# Thread: Analysis problem on Sequences

1. ## Analysis problem on Sequences

Hi guys, just got a question of Analysis on Sequences.

a sequence e(n) (sry dun know how to type (en) little n.
ok....

e(n)=(1+1/n)^n

(a). Show that e(n) is bounded above

(b). Use The Monotone Convergence Theorem to conclude that e(n) converges to a limit "e", and give an interval of length less than one in which "e" lies. The number "e" is called the Euler Number.

Appreciate!!!

2. This may not be the neatest way to do this.

$\displaystyle e_{n} = \left(1 + \frac{1}{n} \right)^n$

binomial theorem
$\displaystyle = 1 + \frac{n}{n} + \frac{n(n-1)}{2 n^2} + \frac{n(n-1)(n-2)}{3! n^3} + \cdots +\frac{n!(n-1)(n-2) \cdots 3 \cdot 2\cdot 1}{n n^n}$

$\displaystyle = 1 + 1 + \frac{1}{2} \left(1 - \frac{1}{n} \right) + \frac{1}{3!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) + \ \cdots \ + \frac{1}{n!} \left(1 - \frac{1}{n} \right) \cdots \left(1 - \frac{n-1}{n} \right)$

bracketed terms are positive and all less than 1

$\displaystyle < 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}}$

less than infinite geometric sum $\displaystyle \therefore <3$

the rest should be easy.

Bobak

3. thx man

4. but what about part (b).

5. Well i showed you the limit is less than 3, can you show it is greater than 2 ?

Bobak

6. sure,thx man