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How to compute the following limit which is related to a series?

$ \lim_{N\rightarrow\infty}N^2\sum^{N-1}_{k=1}\left(\frac{k}{N}\right)^{N\ln N}$

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    Yes, jany, the details of the solutions are different, but my question remains: have you not learned anything from the other four that could at least get you started working on this one, instead of offering it to us with no sign that you have put even the least bit of thought into it? If you aren't learning anything usable, what is the purpose of asking all the questions?2012-08-31

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$\lim_{N\rightarrow \infty }N^{2}\sum_{k=1}^{N-1}\left( \frac{k}{N} \right) ^{N\ln N}=\lim_{N\rightarrow \infty }\sum_{k=1}^{N-1}N^{2}\left( \frac{k}{N}\right) ^{N\ln N}$

All the terms from $k=1$ to $k=N-3$ tend to $0$. The $k=N-2$ term tends to $1$ and the $k=N-1$ term tends to infinity. So the limit doesn't exist.

$\begin{eqnarray*} &&\lim_{N\rightarrow \infty }N^{2}\left( \frac{k}{N}\right) ^{N\ln N}=0,\qquad 1\le k\le N-3, \\&& \lim_{N\rightarrow \infty }N^{2}\left( \frac{N-2}{N}\right) ^{N\ln N}=1\\&&\lim_{N\rightarrow \infty }N^{2}\left( \frac{N-1}{N}\right)^{N\ln N}=+\infty . \end{eqnarray*} $

Detailed computation. Since for $p\le N$ $\begin{equation*} \lim_{N\rightarrow \infty }\left( 1-\frac{p}{N}\right) ^{N}=e^{-p} \end{equation*}$ we have $\begin{eqnarray*} \lim_{N\rightarrow \infty }N^{2}\left( \frac{N-p}{N}\right) ^{N\ln N} &=&\lim_{N\rightarrow \infty }N^{2}\left( \left( 1-\frac{p}{N}\right) ^{N}\right) ^{\ln N} \\ &=&\lim_{N\rightarrow \infty }N^{2}e^{-p\ln N} \\ &=&\lim_{N\rightarrow \infty }N^{2}N^{-p},\qquad N=e^{\ln N} \\ &=&\lim_{N\rightarrow \infty }N^{2-p} \end{eqnarray*}$ and

$\lim_{N\rightarrow \infty }N^{2-p} =\begin{cases} 0, & \text{if $p>2$ } \\ 1, & \text{if $p=2$ } \\ +\infty, & \text{if $p<2.$ } \\ \end{cases}$

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The last term in the sum is $\left(1-\frac{1}{N}\right)^{N\log N}$. Since $(1-1/N)^N$ is roughly $1/e$, the last term is roughly of size $1/N$. Multiply by $N^2$.

More precisely, $\lim_{N\to\infty}(1-1/N)^N=e^{-1}$. Thus if $N$ is large enough, then $(1-1/N)^N\gt e^{-1.1}$, and therefore $(1-1/N)^{N\log N}\gt N^{-1.1}$. So the $N$-th term of our sequence is $\gt N^{0.9}$. Thus, depending on taste, the limit doesn't exist or is $+\infty$.

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    @SeyhmusGüngören: The other terms are each "small" but of course there are lots of them. So we do have to be careful. In this case, as we go backwards, they go down rapidly enough for convergence. But one has to do the details, funny things can happen: after all, $\sim\frac{1}{n}$ diverges.2012-08-31