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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.

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    Hint: $a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$2012-05-26

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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 $\mathrm e$.

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    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](http://math.stackexchange.com/questions/28476/finding-the-limit-of-frac-n-sqrtnn).2012-05-26