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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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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: $a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$ – data May 26 '12 at 12:49
up vote 11 down vote accepted

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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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

Take into account that $$\sum_{k=0}^n \frac{{(1+k)}^{k}-{k}^{k}}{k!} = \frac {(n + 1)^n} {n!},$$ then we have

$$\lim_{n\to\infty} \left(\sum_{k=0}^n \frac{{(1+k)}^{k}-{k}^{k}}{k!}\right)^{1/n} = \lim_{n\to\infty} \frac {n + 1} {\sqrt [n] {n!}} = e,$$ by Stirling approximation.

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Why do you deliberately copy 4-years-old answers by others? – Did Feb 13 at 11:17
@Did When somebody that I respect a lot attacks me with childish blames, I become most disappointed. Whom have I copied, sir? – Nima Bavari Feb 13 at 14:01
The other answer (what else?). Re "childish", you will learn that plagiarism (even if widely tolerated in many quarters of our societies at large) is very much frowned upon in mathematics. – Did Feb 13 at 18:01
@Did OMG, now I got it. I'm really sorry, I haven't seen your answer. I was just answering some other question, then I passed to a related question and only read the question itself, not the answers. I swear I haven't taken it from you since I haven't even seen that. I used maple (c) and some heruistics. Anyway, I'm sorry again, and if I have read through the answers, I wouldn't have posted my answer. – Nima Bavari Feb 13 at 19:35

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