TITLE: Limit of $\left(\sum\limits_{k=0}^n \frac{{(1+k)}^{k}-{k}^{k}}{k!}\right)^{1/n}$
QUESTION [6 upvotes]: Calculate the following limit:
  $$\lim_{n\to\infty} \left(\sum_{k=0}^n \frac{{(1+k)}^{k}-{k}^{k}}{k!}\right)^{1/n} $$

First of all, I am just looking for any helping hint that will allow me to solve
it. I thought of Stirling's formula, but I am not convinced that it helps me here. Maybe if I had $n!$ when $n$ goes to infinity it would work, otherwise I doubt I can do
something about it. Not sure how to approach it, yet.

REPLY [11 votes]: Note that $\dfrac{{(1+k)}^{k}-{k}^{k}}{k!}=\dfrac{{(1+k)}^{k+1}}{(k+1)!}-\dfrac{k^{k}}{k!}$, hence,
$$
\sum_{k=0}^n \frac{{(1+k)}^{k}-{k}^{k}}{k!}=\dfrac{(n+1)^{n+1}}{(n+1)!}=\dfrac{(n+1)^{n}}{n!},
$$
and
$$
\left(\sum_{k=0}^n \frac{{(1+k)}^{k}-{k}^{k}}{k!}\right)^{1/n}=(n+1)\cdot(n!)^{-1/n}.
$$
From that point, the usual Stirling's approximation applied to $n!$ shows the limit is $\mathrm e$.