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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.

Did
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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 $k \to \infty$. –  May 26 '12 at 12:45
  • Hint: $a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$ – borg May 26 '12 at 12:49

1 Answers1

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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 $\mathrm e$.

Did
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    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
  • Not sure I understand the question. See Edit. – Did May 26 '12 at 13:38
  • 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