Prove that $\lim_{n\to\infty} \sum_{k=1}^{n}$ $(\frac {k}{n^2})^{{k\over n^2}+1} = {1\over 2}$. To simplify the question I imagined what if $k\over n^2$= $x$ then it will look like  $x^{x+1}$.Else I couldn't think on how to proceed ahead.
In one of my previously asked question viz. Prove that $\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = {1\over2}$.. Can I relate both of these questions?
 A: Both questions are related, indeed: $x^x$ is monotone decreasing for $0<x<e^{-1}$, so (at least for $n\ge3$) we have
$$n^{-1/n}\le x^x<1$$ for $0<x\le1/n$. So for this question, since $0<k/n^2\le1/n$,
$$n^{-1/n}\sum^n_{k=1}\frac{k}{n^2}\le\sum^n_{k=1}\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}<\sum^n_{k=1}\frac{k}{n^2}.$$ Now, use $$\sum^n_{k=1}k=\frac{n(n+1)}2$$ and the squeeze theorem.
A: We will show that
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}=\frac{1}{2}$$
Your desired limit will then follow from $\lim_{n\to\infty}\left(1+1/n\right)=1$ and the product rule for limits.
$$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}=\lim_{n\to\infty}\left[\left(1+\frac{1}{n}\right)\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}\right]=1\cdot\frac{1}{2}=\frac{1}{2}$$
Fix $\varepsilon>0$ arbitrarily. Since $\lim_{x\to 0^+}x^{x+1}/x=1$ (see my answer in your linked post), there is a $\delta>0$ satisfying
$$\left|\frac{x^{x+1}}{x}-1\right|<2\varepsilon\text{ whenever }0<x<\delta$$
which says the same thing as
$$\begin{equation}(1-2\varepsilon)x<x^{x+1}<(1+2\varepsilon)x\text{ whenever }0<x<\delta\end{equation}\tag{$\dagger$}$$
Pick $N\in\mathbb N$ sufficiently large for $1/N\leq \delta$. Then for any $n>N$ and any $k\in\mathbb N$ with $1\leq k\leq n$, we can see from division by $n^2$ that
$$\frac{1}{n^2}\leq\frac{k}{n^2}\leq\frac{n}{n^2}=\frac{1}{n}<\frac{1}{N}\leq\delta$$
so $0<k/n^2<\delta$ for every $n>N$ and $k\in\{1,\dots,n\}$. From $(\dagger)$, we get
$$(1-2\varepsilon)\frac{k}{n^2}<\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}<(1+2\varepsilon)\frac{k}{n^2}\text{ whenever }n>N\text{ and }1\leq k\leq n$$
Summing from $k=1$ to $n$ gives
$$\sum_{k=1}^n(1-2\varepsilon)\frac{k}{n^2}<\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}<\sum_{k=1}^n(1+2\varepsilon)\frac{k}{n^2}\text{ whenever }n>N$$
and, after evaluating the left and rightmost sums with $\sum_{k=1}^n k=n(n+1)/2$ and doing some algebra,
$$\left(\frac{1}{2}-\varepsilon\right)\left(1+\frac{1}{n}\right)<\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}<\left(\frac{1}{2}+\varepsilon\right)\left(1+\frac{1}{n}\right)\text{ whenever }n>N$$
Dividing through by $1+1/n$ and then subtracting by $1/2$ shows that
$$\left|\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}-\frac{1}{2}\right|<\varepsilon\text{ whenever }n>N$$
and since $\varepsilon>0$ was arbitrary, this implies
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}=\frac{1}{2}$$
