Something around the number e
1
n
1. Show that the sequence 1
converges, and denote the limit by e.
Proof: Since
n
1 1
n
nk
k0
1
n
nn 1 1 2
nn 1 1 1 n
1n 1
. .
n
n
n
2!
n!
1
1
n1
11 1 1 1
n . . . n! 1 n 1 n
2!
1 . . .
1 1 1 12 . . n1
2
2
2
3,
and by (1), we know that the sequence is increasing. Hence, the sequence is convergent.
We denote its limit e. That is,
n
lim
1 1
e.
n
n
Remark: 1. The sequence and e first appear in the mail that Euler wrote to Goldbach.
It is a beautiful formula involving
e i 1 0.
2. Use the exercise, we can show that k0
1
k!
e as follows.
Proof: Let x n 1 1n , and let k n, we have
1 1 1 1 1 . . 1 1 1 1 n 1
n!
2!
k
k
k
which implies that ( let k )
xk
i!1
y n :
e.
i0
On the other hand,
xn yn
So, by (2) and (3), we finally have
k!1
e.
k0
3. e is an irrational number.
Proof: Assume that e is a rational number, say e p/q, where g.c.d. p, q 1. Note
that q 1. Consider
q!e q!
k!1
k0
q
q!
k0
and since q!
integer. However,
q
1
k0 k!
1
k!
q!
kq1
1
k!
and q!e are integers, we have q! kq1
1
k!
is also an
q!
kq1
1
k!
kq1
q!
k!
1
1
. . .
q1
q 1q 2
2
1
1
. . .
q1
q1
1
q
1,
a contradiction. So, we know that e is not a rational number.
4. Here is an estimate about e k0
that e 2. 71828 18284 59045 . . . . )
Proof: Since e k0
1
k!
nn!
, where 0 1. ( In fact, we know
, we have
0 e xn
kn1
1
k!
1 , where x n
k!
k!1
k0
1
1
1 1
. . .
n2
n 1!
n 2n 3
1
1
1 1
. . .
n2
n 1!
n 2 2
1
n2
n 1! n 1
1 since n 2 2 1
n.
nn!
n 1
So, we finally have
n
k!1
k0
, where 0 1.
nn!
Note: We can use the estimate dorectly to show e is an irrational number.
2. For continuous variables, we have the samae result as follows. That is,
x
lim 1 1x
e.
x
n
Proof: (1) Since 1 1n e as n , we know that for any sequence a n N,
with a n , we have
an
lim
1 a1n
e.
n
(2) Given a sequence x n with x n , and define a n x n , then
a n x n a n 1, then we have
an
xn
a n 1
1
1
1 x1n
1 a1n
.
an 1
Since
an
a n 1
1
1
e and 1 a1n
e as x by (5)
an 1
we know that
1 x n e.
lim
1
xn
n
Since x n is arbitrary chosen so that it goes infinity, we finally obtain that
1 x e.
lim
1
x
x
1 x
(3) In order to show 1 x e as x , we let x y, then
x
1 y
1 1x
1 y
y
y
y1
y1
1
1 1
.
y1
y1
Note that x y , by (6), we have shown that
1
y1
x
x
lim 1 1x .
e y
lim 1
3. Prove that as x 0, we have1
dstrictly ecreasing.
1
x
y1
1
y1
is strictly increasing, and 1
1
x
x1
is
Proof: Since, by Mean Value Theorem
1 log 1 1 logx 1 logx 1 1 for all x 0,
x
x
x1
we have
x log 1 1x
log 1 1x 1 0 for all x 0
x1
and
log 1 1x 1x 0 for all x 0.
x 1 log 1 1x
Hence, we know that
x log 1 1x is strictly increasing on 0,
and
x 1 log 1 1x is strictly decreasing on 0, .
It implies that
x
x1
1 1x
is strictly increasing 0, , and 1 1x
is strictly decreasing on 0, .
Remark: By exercise 2, we know that
1 x e lim 1 1
1
lim
x
x
x
x
x1
4. Follow the Exercise 3 to find the smallest a such that 1
decreasing for all x 0, .
Proof: Let fx 1
1
x
xa
1
x
, and consider
log fx x a log 1 1x
Let us consider
: gx,
xa
e and strictly
x2 a
x x
log1 y y 1 ay 2
g x log 1 1x
1 , where 0 y 1 1
1x
1y
yk
y 1 ay 2 y k
k
k0
k1
1 a y 2 1 a y 3 . . . 1 a y n . . .
n
2
3
It is clear that for a 1/2, we have g x 0 for all x 0, . Note that for a 1/2,
if there exists such a so that f is strictly decreasing for all x 0, . Then g x 0 for
all x 0, . However, it is impossible since
n
g x 1 a y 2 1 a y 3 . . . 1
n a y . . .
2
3
1 a 0 as y 1 .
2
So, we have proved that the smallest value of a is 1/2.
x1/2
Remark: There is another proof to show that 1 1x
is strictly decreasing on
0, .
Proof: Consider ht 1/t, and two points 1, 1 and 1 1x , 11 1 lying on the graph
x
From three areas, the idea is that
The area of lower rectangle The area of the curve The area of trapezoid
So, we have
1 1
x
1x
1
1
log 1 1x
1
x
1 1 1
2x
1
1
x
x 1
2
1
.
xx 1
Consider
1 1x
x1/2
1 1x
x1/2
log 1 1x
x 1
2
1
xx 1
0 by (7);
x1/2
hence, we know that 1 1x
is strictly decreasing on 0, .
x
Note: Use the method of remark, we know that 1 1x is strictly increasing on
0, .
