Sketch of solutions 2019 Calculus 1-3

1. Find all real values of p for which the series is convergent.

∞
X (n + ln n) tanh n
√
n=1
np ( n + ln n)p

(n+ln n) tanh n n 1
solution. Let an = √
np ( n+ln n)p
and take bn = √
np ( n)p
= n3p/2−1
,
√
an (n + ln n) tanh n np ( n)p
lim = p √ · = 1,
n→∞ bn n ( n + ln n)p n
√
( n)p
since tanh n → 1, n+ln n
P
n → 1, √
( n+ln n)p
→ 1 as n → ∞. Therefore bn =
1
P P
n3p/2−1
converges if and only if 3p/2 − 1 > 1 (as a p-seires), hence an
converges if and only if p > 4/3.(by Limit Comparison Test)
2. (multiple choice)
∞ √ ∞ ∞
X n+4 X n! X n2
ㄱ. ㄴ. ㄷ. (−1)n
n=1
n2 n=1
n n
n=1
n2 +n+1
∞
X ln n ∞  
X 1
ㄹ. ㅁ. sin
n=2
n2 n=1
n

Which of the above series are convergent?

solution. The answer is ㄱ,ㄴ,ㄹ.
√
ㄱ. By Limit Comparson with bn = n2n = n3/21
P
, where bn converges
√
P n+4
by p-series test. Hence n2 converges by Limit comparison test.
ㄴ.
n! 1 · 2···n 2 3···n 2
0< = ≤ 2 ≤ 2
nn n · n···n n n···n n
since nk ≤ 1 for all k = 3, 4, . . . , n. Since
P 2
P n! n2 converges by p-series
test, nn converges by Comparison Test.
ㄷ.
n2 n2
lim = 1(̸= 0) ⇒ lim (−1)n 2 ̸= 0
n→∞ n2 +n+1 n→∞ n +n+1
the series diverges by test for divergence.
1
ㄹ. Using ln n ≤ nc , all n > N for some N , take c = 2 so that we have

∞ N ∞ N ∞
X ln n X ln n X ln n X ln n X n1/2
0≤ = + ≤ +
n=2
n2 n=2
n2 n2 n=2
n2 n2
n=N +1 n=N +1

n1/2
P∞ P∞ 1
, where n=N +1 n2 = n=N +1 n3/2 converges by p-series test.
Hence the series converges by Comparison Test.
1
P
ㅁ. By Limit Comparison
P∞ with

bn = n , where bn diverges by p-series
1
test. Hence n=1 sin n diverges by Limit comparison test.

1
 ∞
X
3. Let {an } be a sequence of real numbers. If an converges absolutely,
n=1
then does the series
∞
X
acrtan(a2n )
n=1

converge or not? Justify your answer.

∞
X ∞
X
solution. Since an converges absolutely, |an | converges so we have
n=1 n=1
limn→∞ |an | = 0. Hence for some N > 0, we have |an | ≤ 1 for n > N .
Check
arctan(a2n ) ≤ a2n = |an |2 ≤ |an |
P∞
for all n > N . Consequently n=1 acrtan(a2n ) converges by Comparison
Test.
4. (multiple choice)
∞ 2 ∞ ∞
X 2n X n100 100n X cos(nπ/3)
ㄱ. ㄴ. ㄷ.
n=1
n! n=1
n! n=1
n!
∞ ∞
X X (−1)n−1
ㄹ. (arctan(n))n ㅁ.
n=2 n=1
(ln n)n

Which of the above series are convergent?

solution. The answer is ㄴ,ㄷ,ㅁ.
ㄱ. Check 2n > n for n ≥ 1.
2
2n nn
≥ ≥1
n! n!
2
2n
Hence limn→∞ n! ̸= 0, the series diverges by test for divergence.
ㄴ. Use ratio test
(n+1)100 100n+1  100
an+1 (n+1)! n+1 100
= n100 100n
= → 0(< 1)
an n!
n n+1

as n → ∞. Hence the series converges absolutely by ratio test.

ㄷ.
cos(nπ/3) 1
≤
n! n!
P 1 1 2
where n! converges by ratio test or comparison test( n! ≤ n2 ).
Hence the series converges by comparison test.
ㄹ. Use root test
1/n π
|an | = arctan(n) →
(> 1)
2
as n → ∞. Hence the series diverges by root test.
1
ㅁ. (ln n)npositive, decreasing and (ln1n)n → 0 as n → ∞ since (ln n)n is
increasing and (ln n)n → ∞ as n → ∞. Hence the series converges
by Alternating Series Test.

2
5. Let α be a positive real number. Find the radius of convergence and the
interval of convergence of the following power series.
∞
X xn
n=2
n(ln n)α

solution. Use ratio test

α
xn+1 . xn 
an+1 n ln(n)
= α
= |x| → |x|
an (n + 1)(ln(n + 1)) n(ln n)α n+1 ln(n + 1)
as n → ∞. Hence the series absolutely converges for |x| < 1 and diverges
for |x| > 1. P
∞
For x = 1, n=1 n(ln1n)α we can use integral test since f (x) = x(ln1x)α is
continuous, positive and decreasing for x > 2(check x(ln(x))α is increasing
for α > 0.) And Z ∞ Z ∞
1 1
α
= α
dt
2 x(ln x) ln 2 t
by substitution ln x = t. Hence the integralPconverges for α > 1 and
∞
diverges for α ≤ 1 by p-test. Hence the series n=1 n(ln1n)α converges for
α > 1 and diverges for α < 1 by Integral Test.
P∞ (−1)n 1
For x = −1, n=1 n(ln n)α is a alternating series where n(ln n)α is positive,
decreasing and
1
lim =0
n→∞ n(ln n)α
P∞ (−1)n
for α > 0. Hence the series n=1 n(ln n)α converges by A.S.T.
Consequently we have the radius of the convergence R = 1 and the interval
of the convergence [−1, 1] for α > 1 and [−1, 1) for α ≤ 1.
6. Find the sum of the series
∞
X
n(n − 1)3−n
n=2

solution. From the series

∞
1 X
= xn
1 − x n=0

we have
∞
1 X
= nxn−1
(1 − x)2 n=0
by differentiation of both side. Also
∞
2 X
= n(n − 1)xn−2
(1 − x)3 n=0

by differentiation of both side again, moreover multiply both sides by x2

∞ ∞
2x2 X
n
X
= n(n − 1)x = n(n − 1)xn .
(1 − x)3 n=0 n=2

3
 Consequently
∞
2
X
−n 2 31 3
n(n − 1)3 = = .
n=2
(1 − 31 )3 4

7. Let
∞
X x3n
f (x) =
n=0
(3n)!

Find the value of f (7) (1) + f (8) (1) + f (9) (1).

solution. Check
∞
X x3n
f (x) = = f (3) (x) = f (6) (x) = · · ·
n=0
(3n)!

and
∞
X x3n−1
f ′ (x) = = f (4) (x) = f (7) (x) = · · ·
n=1
(3n − 1)!
and
∞
X x3n−2
f ′′ (x) = = f (5) (x) = f (8) (x) = · · ·
n=1
(3n − 2)!
also
ex = f (x) + f ′ (x) + f ′′ (x) = f (2) (x) + f (3) (x) + f (4) (x)
= f (3) (x) + f (4) (x) + f (5) (x) = · · · = f (7) (x) + f (8) (x) + f (9) (x) = · · ·
Consequently f (7) (1) + f (8) (1) + f (9) (1) = e1 .
8. Z 1
2 2
x2 e−x dx (|error| < 0.001).
0
Which of the following values is the best approximation to the above
definite integral to within the indicated accuracy.
solution. Using the Maclaurin series of ex
1 1
ex = 1 + x + x2 + · · · + xn + · · · ,
2 n!
2 1 1
e−x = 1 − x2 + x4 + · · · + (−x2 )n + · · ·
2 n!
2 1 1
x2 e−x = x2 − x4 + x6 + · · · + (−1)n x2n+2 + · · · .
2 n!
Hence
Z 21  3  21
2 −x2 x x5 1 1 7 1 (−1)n x2n+3
x e dx = − + · x + ··· + + ···
0 3 5 2 7 n! 2n + 3 0
1 1 1 1 1 1 1 1 (−1)n 1
= · 3 − · 5 + · · 7 + ··· + · + ···
3 2 5 2 2 7 2 n! 2n + 3 22n+3
is the alternating series satisfying the conditions in A.S.T. Therefore we
need to find n so that
1 1 1
|error| = |Rn | ≤ · 2n+5 ≤ 0.001
(n + 1)! 2n + 5 2

4
 1 1
Take n = 1 so that 2!·7·27 = 1792 = 0.000558 · · · < 0.001. The aprroxima-
tion is Z 1
2 2 1 1 1 1 17
x2 e−x d ∼
= · 3− · 5 =
0 3 2 5 2 480
9. In Newton’s method for approximating a root r of the equation f (x) = 0,
from an initial approximation x1 , we obtain successive approximations
x2 , x3 , . . ., where
f (xn )
xn+1 = xn − ′
f (xn )
Use Taylor’s Inequality with n = 1, a = xn , and x = r to show that if
f ′′ (x) exists on an interval I containing r, xn , and xn+1 , and |f ′′ (x)| ≤ M ,
|f ′ (x)| ≥ K for all x ∈ I, then
M
|xn − r|2
|xn+1 − r| ≤
2K
solution. We will assume K > 0. By Taylor’s Inequality we have
M
|Rn (x)| ≤ |x − a|n+1
(n + 1)!
where Rn (x) = f (x) − Tn (x), Tn (x) is a Taylor polynomial of degree n,
and |f (n+1) (x)| ≤ M for all |x − a| ≤ d. for some d. (we choose d so that
all x satisfying |x − a| ≤ d will contained in the interval I.)

For n = 1, a = xn and x = r, we have

|R1 (r)| = |f (r) − T1 (r)| = |f (r) − f (xn ) − f ′ (xn )(r − xn )|
M
= |f (xn ) + f ′ (xn )(r − xn )| ≤ |r − xn |2
2
since f (r) = 0. From
f (xn )
xn+1 = xn − ′
f (xn )
f (xn ) = f ′ (xn )(xn − xn+1 ), hence we have
M
|f (xn ) + f ′ (xn )(r − xn )| = |f ′ (xn )(r − xn+1 )| ≤ |r − xn |2
2
Since |f ′ (x)| ≥ K for all x ∈ I, we have
M
K|r − xn+1 | ≤ |f ′ (xn )||(r − xn+1 )| ≤ |r − xn |2
2
Divide the both sides by K, we have the result.
10. Let v be the vector with the smallest length such that
⟨1, 2, 1⟩ × v = ⟨3, 1, −5⟩
Find |v|2 .
solution.
|⟨1, 2, 1⟩ × v| = |⟨1, 2, 1⟩||v| sin θ
√ √
= 6|v| sin θ = |⟨3, 1, −5⟩| = 35
√ q
35 35 π 35
Hence |v| = √6 sin θ
which has the smallest 6 when θ = 2. |v|2 = 6 .

5
11. Let a and b be two vectors such that
√
a · b = 5, a × b = ⟨2, 5, 4⟩.

(a) Find Proja b × Projb a.

(b) Find the distance from the point P (1, 2, 3) to the plane through the
point Q(1, 1, 1) and with normal vector a × b.
solution.

(a)

a · b = |a||b| cos θ = 5,
√ √
|a × b| = |a||b| sin θ = |⟨2, 5, 4⟩| = 22 + 5 + 42 = 5
π
Hence we have tan θ = 1, that is θ = 4. Therefore

(a · b) (a · b) (a · b)2
Proja b × Projb a = 2
a× 2
b= (a × b)
|a| |b| |a|2 |b|2
1 √
= cos2 θ(a × b) = ⟨2, 5, 4⟩
2

(b) By the formula, the distance is

√
|QP · (a × b)| 8+ 5
=
|a × b| 5
√
where QP = ⟨0, 1, 2⟩ and a × b = ⟨2, 5, 4⟩.
12. Find the volume of the parallelepiped with adjacent edges P Q, P R, and
P S.
P (−2, 1, 0), Q(2, 3, 2), R(1, 4, −1), S(3, 6, 1).
solution. P Q = ⟨4, 2, 2⟩, P R = ⟨3, 3, −1⟩, P S = ⟨5, 5, 1⟩ hence the triple
product is
4 2 2
P Q · (P R × P S) = 3 3 −1 = 16
5 5 1
and the volume of the parallelepiped is |P Q · (P R × P S)| = 16.