how would you prove this little inequality?

[stunner5000pt]

i'm stuck trying to prove this little inequality:

(1+ 1/n)^n <= e <= (1+1/n)^n+1
is there a way to prove this without without resorting to power series for e (because we're not allowed to, and we don't know this yet) and also note that n is a natural number, (positive integer).

[cookiemonster]

Here's a hint:

\lim_{n \to \infty} \Big( 1 + \frac{1}{n} \Big)^n = e

cookiemonster

[stunner5000pt]

ok i know tha already i just dont know how to prove it give me hint on hwo to prove it

[cookiemonster]

Just how formally do you want to prove it?

It's pretty easy to notice that for n<\infty, the left side is less than e. When n = \infty, it is exactly equal to e.

The same holds true for the right side, except that it's always greater than e except when n = \infty.

cookiemonster

[matt grime]

You do it by power series for (1+x)^n valid when |x|<1 (ie x=1/n)

[Wooh]

Originally posted by stunner5000pt
i'm stuck trying to prove this little inequality:

(1+ 1/n)^n <= e <= (1+1/n)^n+1
is there a way to prove this without without resorting to power series for e (because we're not allowed to, and we don't know this yet) and also note that n is a natural number, (positive integer).
Why can you not just reason that (1+1/n)^n has to be less than (1+1/n)^n * (1+1/n), as, until n => infinity, 1+1/n will always be a positive value above * a positive value that will always make the right hand larger.