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How to compute the series which seems different with the questions I raised before? $\lim_{N\rightarrow\infty}N\sum^{N}_{k=2}\left(\frac{k-1}{N}\right)^{N^2}$

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Do you think people keeps track of who asks what and when?? – DonAntonio Aug 30 '12 at 3:36
up vote 2 down vote accepted

The biggest term in the sum $\sum_{k=2}^N((k-1)/N)^{N^2}$ is the one with $k=N$, and that's $$((N-1)/N)^{N^2}=((1-(1/N))^N)^N$$ Now $(1-(1/N))^N\to(1/e)$ as $N\to\infty$, so the biggest term looks like $e^{-N}$. If you can get some bound for $(1-(1/N))^N$ in terms of $1/e$ then you should be able to relate your whole problem to $N^2e^{-N}$, which goes to 0 as $N\to\infty$.

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Thank you so much – jany Aug 30 '12 at 4:01

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