# The limit of an infinite sum …

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

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One idea: Write out what $(1 + k)^k$ is in terms of binomial coefficients. –  Thomas May 26 '12 at 12:42
It doesn't hurt to try. It's probably important to work out how big the terms $k=n$ is, as compared to $k=n-1$, as well as the terms near $k=n$ are as compared to much smaller terms, as that might give you ideas. Among other things, this means determining the asymptotic behavior of your summand as $k \to \infty$. –  Hurkyl May 26 '12 at 12:45
Hint: $a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$ –  m.woj May 26 '12 at 12:49
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 $\mathrm e$.
This assumes that one interprets $0^0$ as $0$ but the argument is easily modified if one considers that $0^0=1$, leading to the same limit. –  Did May 26 '12 at 13:27
Several approaches how to find the last limit $\lim_{n\to\infty} \frac{n+1}{\sqrt[n]{n!}}=\lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}}$ can be found here. –  Martin Sleziak May 26 '12 at 13:59