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How to compute this limit of the series?


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Reparametrize the sum as $m=N-k+1$ so you have


If you take $N$ to be large, the terms of the sum each tend to $e^{-m}$ and you're left with a geometric sum, giving the result $1/(e-1)$. The details can, of course, be made more rigorous.

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Thank you. What I want is just the rigrous detail. Could you please show us? – jany Aug 31 '12 at 3:11
@Jany: By the dominated convergence theorem, we can switch the order of the summation and the limit. This proves the result. – Eric Naslund Aug 31 '12 at 3:45
To make Eric's argument explicit, use $\left(1-\frac{m}{N}\right)^N < \mathrm{e}^{-m}$ for all $N \geqslant m$. – Sasha Aug 31 '12 at 3:50

Here is another way to see that the limit is $\frac{1}{e-1}$ based on the Bernoulli numbers and Bernoulli polynomials. The following is an outline, but as with Jonathan's answer, the details can be made rigorous.

If $B_p(x)$ is the $p^{th}$ Bernoulli polynomial, we have the formula $$\sum_{k=0}^{N-1}k^{p}=\frac{B_{p+1}\left(N\right)-B_{p+1}(0)}{p+1}, $$ and so the above limit is equal to $$\lim_{N\rightarrow\infty}\frac{1}{N^{N}}\frac{B_{N+1}\left(N\right)}{N+1}. $$ Using the expansion for the Bernoulli polynomial, $B_N(x)=\sum_{k=0}^N \binom{N}{k}B_k x^{N-k}$, we are looking at $\sum_{k=0}^{N+1}\binom{N+1}{N+1-k}B_{k}N^{N+1-k}, $ and so our limit is $$\lim_{N\rightarrow \infty}\sum_{k=0}^{N+1}\frac{B_{k}}{k!} \left(\frac{(N+1)!}{(N+1-k)!N^{k}}\right).$$ Being careful with bounds, the right most factor does not deviate far from $1$ when $k$ is small, and so it is possible to show that the above limit converges to $$\sum_{k=0}^\infty \frac{B_k}{k!}.$$ Recalling that $\frac{t}{e^t-1}$ is the generating function, we see that the above sum equals $$\frac{1}{e-1}.$$

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