Given $$x_n=\frac{1}{n^2+1}+\frac{1}{n^2+2}+\frac{1}{n^2+3}+\cdots+\frac{1}{n^2+n}$$
Verify if there is or no a limit. Find it if affirmative.
Let $a_n=\frac{n}{n^2+1}$ (The biggest portion of the sum $n$ times) and $b_n=\frac{n}{n^2+n}$ (the smallest portion of the sum $n$ times) then $$b_n\le x_n \le a_n$$ since$$\lim \frac{n}{n^2+1}=\lim \frac{n}{n^2+n}=0,$$ we have that $$\lim x_n=0.$$
Is this wrong? why? If it is, any tips on how to find $\lim x_n$? Grateful for any help.
**Edited
Your reasoning is true, but I think the following a bit of better.
$$0<x_n<n\cdot\frac{1}{n^2}=\frac{1}{n}$$ Thus, $$0\leq\lim_{n\rightarrow+\infty}x_n\leq\lim_{n\rightarrow+\infty}\frac{1}{n}=0,$$ which says $\lim\limits_{n\rightarrow+\infty}x_n=0$.