# Find the limit $\lim\limits_{n\to \infty}\sum\limits_{k=1}^{n}\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}$

Find the limit $$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{k}{n^2}\right )^{\frac{k}{n^2}+1}$$ Have no idea.

• In which context did this series araised ? Are you sure it converges ? – Renato Faraone Sep 1 '15 at 21:32
• I have computed partial sums on my computer and i think answer is $\frac{1}{2}$. I don't know original context, but i have found it in olympiad. – uijok bijk Sep 1 '15 at 21:34

Hints: $x^x$ is a continuous and decreasing function on $\left(0,\frac{1}{3}\right)$, hence for any sufficiently large $n$, $\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}}$ is between $1$ and $\frac{1}{\sqrt[n]{n}}$. On the other hand, $$\sum_{k=1}^{n}\frac{k}{n^2}=\frac{n(n+1)}{2n^2}=\frac{1}{2}+\frac{1}{2n},$$ hence the limit is $\color{red}{\frac{1}{2}}$ by squeezing.
• Also, $n^{1/n} < 1+c\log(n)/n$, so $1/n^{1/n} > 1-d\log(n)/n$ for some real $c$ and $d$ for large enough $n$. – marty cohen Sep 1 '15 at 22:15
• @martycohen: we may also use the AM-GM inequality, since $$\sqrt[n]{n}=\sqrt[n]{\left(1+\frac{1}{n-1}\right)\cdot\ldots\left(1+\frac{1}{1}\right)\cdot 1}\leq 1+\frac{H_{n-1}}{n}.$$ – Jack D'Aurizio Sep 1 '15 at 22:32
• And by grouping into groups of size $2^n$, $H_n < \log_2(n)$. – marty cohen Sep 1 '15 at 23:36
• @martycohen Even simply perhaps, $\frac{1}{1+\sqrt{\frac{2}{n-1}}}<\frac{1}{n^{1/n}}<1$ – Mark Viola Sep 2 '15 at 2:05