Trying to transform a Riemann-sum-like expression into an integral This question comes from curiosity. I know that, in general:
$$\lim_{n\to\infty}\frac1n\sum_{k=0}^n f(k/n)=\int_0^1 f(x)\,\mathrm dx\tag1$$
Then I have this expression
$$\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^nk^{1+\frac1k}\tag2$$
I know, by other methods, that the value of $(2)$ is $1/2$, but Im interested to know if its possible (or someone know how) to transform this Riemann-sum-like in an integral of Riemann for some appropriated $f$.
Indeed Im interested too to know some reference about this topic (if it exists).
 A: You may take a look at Theorem 1 given in A Generalization of Riemann Sums by Omran Kouba. In your case, take $a_k=k^{1/k}$, $f(x)=x$, $\alpha=1$, and $L=1$ (see Showing $ \sum_{k=1}^{n} k^{1/k}\sim n$), then
$$\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^nk^{1+\frac1k}=
\lim_{n\to\infty}\frac1{n^{\alpha}}\sum_{k=1}^n f(k/n)a_k=L\int_0^1 \alpha x^{\alpha-1}f(x)\,dx=\int_0^1 xdx=\frac{1}{2}.$$
More generally, by taking $f(x)=x^m$ we have that
$$\lim_{n\to\infty}\frac1{n^{m+1}}\sum_{k=1}^nk^{m+\frac1k}=
\lim_{n\to\infty}\frac1{n^{\alpha}}\sum_{k=1}^n f(k/n)a_k=L\int_0^1 \alpha x^{\alpha-1}f(x)\,dx=\frac{1}{m+1}.$$ 
A: Yet another proof:
By the Cesaro-Stolz theorem, we are led to consider the limit
$$
\lim_n \frac{n^{1+1/n}}{n^2-(n-1)^2} = 
\lim_n \frac{n}{2n-1}\, n^{1/n} = \frac{1}{2}.
$$
Since this limit exists, we can conclude that also the original limit exists and it is $1/2$.
A: I propose a direct proof.
Since $k^{1/k} > 1$ and $\lim_k k^{1/k} = 1$, given $\varepsilon > 0$ there exists $N\in\mathbb{N}$ such that $1 < k^{1/k} < 1+\varepsilon$ for every $n>N$.
Hence, for every $n>N$, we have that
$$
\begin{split}
0 & \leq \frac{1}{n^2}\sum_{k=1}^n k^{1+1/k} - \frac{1}{n^2}\sum_{k=1}^n k
= \frac{1}{n^2}\sum_{k=1}^N (k^{1+1/k} - k)
+ \frac{1}{n^2}\sum_{k=N+1}^n k(k^{1/k} - 1)
\\ & \leq
\frac{1}{n^2}\sum_{k=1}^N (k^{1+1/k} - k)
+ \frac{\varepsilon}{n^2}\sum_{k=N+1}^n k
\leq
\frac{1}{n^2}\sum_{k=1}^N (k^{1+1/k} - k)
+ \frac{\varepsilon}{2}\,,
\end{split}
$$
so that
$$
0\leq \limsup_{n\to +\infty}
\left(\frac{1}{n^2}\sum_{k=1}^n k^{1+1/k} - \frac{1}{n^2}\sum_{k=1}^n k\right)
\leq \frac{\varepsilon}{2}\,.
$$
It follows that
$$
\lim_{n\to+\infty}
\frac{1}{n^2}\sum_{k=1}^n k^{1+1/k}
=
\lim_{n\to+\infty}
\frac{1}{n^2}\sum_{k=1}^n k
=
\lim_{n\to+\infty}\frac{n(n+1)}{2n^2} = \frac{1}{2}\,.
$$
