2
$\begingroup$

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

  • 1
    Do you think people keeps track of who asks what and when??2012-08-30

1 Answers 1

2

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 $N\to\infty$.