# Analysis problem on Sequences

Printable View

• Apr 1st 2009, 01:44 AM
luoginator
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!!!
• Apr 1st 2009, 03:42 AM
bobak
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
• Apr 1st 2009, 03:06 PM
luoginator
thx man
• Apr 1st 2009, 03:06 PM
luoginator
but what about part (b).
• Apr 1st 2009, 03:10 PM
bobak
Well i showed you the limit is less than 3, can you show it is greater than 2 ?

Bobak
• Apr 1st 2009, 07:25 PM
luoginator
sure,thx man:)