# 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?

• Could you define $e'$? Nov 14, 2011 at 16:47
• I believe it is defined int his homework as just $e$, I will change the OP. Nov 14, 2011 at 16:54
• use $\lim_{n \to \infty} (1+\frac{1}{n})^{n}=e$
– Peđa
Nov 14, 2011 at 16:54
• @pedja: I think you need the stronger $\lim_{n \rightarrow \infty} (1 + \frac{x}{n})^n = e^x$. (Applied with $x = -1$.) Nov 14, 2011 at 16:59
• Assuming you are to establish the limit, L' Hospital rule may be a good choice.
– Tapu
Nov 14, 2011 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 $e$. Nov 14, 2011 at 17:17