Limit of the sequence $\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$ As in the title, we have to calculate the limit $$\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$$ when $n\rightarrow+\infty$. It is easy to see that $$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}=0$$ and $$\lim_{n\to\infty} \frac{n}{n^2+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n}=1$$
I don't know how to calculate the limit. I am  pretty sure that we somehow apply the sandwich rule, but I am afraid that there is something similary with $$\lim_{n\to\infty} \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}=\frac{\pi^2}{6}$$    

Another idea is to identify the sum
$$s = \sum _{k=1}^n \frac{k}{k+n^2}$$
in the limit $n\to \infty$ as a Riemann integral.
In fact we can write
$$s = \frac{1}{n} \sum _{k=1}^n \frac{k}{n \left(\frac{k}{n^2}+1\right)}$$
For $n\to\infty$ we can neglect $\frac{k}{n^2}$ against unity:
$$s = \frac{1}{n} \sum _{k=1}^n \frac{k}{n} $$
which asymptotically goes to
$$s \to \int_0^1 x \, dx = 1/2$$
An obvious generalization of the sum gives immediately
$$s_3 = \sum _{k=1}^n \frac{k^2}{k^2+n^3}\to \int_0^1 x^2 \, dx = 1/3 $$
and more generally
$$s_m = \sum _{k=1}^n \frac{k^{m-1}}{k^{m-1}+n^m}\to \int_0^1 x^{m-1} \, dx = 1/m $$
The latter relation holds even for any real $m \gt 0$.
