# Prove that $(1 - \frac{1}{n})^{-n}$ converges to $e$

This is a homework question and I am not really sure where to go with it. I have a lot of trouble with sequences and series, can I get a tip or push in the right direction?

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Could you define $e'$? –  Jacob Nov 14 '11 at 16:47
I believe it is defined int his homework as just $e$, I will change the OP. –  Matt Nashra Nov 14 '11 at 16:54
use $\lim_{n \to \infty} (1+\frac{1}{n})^{n}=e$ –  pedja Nov 14 '11 at 16:54
@pedja: I think you need the stronger $\lim_{n \rightarrow \infty} (1 + \frac{x}{n})^n = e^x$. (Applied with $x = -1$.) –  Michael Joyce Nov 14 '11 at 16:59
Assuming you are to establish the limit, L' Hospital rule may be a good choice. –  Tapu Nov 14 '11 at 17:04

You have: $$x_n:=\left(1-\frac1n\right)^{-n} = \left(\frac{n-1}n\right)^{-n} = \left(\frac{n}{n-1}\right)^{n}$$ $$= \left(1+\frac{1}{n-1}\right)^{n} = \left(1+\frac{1}{n-1}\right)^{n-1}\cdot \left(1+\frac{1}{n-1}\right) = a_n\cdot b_n.$$ Since $a_n\to \mathrm e$ and $b_n\to 1$ you obtain what you need.
Nice solution, uses almost nothing other than the definition of $e$. –  André Nicolas Nov 14 '11 at 17:17