We see that (xn) is an increasing sequence

1. A note on e.

Theorem 1.1. If (a_n) is increasing and bounded above, then (a_n) is convergent.

Let (xn) be the sequence of numbers defined by xn= 1 + 1

1!+ 1

2!+ · · · + 1

n!, n ≥ 1.

Proposition 1.1. The sequence (xn) is convergent.

Proof. For each n ≥ 1,

x_n+1= x_n+ 1

(n + 1)! > x_n.

We see that (xn) is an increasing sequence. For each n ≥ 4, n! > n(n − 1). Then 1

n! < 1

(n − 1)n, n ≥ 3.

Hence we see that

xn= 1 + 1 1!+ 1

2! + 1 3!+ 1

4!+ · · · + 1 n!

< 1 + 1 1!+ 1

1 · 2 + 1

2 · 3 + 1

3 · 4 + · · · + 1 (n − 1) · n. We know that

1

1 · 2 + 1

2 · 3 + · · · + 1

(n − 1)n = 1 − 1 n. Hence

x_n< 1 + 1

1!+ 1 − 1

n = 3 − 1 n < 3.

This shows that (x_n) is bounded above. Hence by Theorem 1.1, we see that (x_n) is conver-

gent. 

Let us denote lim

n→∞x_n= x.

Let (yn) be the sequence of numbers defined by y_n=

 1 + 1

n

n

, n ≥ 1.

Proposition 1.2. (yn) is convergent.

Before proving Proposition 1.2, let us review the following two important facts.

Theorem 1.2. (A.G. inequality) Let a₁, · · · , a_n be any nonnegative real numbers. Then a₁+ · · · + a_n

n ≥ √ⁿ

a₁· · · a_n. Theorem 1.3. Let x, y be real numbers. Then

(x + y)ⁿ=

n

X

k=0

n k



x^ky^n−k.

Here n k



= n!

k!(n − k)!.

1

2

Let us choose a1= 1 and ak= 1 + 1

n for 2 ≤ k ≤ n + 1. Then a₁+ · · · + a_n+1

n + 1 = 1 + n · 1 +_n¹

n + 1 = n + 2

n + 1 = 1 + 1 n + 1. and

a1a2· · · a_n+1=

 1 +1

n

n

Using A.G. inequality, we see that 1 + 1

n + 1 > ⁿ⁺¹ s

 1 + 1

n

n

. Taking n + 1-th power of both side of the inequality, we get

y_n+1=



1 + 1 n + 1

n+1

>

 1 + 1

n

n

= y_n.

Hence (yn) is an increasing sequence. Using the binomial theorem, we see that y_n=

 1 +1

n

n

= 1 +n 1

 1 n +n

2

 1

n² + · · · +n n

 1 nⁿ

= 1 + n · 1

n+n(n − 1) 2!

1 n² + · · ·

= 1 + 1 + 1 2!

 1 − 1

n

 + 1

3!

 1 −1

n

  1 − 2

n



+ · · · + 1 n!

 1 −1

n



· · ·



1 −n − 1 n

 . Here we use the following simple equality:

n k

 1

n^k = n(n − 1) · · · (n − k + 1)

k! · 1

n^k = 1 k!

 1 −1

n



· · ·



1 −k − 1 n

 . Notice that 1 − j

n < 1 for all 1 ≤ j ≤ n. Hence

n k

 1 n^k = 1

k!

 1 − 1

n



· · ·



1 −k − 1 n



< 1 k!. Therefore

yn= 1 +n 1

 1 n+n

2

 1

n² + · · · +n n

 1

nⁿ < 1 + 1 1! + 1

2!+ · · · + 1 n! = xn. We have already seen that x_n< 3 for all n ≥ 1, and hence

yn≤ x_n< 3

for all n ≥ 1. This shows that (y_n) is bounded. We conclude that (y_n) is convergent by Theorem 1.1. We let us denote y = lim

n→∞yn.

Theorem 1.4. Let (xn) and (yn) be as above and x and y be their limits respectively.

Then x = y.

To prove this theorem, we need the following lemma.

3

Lemma 1.1. Let (an) and (bn) be two convergent sequence of real numbers. Suppose that a_n≤ b_n for all n ≥ 1 and lim

n→∞a_n= a and lim

n→∞b_n= b. Then a ≤ b.

Proof. Given  > 0, there is N > 0 so that for any n ≥ N,

|a_n− a| < , |b_n− b| < .

Then a −  < an and bn< b +  for any n ≥ N. Since an≤ b_n for all n ≥ 1, an≤ b_n for all n ≥ N. We see that

a −  < an< bn< b + 

for all n ≥ N. We see that a −  < b +  for all  > 0. The only possibility is a ≤ b.

 Corollary 1.1. Let (a_n) be a convergent sequence whose limit is a. Suppose that a_n≤ M for all n ≥ 1. Then a ≤ M.

Proof. Let us take bn= M for all n ≥ 1. Using Lemma 1.1, we see that a ≤ M.

 Since y_n≤ x_n for all n ≥ 1, using Lemma 1.1, we see that y ≤ x. On the other hand, for each 1 ≤ k ≤ n,

y_n= 1 + 1 +n 2

 1

n² + · · · +n k

 1 n^k +

 n k + 1

 1

n^k+1 + · · · +n n

 1 nⁿ

> 1 + 1 +n 2

 1

n² + · · · +n k

 1 n^k. Here we use the fact that

 n k + 1

 1

n^k+1 + · · · +n n

 1 nⁿ > 0.

Since we know that

n j

 1 n^j = 1

j!

 1 −1

n



· · ·



1 −j − 1 n

 , we see that

n→∞lim

n j

 1 n^j = 1

k!. Hence taking n → ∞ of the following inequality

y_n> 1 + 1 +n 2

 1

n² + · · · +n k

 1 n^k, we see that by Lemma 1.1,

y ≥ 1 + 1 + 1

2! + · · · + 1 k! = xk

for all k ≥ 1. Using Corollary 1.1, we see that x = lim

k→∞xk ≤ y.

We obtain x ≤ y. Hence x ≤ y and y ≤ x. This implies x = y.