1. ## tough limit

Does anyone know how to show this limit?

limit as n approaches infinity of the summation from k = 0 to n of [(n-k)/n]^n.

The summation is just the gamma symbol with k = 0 on the bottom and n on the top. Apparently, the limit equals e/(e-1), but how would you show it with algebra?

2. What it amounts to is whitttling it down to:

$\displaystyle \sum_{k=0}^{\infty}\frac{1}{e^{k}}=\frac{e}{e-1}$

Take note:

$\displaystyle \left(\frac{n-k}{n}\right)=\left(1-\frac{k}{n}\right)$

A take on the famous limit of e:

$\displaystyle \lim_{n\rightarrow{\infty}}\left(1-\frac{k}{n}\right)^{n}=\frac{1}{e^{k}}$

Now, if you take the sum of $\displaystyle \frac{1}{e^{k}}$

$\displaystyle \sum_{k=0}^{\infty}\frac{1}{e^{k}}=\frac{e}{e-1}$