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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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Hint: $n\ln n$ is asymptotically the $n$th prime number. – dot dot Aug 31 '12 at 10:42
The 5th question of this type you have asked. Have you not learned something from the other four that you could at least try to apply to this one? – Gerry Myerson Aug 31 '12 at 12:17
@dot, what do primes have to do with it? – Gerry Myerson Aug 31 '12 at 12:18
the solutions of each one is different. please see the corresponding answer. – jany Aug 31 '12 at 13:13
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? – Gerry Myerson Aug 31 '12 at 13:35

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

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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If instead of $N^2$ we had $N$ would we have also $\infty$ as the limit? ass it seems then we would have again the summation of infinitely many number of constants? – Seyhmus Güngören Aug 31 '12 at 12:02
If instead of $N^2$ we have $N^{1.00001}$, the limit is still infinite. But if instead of $N^2$ we have $N$, then we need to look at the other terms in the sum. An informal quick calculation makes me fairly confident that with $N$ instead of $N^2$ the limit is finite. – André Nicolas Aug 31 '12 at 12:31
@SeyhmusGüngören If we had $N$ the limit would be $1$. All terms would tend to $0$, but the last one, which would tend to $1$. – Américo Tavares Aug 31 '12 at 12:31
ahh okay I never checked the other terms I just thought that all the terms would have some constant and infintely many summation of constants would yield $\infty$. Okay I see this also from Americo's answer. Thanks alot. – Seyhmus Güngören Aug 31 '12 at 12:35
@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. – André Nicolas Aug 31 '12 at 12:40

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