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 $$\lim_{n\to\infty} \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}=\frac{\pi^2}{6}$$    

Convince yourself that
$$\frac1{n^2+n}+\frac2{n^2+n}+\cdots+\frac n{n^2+n}\le \frac1{n^2+1}+\frac2{n^2+2}+\cdots+\frac n{n^2+n} \le$$
$$\le \frac1{n^2}+\frac2{n^2}+\cdots+\frac n{n^2},$$
and remember that $1+2+\cdots+n=\tfrac {n(n+1)}2$. Can you apply the Sandwich rule now?
