
n=1
x −p
q > 0. Now, we suppose that p 6= 1. Let f (x) = 1+x p and let bn = an . Since

f (x) = x(1 − xp + x2p + o(x2p )) as x → 0+ , we have, as n → ∞,

bn+1 = (f (an ))−p = (an − ap+1

n + a2p+1
n + o(an2p+1 ))−p
p(p + 1) 2p
= a−p p 2p
n (1 + pan − pan + an + o(a2p
n ))
2
p(p − 1) p
= bn + p + an + o(apn ).
2

We deduce bn+1 − bn ∼ p and bn+1 − bn − p ∼ p(p−1) 2 apn . At this point, we

shall use the following form of Stolz’s Theorem: if un ∼ vn > 0 as n → ∞
P n
P Pn
and the series n≥1 vn is divergent, then uk ∼ vk as n → ∞. From
k=1 k=1
1
bn+1 − bn ∼ p, we then deduce bn ∼ np, that is, apn ∼ np as n → ∞. It follows
p−1 1
that bn+1 − bn − p ∼ 2 · n and a second application of Stolz’s Theorem gives
n
bn − np ∼ p−1 1 p−1
P
2 k ∼ 2 ln(n). It follows that
k=1

 −1/p
p−1
an = b−1/p
n = np + ln(n) + o(ln(n))
2
 
−1/p p − 1 ln(n)
= (np) 1− · + o((ln(n))/n)
2p2 n
ln(n) 1
= (np)−1/p − α · 1+ 1 + o((ln(n))/n1+ p )
n p
540

p−1
where we set α = 2+ 1
. As a result,
2p p

1 q (ln(n))q
(pn)− p − an ∼ |α| 1
nq(1+ p )
as n → ∞. From known results about Bertrand’s series, we deduce that the series
∞ 1 q
(pn)− p − an is convergent if and only if q(1 + p1 ) > 1 i.e. q > p+1
p
P
. In
n=1
conclusion, convergence occurs if and only if (p = 1 and q > 0) or (p 6= 1 and
p
q > p+1 ).
97.Let n ∈ N∗ , and for an integer k such that 1 ≤ k ≤ n let nk be the remainder on
euclidean division of n by k. Finally, define pn to be the probability that nk ≥ k2 .
Calculate pn and find lim pn .
n→∞

(Jozsef Wildt IMC 2016)

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and
Technology, Damascus, Syria.
Let q = nk so that n = qk + nk with 0 ≤ nk < k. Now,
 
 2nif nk ≥ k/2 then
2n
2n
=  + 2nk − k with 0 ≤ 2nk − k < k so 2q + 1 = k or equivalently
 (2q +n1)k
k − 2 k = 1. On
 the
 other hand, if 2nnk < k/2
 n then 2n = 2qk + 2nk with
0 ≤ 2nk < k so 2q = 2nk or equivalently k − 2 k = 0. We have proved that
  (
2n jnk 1 if nk ≤ k/2
−2 =
k k 0 if nk < k/2
It follows that
n   j n k n  n o  
X 2n X n 2n
Card{1 ≤ k ≤ n : nk ≥ k/2} = −2 = 2 −
k k k k
k=1 k=1

(where {x} is the fractional part of x). Consequently,

n  n o   n  
1X n 2n 1X k
pn = 2 − = f
n k k n n
k=1 k=1
1 2
where f (x) = 2 x − x . So, pn is a Riemann sum of a Riemann integrable
function, (because f is bounded and continuous on (0, 1) \ { 1j , j ∈ N∗ }). It follows
that
Z 1 Z ∞ Z ∞ Z ∞
2{t} − {2t} {t} {2t}
lim pn = f (x)dx = 2
dt = 2 2
dt − dt
n→∞ 0 1 t 1 t 1 t2
Z ∞ Z ∞ Z 2 Z 2
{t} {u} {t} t−1
=2 2
dt − 2 2
du = 2 2
dt = 2 dt
1 t 2 u 1 t 1 t2
 2
2
= 2 log t + = 2 log 2 − 1 ≈ 0.386294.
t 1

Solution 2 by Michel Bataille, Rouen, France. Let bac denote the integer
part of the real number a and {a} = a−bac its fractional part. Since nk = n− nk k,
 

it is readily seen that the condition nk ≥ k2 is equivalent

n
 n to 2 k ≥ 1. Observing
that 0 ≤ {a} < 1 for all positive real a (so that 0 ≤ 2 k < 2), we finally obtain
 541

k
when 2 nk when 2 nk k
     
that nk ≥ 2 = 1 (and nk < 2 = 0). It follows that
n  
the number of k ∈ {1, 2, . . . , n} such that nk ≥ k2 is equal to 2 nk
P 
and so
k=1

n
1 X j n n ok
pn = 2 .
n k
k=1

To evaluate lim pn , we consider the function f defined on [0, 1] by f (0) = 0 and

  n→∞
f (x) = 2 x1

if x ∈ (0, 1]. The number pn is a Riemann sum attached to this
1 1
function f . If m is any positive integer, we have f (x) = 0 for x ∈ ( m+ 1 , m ] and
2
1 1
f (x) = 1 for x ∈ ( m+1 , m+ 1 ]. Thus the points of discontinuity of f are all in the
2
1 1
set formed by 0 and the numbers m , m+ 1 (m ∈ N). Thus, f is continuous almost
2
everywhere and bounded, hence Riemann integrable on [0, 1]. As a result, we have

Z 1 ∞ Z 1/(m+1/2) ∞  
X X 1 1
lim pn = f (x)dx = 1 · dx = 1 − .
n→∞ 0 m=1 1/(m+1) m=1
m+ 2
m+1

Now, if N is a positive integer, we have

N   N N
X 1 1 X 2 X 1
1 − = −
m=1
m+ 2
m+1 m=1
2m + 1 m=1
m +1

1 1
= 2(H2N +1 − HN − 1) − (HN +1 − 1) = 2H2N +1 − 2HN − 1 −
2 N +1

N
1
P
where HN = k is the N th harmonic number. Since HN = ln(N ) + γ + o(1) as
k=1
N → ∞ (where γ is the Euler constant), we obtain
 
1
lim pn = lim 2[ln(2N + 1) + γ − ln(N ) − γ + o(1)] − 1 − = 2 ln(2)−1.
n→∞ N →∞ N +1

Also solved by Albert Stadler, Herrliberg, Switzerland; Paolo Perfetti,

Department of Mathematics, University Tor Vergata, Rome, Italiy. and
Arkady Alt, San Jose, California, USA.
98. Let (xn )n≥0 be the sequence defined inductively by xn+2 = xn+1 − 21 xn with
P∞ xn
initial terms x0 = 2 and x1 = 1. Find .
n=1 n + 2

(Jozsef Wildt IMC 2016)

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and
Technology, Damascus, Syria. A simple verification, by mathematical induction
542

shows that xn = 4 · 2−(n+2)/2 sin (2+n)π

4 for n ≥ 0. It follows that
∞ ∞ ∞ n !
1 eiπ/4

X xn X sin(nπ/4) X
=4· √ =4·= √
n=0
n+2 n=2
n( 2)n n=2
n 2
∞
!
  n
1+i X 1 1+i
=4·= − +
2 n=1
n 2
  
1+i
= −2 − 4 · = Log 1 −
2
 
iπ
= −2 − 4 · = − = π − 2,
4
∞
X xn
where Log is the principal branch of the logarithm. Finally, = π − 3.
n=1
n +2
Solution 2 by Julian Spahr (student) and Ángel Plaza, University of Las
Palmas de Gran Canaria, Spain. Since the roots of the characteristic equation
are a = 21 − 2i and b = 12 + 2i and taking into acount the initial values x0 = 2 and
x1 = 1, it follows that xn = an + bn .
Since |a| = |b| = √12 , the series expansions n≥0 (az)n = 1−az1
and n≥0 (bz)n =
P P
1
√ √
1−bz are valid for |z| < 2, and for |z| < 2 we may add and integrate so
∞ ∞
1X 1
an + bn
Z Z  
X
n n n+1 z z
= (a + b ) z dz = + dz
n=0
n+2 0 n=0 0 1 − az 1 − bz
Z 1 
4 1
= −2 + 2 dz = [−2z − 4 arctan(1 − z)]0
0 z − 2z + 2
= −2 + π.
∞ ∞
X xn X an + bn
Therefore, the proposed sum is = −1 + = π − 3. Let us
n=1
n+2 n=0
n+2
∞
X zn
consider the sum f (z) = for z ∈, with |z| < 1. Then
n=1
n+2

∞ ∞ Z
1 X z n+2 1 X z n+1
f (z) = = 2 w dw
z 2 n=1 n + 2 z n=1 0
z2
2 + z + ln(1 − z)
= − .
z2
√
   
1+i 1 π
Since ln(c + di) = ln c2 + d2 + i arctan(d/c), then ln = ln √ + i,
    2 2 4
1−i 1 π
and ln = ln √ − i. Therefore
2 2 4
∞    
X xn 1+i 1−i
=f +f = π − 3.
n=1
n+2 2 2
 543

Also solved by Albert Stadler, Herrliberg, Switzerland; Michel Bataille,

Rouen, France; Paolo Perfetti, Department of Mathematics, University
Tor Vergata, Rome, Italiy.
99. Find all functions f : N → N such that
mn · (f (m) − nm) · (n − f (n2 ))
is a square for all m, n ∈ N.
(Proposed by Valmir Krasniqi, Departament of Mathematics, University of
Prishtina, Republic of Kosova.)
Solution by PCO - AOPS. Let g(m, n) = mn(f (m) − mn)(n − f (n2 )), then
g(1, 1) = −(f (1) − 1)2 can only be a perfect square when f (1) = 1. If f (n2 ) = n
∀n ∈ N, we get a first trivial solution : f (n2 ) = n, ∀n ∈ N and f (m) is any value
when m is not a perfect square.
Suppose now that ∃a such that f (a2 ) 6= a (a − 1)2 g(m, a) = (1 − a)m(f (m) −
ma)g(1, a) and, since both g(m, a) and g(1, a) 6= 0 are perfect squares, we get
4(a − 1)m(am − f (m)) is a perfect square. This may be written (2ma − m −
f (m))2 − (f (m) − m)2 is a perfect square ∀m. If infinitely many such a exist,
this implies f (n) = n∀n which indeed is a solution. Suppose now that ∃ finitely
many a such that f (a2 ) 6= a. So f (n2 ) = n ∀n > M for a given M. Let then such
n > M, then (a−1)2 g(n2 , a) = n3 (a−1)(an−1)g(1, a). And since both g(n2 , a) and
n2 g(1, a) 6= 0 are perfect squares, we get n(a − 1)(an − 1) perfect square ∀n > M.
Which is clearly impossible, since a 6= 1. So no more solutions.
Also solved by the proposer.
544

——————————————————————————————————-
MATHNOTES SECTION
——————————————————————————————————-
The evaluation of a special fractional part integral with an
integrand raised to positive integer powers

Cornel Ioan Vălean

Abstract. The present paper is about calculating in closed-form the following

class of special fractional part integrals
Z 1 Z 1 Z 1    n
x y z
dxdydz
0 0 0 y z x
where {x} is the fractional part of x and n is a positive integer. We show the
value of the integral reduces to a sum involving specific values of the Riemann zeta
function.
Keywords: Fractional part integrals, Leibniz integral rule, Riemann zeta function.

1. Introduction and the main result

Let n be positive integers and In denotes the integral with fractional part of power
n
Z 1 Z    n
1 Z 1
x y z
In = dxdydz,
0 0 0 y z x
where {x} denotes the fractional part of x.
The aim of the present paper is to calculate in closed-form the proposed triple
fractional part integral with integer powers. We will show that its closed-form can
be expressed in terms of specific values of the Riemann zeta function.
We state below the theorem we are going to prove.
Let n ≥ 1 be a positive integer. Then the following equality holds:
Z 1 Z 1 Z 1    n
x y z
In = dxdydz
0 0 0 y z x
n n
! n !
3 X 1 X X
= 1− ζ(i + 1) + ζ(i + 1) (i + 1)ζ(i + 2) ,
2(n + 1) i=1 (n + 1)2 (n + 2) i=1 i=1

where ζ(k), k ≥ 2 denotes the Riemann zeta function at positive integer values, and
it is defined by
∞
X 1 1 1 1
ζ(k) = k
= 1 + k + k + ··· + k + ··· .
n=1
n 2 3 n
Before we prove Theorem 1 we collect some results we need in our analysis.
Next we prove the following lemma which is used in the proof of Theorem 1.
Lemma 1. Two special fractional part integrals
Let n ≥ 1 be an integer. The following equalities hold :
 545

Z  n
1 n
1 n 1 X
(a) In,n = x dx = 1 − ζ(i + 1);
0 x n + 1 i=1
Z 1  n n
1 1 1 X
(b) In+1,n = xn+1 dx = − (i + 1)ζ(i + 2).
0 x 2 (n + 2)(n + 1) i=1

Proof. (a) We start with the change of variable x = 1/y, and then we have that
Z 1  n Z ∞
1
x n
dx = y −n−2 {y}n dy
0 x 1
X∞ Z k+1
= y −n−2 {y}n dy
k=1 k
∞ Z k+1
X (27)
= y −n−2 (y − byc)n dy
k=1 k
X∞ Z k+1
= y −n−2 (y − k)n dy
k=1 | k
{z }
Jn,n,k
R c+1
where we denote the last integral by Jn,n,k , where Ja,b,c = c
x−a−2 (x − k)b dx.
Integrating by parts, we obtain that
k+1 0 y=k+1 Z k+1
y −n−1 y −n−1
Z 
n
Jn,n,k = − (y − k)n dy = − (y − k)n + y −n−1 (y − k)n−1 dy
k n+1 n+1 y=k n + 1 k
1 n
=− + Jn−1,n−1,k
(n + 1)(k + 1)n+1 n+1
whence
1 n
Jn,n,k = − n+1
+ Jn−1,n−1,k
(n + 1)(k + 1) n+1
or rearranging, we get that
1
(n + 1)Jn,n,k − nJn−1,n−1,k = − .
(k + 1)n+1
If replacing n by i in the relation above
1
(i + 1)Ji,i,k − iJi−1,i−1,k = − ,
(k + 1)i+1
and then give values from i = 1 to n, we get by the telescoping process that
n
1 X 1
(n + 1)Jn,n,k = −
k(k + 1) i=1 (k + 1)i+1

or
n
1 1 X 1
Jn,n,k = − . (28)
k(k + 1)(n + 1) n + 1 i=1 (k + 1)i+1
546

Now, we use (28) in (27), and then we get that

Z 1  n ∞ Z k+1
1 X
In,n = xn dx = y −n−2 (y − k)n dy
0 x k
k=1
X∞
= Jn,n,k
k=1
∞ n
!
X 1 1 X 1
= −
k(k + 1)(n + 1) n + 1 i=1 (k + 1)i+1
k=1
∞ ∞ n
1 X 1 1 XX 1
= − (29)
n+1 k(k + 1) n + 1 i=1
(k + 1)i+1
k=1 k=1
n ∞
1 1 XX 1
= −
n + 1 n + 1 i=1 (k + 1)i+1
k=1
n n
1 1 X 1 X
= − ζ(i + 1) + 1
n + 1 n + 1 i=1 n + 1 i=1
n
1 X
=1− ζ(i + 1).
n + 1 i=1
and the part (a) of the lemma is proved.
(b) Following the same steps as in the part (a) of the lemma, we get that
Z 1  n Z ∞
1
In+1,n = xn+1
dx = y −n−3 {y}n dy
0 x 1
X∞ Z k+1
= y −n−3 {y}n dy
k=1 k
∞ Z k+1
X (30)
= y −n−3 (y − byc)n dy
k=1 k
X∞ Z k+1
= y −n−3 (y − k)n dy
k=1 | k
{z }
Jn+1,n,k
R c+1
where we denote the last integral by Jn+1,n,k , where Ja,b,c = c x−a−2 (x − k)b dx.
Integrating by parts, we obtain that
Z k+1  −n−2 0 y=k+1 Z k+1
y y −n−2 n
Jn+1,n,k = − n
(y − k) dy = − (y − k)n
+ y −n−2 (y − k)n−1 dy
k n+2 n+2 y=k n+2 k
1 n
=− + Jn,n−1,k
(n + 2)(k + 1)n+2 n+2
whence
1 n
Jn+1,n,k = − + Jn,n−1,k
(n + 2)(k + 1)n+2 n+2
or rearranging, we get that
n 1
Jn+1,n,k − Jn,n−1,k = − .
n+2 (n + 2)(k + 1)n+2
 547

Multiplying both sides by (n + 1)(n + 2), we get that

n+1
(n + 2)(n + 1)Jn+1,n,k − (n + 1)nJn,n−1,k = − .
(k + 1)n+2
If replacing n by i in the relation above
i+1
(i + 2)(i + 1)Ji+1,i,k − (i + 1)iJi,i−1,k = −
(k + 1)i+2
and then give values from i = 1 to n, we get by the telescoping process that
n
1 1 X i+1
(n + 2)(n + 1)Jn+1,n,k = 2 − −
k (k + 1)2 i=1 (k + 1)i+2
or
  n
1 1 1 1 X i+1
Jn+1,n,k = − − .
(n + 2)(n + 1) k2 (k + 1)2 (n + 2)(n + 1) i=1 (k + 1)i+2
(31)
Now, we use (31) in (30), and we get
Z 1  n
n+1 1
In+1,n = x dx
0 x
X∞ Z k+1
= y −n−3 (y − k)n dy
k=1 k

X∞
= Jn+1,n,k
k=1
∞   n
!
X 1 1 1 1 X i+1
= − −
(n + 2)(n + 1) k 2 (k + 1)2 (n + 2)(n + 1) i=1 (k + 1)i+2
k=1
∞   ∞ X n
1 X 1 1 1 X i+1
= − −
(n + 2)(n + 1) k2 (k + 1)2 (n + 2)(n + 1) i=1
(k + 1)i+2
k=1 k=1
n ∞
1 1 XX i+1
= −
(n + 2)(n + 1) (n + 2)(n + 1) i=1 (k + 1)i+2
k=1
n
1 1 X
= − (i + 1)(ζ(i + 2) − 1)
(n + 2)(n + 1) (n + 2)(n + 1) i=1
n n
1 1 X 1 X
= − (i + 1)ζ(i + 2) + (i + 1)
(n + 2)(n + 1) (n + 2)(n + 1) i=1
(n + 2)(n + 1) i=1
n
1 1 X
= − (i + 1)ζ(i + 2).
2 (n + 2)(n + 1) i=1
and the second part of the lemma is proved. 
Now we are ready to prove Theorem 1.
Proof. By letting the variable change x/y = u, we get that
Z 1 Z 1 Z 1    n   Z 1 Z 1 Z 1/y    n ! !
x y z y z
In = dx dy dz = y {u} du dy dz,
0 0 0 y z x 0 0 0 z uy
548

and after changing the integration order

Z 1 Z 1/y Z 1    n  !
y z
In = y {u} dz du dy,
0 0 0 z uy
we make the change of variable z/y = t, and we get that
Z 1 Z 1/y Z 1/y    n ! !
1 t
In = y2 {u} dt du dy. (32)
0 0 0 t u
Now, recall the Leibniz integral rule (see [1])
Z b(z) Z b(z)
∂ ∂f ∂b ∂a
f (x, z)dx = dx + f (b(z), z) − f (a(z), z) ,
∂z a(z) a(z) ∂z ∂z ∂z
R b(x)
where applying the mentioned rule for a function of the form g(z, x) = 0 f (y, z) dy,
we obtain that
Z a(x) ! Z a(x) Z b(x) ! !
∂ ∂
g(z, x)dz = f (y, z)dy dz
∂x 0 ∂x 0 0
Z a(x) Z b(x)
∂ ∂
= (b(x)) f (b(x), z)dz + (a(x)) f (y, a(x))dy.
∂x 0 ∂x 0
Using this result in (32) where we apply the integration by parts, we have that
Z 1  3 0 Z 1/y Z 1/y    n ! !
y 1 t
In = {u} dt du dy
0 3 0 0 t u
  n ! y=1   n ! !
y 3 1/y
Z 1/y 
1 1 3 ∂
Z Z Z 1/y Z 1/y 
1 t 1 t
= {u} dt du − y {u} dt du dy
3 0 0 t u y=0 3 0 ∂y 0 0 t u
  n 
1 1
Z Z 1 
1 t
= {u} dt du
3 0 0 t u
  n !
y 3 1/y
Z Z 1/y 
1 t
− lim {u} dt du
y→0 3 0 0 t u
 n !   n !
1 1
Z 1/y 
1 1
Z Z Z 1/y 
1 1 1
+ y {u}{y} du dy+ y {ty} dt dy
3 0 0 uy 3 0 0 y t
  n 
1 1
Z Z 1
1 t
= un dt du
3 0 0 t u

1 1/y  n !   n !

1 1 1/y
Z Z  Z Z 
1 1 1 1
+ y {u}{y} du dy+ y {ty} dt dy, (∗)
3 0 0 uy 3 0 0 y t
where weused that
 the limit 
  tends
n
to
 0 since
R 1/y R 1/y 1 t
0≤ 0 0
{u} t u dt du ≤ y12 , and then

1/y 1/y   n !

y3
Z Z 
1 t y
0≤ {u} dt du ≤ .
3 0 0 t u 3
 549

Now we make the changes of variable uy = w in the second integral and ty = z in

the third integral, and then we get, for the second integral, that
Z 1 Z 1/y   n ! Z 1 Z 1     n 
1 w 1
y {u}{y} du dy = {y} dw dy
0 0 uy 0 0 y w
Z 1 Z 1    n 
w 1
= yn dw dy,
0 0 y w
(33)
and for the third integral we obtain that
Z 1 Z 1/y    n ! Z 1 Z 1    n 
1 1 1 y
y {ty} dt dy = {z} dz dy
0 0 y t 0 0 y z
Z 1 Z 1   n 
1 y
= zn dz dy.
0 0 y z
(34)
Using (33) and (34) in (*), we get
  n  Z Z 1    n
1 1
Z Z 1
1 1

n 1 t n w 1
In = = u dt du + y dw dy
3 0 0 t u 3 0 0 y w
Z Z 1   n 
1 1 1 y
+ zn dz dy
3 0 0 y z
Z 1 Z 1   n 
1 t
= un du dt,
0 0 t u
where above we used the fact that all three integrals are equal, and to see that it’s
enough to change the integration order in the first and second integrals. 

Next, we split the last integral, and we have that

Z 1 Z 1   n 
n 1 t
In = u du dt
0 0 t u
Z 1 Z t   n  Z 1 Z 1   n 
1 t 1 t
= un du dt + un du dt
0 0 t u 0 t t u
(35)
If making the change of variable t/u = v in the first integral from (35), we get that
Z 1 Z t   n  Z 1 Z ∞ n+1   n 
1 t t 1
un du dt = n+2
v dv dt
0 0 t u 0 1 v t
Z 1 Z 1   n 
v=1/s n+1 n 1 1
= t s ds dt
0 0 t s
Z 1   n Z 1  n
1 1
= tn+1 dt sn ds
0 t 0 s
n
! n
!
1 X 1 1 X
= 1− ζ(i + 1) − (i + 1)ζ(i + 2) ,
n + 1 i=1 2 (n + 2)(n + 1) i=1
(36)
where we used both results of the Lemma 2.
550

For the other integral in (35), we have that

Z 1 Z 1   n  Z 1 Z 1  n 
n 1 t 1
u du dt = tn du dt
0 t t u 0 t t
Z 1  n
1
= (1 − t)tn dt
0 t
Z 1  n Z 1  n
1 1
= tn dt − tn+1 dt
0 t 0 t
n n
1 1 X 1 X
= + (i + 1)ζ(i + 2) − ζ(i + 1).
2 (n + 2)(n + 1) i=1 n + 1 i=1
(37)
Hence, plugging (36) and (37) in (35), we obtain that
n n
! n !
3 X 1 X X
In = 1− ζ(i+1)+ ζ(i + 1) (i + 1)ζ(i + 2) .
2(n + 1) i=1 (n + 1)2 (n + 2) i=1 i=1

Editor’s comment: Lemma 1, parts a and b are not new. Part (a) is Problem
2.21 on page 103 in [2] and part (b) appears, in a more general form, as part (a) of
problem 2.22 on page 103 in [2].

References
[1] Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with For-
mulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 11, 1972.
[2] Ovidiu Furdui, Limits, Series and Fractional Part Integrals. Problems in Mathematical Anal-
ysis, Springer, New York, 2013.

Cornel Ioan Vălean

Teremia Mare, Nr. 632, Timis, 307405, Romania
E-mail: cornel2001 ro@yahoo.com
 551

——————————————————————————————————-
JUNIOR PROBLEMS
——————————————————————————————————-
Solutions to the problems stated in this issue should arrive before June 19, 2017.

Proposals
61. Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam. Given
a tetrahedron A1 A2 A3 A4 with the volume V, let I and r be incenter and inradius,
respectively. Denote by Si the area of triangle opposite to vertex Ai (i = 1; 2; 3; 4).
Prove that

4
X 2rS1 S2 S3 S4 X
Si IA2i = Ai Aj sin ∠(Ai , Aj ),
n=1
9V 2
1≤i<j≤4

where ∠(Ai , Aj ) is the dihedral angle at edge Ai Aj .

62. Proposed by Daniel Sitaru, Mathematics Department, Colegiul National Eco-

nomic Theodor Costescu, Drobeta Turnu - Severin, Mehedinti, Romania. Let be
A0 , A00 ∈ (BC); B 0 , B 00 ∈ (AC); C 0 , C 00 ∈ (AB) in ∆ABC such that AA0 ∩ BB 0 ∩
CC 0 6= ∅ and AA00 ∩ BB 00 ∩ CC 00 6= ∅. Prove that
3
27[A0 B 0 C 0 ] BA0 CB 0 AC 0

≤ + + ,
[A00 B 00 C 00 ] BA00 CB 00 AC 00
where [ABC] is area of triangle ABC.

63. Proposed by Leonard Giugiuc, National College Traian, Drobeta Turnu Sev-
erin, Romania. Let a, b, c ∈ R. Prove that
√ √
3
9 2(ab(a − b) + bc(b − c) + ca(c − a)) ≤ 3 (a − b)2 + (b − c)2 + (c − a)2 2 .

64. Problem proposed by Arkady Alt, San Jose, California, USA. Let ∆ (x, y, z) :=
2(xy + yz + xz) − (x2 + y 2 + z 2 ) and let a, b, c be sidelengths of a triangle with
area F. Prove that

64F 3
∆ a3 , b3 , c3 ≤ √ .
3

65. Proposed by Dorlir Ahmeti, University of Prishtina, Department of Mathemat-

ics, Republic of Kosova. Find all function f : N → N such that mf (n) + f (m) is
divisible by f (m)(f (n) + 1) for all m, n ∈ N.

Solutions
552

56. Proposed by Valmir Krasniqi, University of Prishtina, Republic of Kosova.

Find all functions f : N → N such that f (m! + n!)|f (m)! + f (n)! and m + n divides
f (m) + f (n) for all m, n ∈ N.

Solution by Dorlir Ahmeti, University of Prishtina, Department of Math-

ematics, Republic of Kosova. Taking m = n in second condition we have
n|f (n). Let be p very large prime number. Now, from Wilson theorem, we have
(p − 1)! + 1 ≡ 0(modp). Hence from first relation and last relation we have

p|f ((p − 1)! + 1)|f (p − 1)! + f (1)!

Now, since p is very large, then we have that gcd(f (1), p) = 1, so we can’t have
f (p − 1) ≥ p which means f (p − 1) ≤ p − 1.
Now taking m = p − 1 in second relation and using last relation we have

p − 1 + n|p − 1 + f (n) ⇒ p − 1 + n|f (n) − n

Since p is very large the only possible last raltion to be true is f (n) − n = 0,
so f (n) = n for every positive integer n. We prove to two relation and we see this
is only solution hence done.
Also solved by the proposer.
57. Proposed by Angel Plaza, University of Las Palmas de Gran Canaria, Spain.
Let x1 , x2 , . . . , xn > 0, with the assumption xn+1 = x1 . Prove that
 Pn 2 n Pn
k=1 xk 1 X x2k + xk xk+1 + x2k+1 x2
≤ ≤ k=1 k
n n 3 n
k=1

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and

Technology, Damascus, Syria.
We consider Cn equipped with the usual scalar product h·, ·i and the corresponding
norm k · k defined by
n
X n
X 2
hu, vi = uk vk , kuk2 = |uk | .
k=1 k=1

Now, consider the real vectors

     
x1 x2 1
 x2   x3  1
     
X =  ...  , Y =  ...  , 1 =  ...  .
     
     
xn−1  xn  1
xn x1 1
√

Pn let Z = X − jY with j = e
and 2iπ/3
= − 12 + i 23 . Clearly we have h1, Zi =
( k=1 xk ) (1 − j), hence
n
!2
2
X
|h1, Zi| = 3 xk .
k=1
 553

2
Moreover, since |xk − jxk+1 | = x2k + xk xk+1 + x2k+1 we conclude that

n
X
2
kZk = (x2k + xk xk+1 + x2k+1 )
k=1

2
Therefore, the Cauchy-Schwarz inequality |h1, Zi| ≤ k1k2 kZk2 is equivalent to the
lower inequality.
The upper inequality is easier, since hX, Y i ≤ kXk kY k = kXk2 , so

n
X
(x2k + xk xk+1 + x2k+1 ) = kXk2 + hX, Y i + kY k2 ≤ 3kXk2
k=1

and the is equivalent to the upper inequality.

Remark. We only need the fact that the xk ’s are real. The positivity assumption
is unnecessary.
Solution 2 by Arkady Alt, San Jose, California, USA. For cyclic sum
Pn x2k + xk xk+1 + x2k+1 Pn x2 + x x + x2
1 1 2 2
we will use more compact notation .
k=1 cyclic 3 cyc 3
2
x2 + xy + y 2

x+y 2
Noting that ≤ ⇐⇒ 0 ≤ (x − y) (for any real x, y) we
2 3
n x2 + x x + x2 n
 2
1P 1 1 2 2 1P x1 + x2
obtain ≥ . By Quadratic Mean-Arithmetic
n cyc 3 n cyc 2
n
 2 n
 !2  Pn 2
1 P x1 + x2 1P x1 + x2 k=1 xk
Mean Inequality ≥ = .
n cyc 2 n cyc 2 n
 
x1 + x2 x2 + x3 xn + x1
(Or, applying Cauchy Inequality to , , ..., and (1, 1, ..., 1
2 2 2
n
 2 n
  !2 n
 2
P x1 + x2 P x1 + x2 1P x1 + x2
we obtain n ≥ ·1 ⇐⇒ ≥
cyc 2 cyc 2 n cyc 2
!2  Pn 2
1P n x +x
1 2 k=1 xk x2 + xy + y 2 x2 + y 2
= . Since ≤ ⇐⇒ 0 ≤
n cyc 2 n 3 2
1P n x2 + x x + x2 1P n x2 + x2
2 1 1 2 2 1 2
(x − y) (for any real x, y) we obtain ≤ =
n cyc 3 n cyc 2
Pn 2
k=1 xk
.
n
x2k +x2k+1
Solution 3 by Michel Bataille, Rouen, France. Since xk xk+1 ≤ 2 , we
have x2k + xk xk+1 + x2k+1 ≤ 32 (x2k + x2k+1 ). It follows that

n n n
1 X x2k + xk xk+1 + x2k+1 1 X 1 2 1X 2
≤ (xk + x2k+1 ) = xk ,
n 3 n 2 n
k=1 cyclic k=1 cyclic k=1
554

hence the right inequality holds.

Using the Cauchy-Schwarz inequality, we obtain
!2  2
n n
X X xk + xk+1 
xk =  1·
2
k=1 k=1 cyclic
 
n 2
X (xk + xk+1 ) 
≤ (12 + 12 + · · · + 12 ) 
4
k=1 cyclic
 
n 2
X (xk + xk+1 ) 
= n .
4
k=1 cyclic

(xk +xk+1 )2 x2k +xk xk+1 +x2k+1

Now, the inequality 4 ≤ 3 holds as being equivalent to (xk −
xk+1 )2 ≥ 0. Thus,
!2  
n n
X X x2k + xk xk+1 + x2k+1
xk ≤ n 
3
k=1 k=1 cyclic

and this gives the left inequality at once.

Also solved by Adnan Ali (student), Mumbai, India and the proposer.
58. Corrected. Proposed by Arkady Alt, San Jose, California, USA. Let P be
arbitrary interior point in a triangle ABC and r be inradius. Prove that
a2 b2 c2
+ + ≥ 36r
da (P ) db (P ) dc (P )
if da (P ), db (P ) and dc (P ) are the distances from the point P to the sides BC, CA
and AB respectively.
Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and
Technology, Damascus, Syria. (The original statement has r2 instead of r, and
it is clearly not correct, because it is not homogeneous.)
Let s = (a + b + c)/2. For q ≥ 1 we have
2
! q+1
q+1 q+1 q+1
2s a+b+c a 2 +b 2 +c 2
= ≤ .
3 3 3
Equivalently
 q+1 q+1 q+1
2
31−q 2q+1 sq+1 ≤ a 2 + b 2 + c 2 .
Now, let Fa , Fb , Fc and F represent the areas of the triangles P BC, P CA, P AB and
ABC respectively. Clearly, we have 2Fa = ada (P ), 2Fb = bdb (P ) and 2Fc = cdc (P ).
Consequently, using Cauchy-Schwarz inequality, we have
 q+1 q+1 q+1
2
31−q 2q+1 sq+1 ≤ a 2 + b 2 + c 2
 q+1
bq+1 cq+1

a
≤ + + (2(Fa + Fb + Fc ))
2Fa 2Fb 2Fc
 q
bq cq

a
≤ 2F · + +
da (P ) db (P ) dc (P )
 555

Thus, since F = sr we obtain

aq bq cq
 
31−q 2q sq ≤ r · + +
da (P ) db (P ) dc (P )
√
Finally, using the well-known inequality: s ≥ 3 3r we obtain
aq bq cq
31+q/2 2q rq−1 ≤ + + .
da (P ) db (P ) dc (P )
For q = 1 we get
√ a b c
6 3≤ + +
da (P ) db (P ) dc (P )
and for q = 2 we get
a2 b2 c2
36r ≤ + + .
da (P ) db (P ) dc (P )
which is the announced inequality.
Solution 2 by Michel Bataille, Rouen, France. Let Sa , Sb , Sc denote the areas
of ∆BP C, ∆CP A, ∆AP B, respectively, and let S = Sa + Sb + Sc be the area of
∆ABC. Since for x = a, b, c we have 2Sx = x · dx (P ), the required inequality (1)
rewrites as
a3 b3 c3
+ + ≥ 72r (2).
Sa Sb Sc
To prove (2), we apply Holder’s inequality as follows
 3 !3
b3 c3

a 3 3 3 a 1/3 b 1/3 c 1/3
+ + (Sa +Sb +Sc )(1 +1 +1 ) ≥ 1/3
· Sa · 1 + 1/3 · Sb · 1 + 1/3 · Sc · 1
Sa Sb Sc Sa Sb Sc
that is,
a3 b3 c3
 
+ + · 3S ≥ (a + b + c)3 .
Sa Sb Sc
a+b+c
With s = 2 , the latter yields
a3 b3 c3 8s2
+ + ≥ (3).
Sa Sb Sc 3r
p
But from Heron’s formula (S = rs = s(s − a)(s − b)(s − c)) and AM-GM, we
obtain
3
s3

2 s−a+s−b+s−c
r s = (s − a)(s − b)(s − c) ≤ =
3 27
so that s2 ≥ 27r2 . Back to (3), we conclude
a3 b3 c3 8 · 27r2
+ + ≥ = 72r.
Sa Sb Sc 3r
Also solved by Adnan Ali (student), Mumbai, India and the proposer.
59. Proposed by Marcel Chiriţă, Bucharest, Romania. Solve in real numbers the
system 
2x + 2y = 12
3x + 4z = 11 .
3y − 4z = 25

Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and
556

Technology, Damascus, Syria. Let α = ln 2/ ln 3 ∈ (0, 1) so that the second

equation is equivalent to 2x = (11 − 4z )α and the third one is equivalent to 2y =
(25 + 4z )α the first equation is then equivalent to (11 − 4z )α + (25 + 4z )α = 12.
Now, consider f (t) = (11 − t)α + (25 + t)α for t ∈ [0, 11). Clearly

f 0 (t) = α (25 + t)α−1 − (11 − t)α−1 < 0

because α − 1 < 0 and 11 − t ≤ 11 < 25 + t for t ∈ [0, 11). It follows that f is

strictly decreasing on [0, 11), and since f (2) = 9α + 27α = 22 + 23 = 12 we conclude
that 2 is the only solution of the equation f (t) = 12 that belongs to [0, 11). It
follows that the equation f (4z ) = 12 has z = 1/2 as unique solution. But then
2x = (11 − 4z )α = 22 and 2y = (25 + 4z )α = 23 . Therefore, the proposed system
has a unique real solution which is (x, y, z) = (2, 3, 21 ).
Solution 2 by Adnan Ali (student), Mumbai, India.
Eliminating 4z , we have the system
2x + 2y = 12
3x + 3y = 36
We prove that the only solutions for the above system are {x, y} = {2, 3}. The
proof is based on the following claim: Fix positive constants a, b and k > 1, then
the equation in t:
tk + (a − t)k = b, 0 ≤ t ≤ a
may have at most two solutions. Indeed, we let f (t) = tk + (a − t)k − b and observe
that since k − 1 > 0, the derivative f 0 (t) = k(tk−1 − (a − t)k−1 ) is negative for
t < a2 , vanishes at t = a2 , and is positive for t > a2 . Hence f (t) is strictly decreasing
from 0 to a2 and strictly increasing from a2 to a, and our claim follows.
Now we let t = 2x , r = 2y and k = ln 3
ln 2 > 1 so that the system is now
t + r = 12
tk + rk = 36
Since t, r > 0, we have 0 < t < 12 and by observation {t, r} = {4, 8} are solutions.
Since we get two solutions, by our claim, we cannot have any more of them. Thus
transforming back to x, y, the only solutions of the system are {x, y} = {2, 3}.
Now back to the original system (proposed one), we observe that 3y > 25 ⇒ y =
3 ⇒ x = 2 and z = 21 . Clearly they satisfy the system and so they are the only
solution, i.e. (x, y, z) = (2, 3, 1/2).
Also solved by Albert Stadler, Herrliberg, Switzerland and the proposer.
60. Proposed by Dorlir Ahmeti, University of Prishtina, Department of Mathemat-
ics, Republic of Kosova. Let ABC be an acute triangle. Let D be the foot of the
altitude from A. Let E, F be the midpoints of AC, AB, respectively. Let G 6= B
and H 6= C be the intersection of circumcircle of the triangle ABC with circumcir-
cles of the triangles BF D and CED, respectively. Suppose that A, G, B, H, C are
order in this way on the circle they belong. Show that line EF, HB and CG are
concurrent.
Solution 1 by Michel Bataille, Rouen, France. Let Γ be the circumcircle of
∆ABC and O its centre. Let γb and γc be the circumcircles of ∆BF D and ∆CED,
respectively.We first note that AB 6= AC: otherwise, OD would be perpendicular
to BC and γb would be the circle with diameter OB (since OF ⊥ AB, OF being
 557

the perpendicular bisector of AB). As such, γb would be tangent to Γ at B, con-

tradicting G 6= B. Since AB 6= AC, O is not on AD and so the tangent t to Γ at
A is not parallel to EF . Let K be the point of intersection of t and EF , γa be the
circle with diameter OA (which passes through E and F ) and I the inversion in the
circle with centre K and radius KA (so that I(A) = A). Since γa is tangent to Γ at
A, KA2 is the power of K with respect to Γ and to γa and so I(Γ) = Γ, I(γa ) = γa
and I(E) = F (since E, F ∈ γa and KE · KF = KA2 ). Since AD ⊥ DB, the
midpoint F of AB satisfies F A = F B = F D so that the centre I of γb is on the
perpendicular to BC through F . It follows that IF ⊥ F E and so γb is tangent
to EF at F . Similarly, γc is tangent to EF at E. Also, the line EF being the
perpendicular bisector of AD, we have KD = KA, hence I(D) = D. We deduce
that I(γc ) is a circle which is tangent to EF at F and passes through D, that is,
I(γc ) = γb . Now, I(C) is on I(Γ) = Γ and on I(γc ) = γb , hence I(C) = G or B. But
the latter cannot occur since CB does not pass through K. Thus I(C) = G and so
CG passes through K. In a similar way, I(H) is on Γ and γb and I(H) 6= G (since
G = I(C) and C 6= H), hence I(H) = B and HB passes through K. In conclusion,
EF, HB and GC are concurrent at K.
Solution 2 by Andrea Fanchini, Cantù, Italy. We use barycentric coordinates
and the usual Conway’s notations with reference to the triangle ABC.
As we know, for the remarkable points D, E, F we have the followings coordinates
D(0 : SC : SB ), E(1 : 0 : 1), F (1 : 1 : 0)
• Circumcircle of the triangle BF D.
We impose that the circle passes for the three points B, F, D and we find that the
equation of the circumcircle is
 2 
c
a2 yz + b2 zx + c2 xy − (x + y + z) x + SC z = 0
2
now the center is the point J(2S 2 − a2 c2 : 2S 2 + c2 SC : c2 SB ) and the radical
axis between this circle and the circumcircle of the triangle ABC has equation
BG : c2 x + 2SC z = 0.
• Coordinates of point G. We denote with K the intersection point between the
radical axis BG and the line that passes for J and that is perpendicular

to BG, we
find K 2SC (b2 − c2 ) : 2SC (3b2 − 2c2 ) + a2 c2 : −c2 (b2 − c2 ) .
Then the point G is simmetric of B respect to K, therefore it is
G 2SC (b2 − c2 ) : 2b2 SC : −c2 (b2 − c2 )

• Coordinates of point H.
Following the same procedure and using the cyclicity, we find that the point H has
coordinates
H 2SB (c2 − b2 ) : −b2 (c2 − b2 ) : 2c2 SB

• Equations of lines EF, BH and CG.

We have easy that
EF : x − y − z = 0, BH : c2 x + (b2 − c2 )z = 0, CG : b2 x + (c2 − b2 )y = 0
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Now the three lines EF, BH and CG are concurrent if and only if
1 −1 −1

c2 0 b2 − c2 =0

b2 c2 − b2 0

that it is true as we can easy check.

Also solved by the proposer.