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