Problem with counting limit. I am obliged to count the
$$\lim_{n\rightarrow\infty} a_n$$
where
$$a_n=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}.$$
I know that this type of task, I should use squeeze theorem. I should find two sequences one that is smaller and one that is bigger. So I find $\large b_n=\frac{1}{n^2+1}$, $ b_n \le a_n$, but I have trouble in finding the bigger sequence $c_n \ge a_n$. I will be glad for any help.

We will use the following: $$\sum_{k=1}^n k=\frac{n(n+1)}{2}.$$
Since $n\geq 1$, we have
$$a_{n}=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}\geq\frac{1}{n^2+n}+\frac{2}{n^2+n}+\frac{3}{n^2+n}+\cdots+\frac{n}{n^2+n}=\frac{n(n+1)}{2(n^2+n)}$$
and
$$a_{n}=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}\leq\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\cdots+\frac{n}{n^2}=\frac{n(n+1)}{2n^2}.$$
By the squeeze theorem, $\lim_{n\rightarrow\infty} a_n=\frac{1}{2}$.
