Find $\lim _{n\to \infty }\left(\frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}\right)$ I’m having a problem approaching the question. The fact that it’s a limit of a sum is confusing to me. 
$$\lim _{n\to \infty }\left(\frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}\right)$$

Assuming you nmeant the first term to be $\frac{1}{n^2+1}$ the expression you present is 
$$
\lim_{n\rightarrow \infty} 
\left( H_{n^2+n} - H_{n^2}\right)$$ where $H_n$ is the harmonic sum $\sum_{k=1}^n \frac{1}{k}$.
For large $n$, an asymptotic expansion of $H_n$ is
$$
H_n = \ln n + \gamma + \frac{1}{2n} - \frac {1}{12n^2} + \cdots
$$
So 
$$
\lim_{n\rightarrow \infty} 
\left( H_{n^2+n} - H_{n^2}\right) = \lim_{n\rightarrow \infty} \left[ \ln\frac{n^2+n}{n^2} + \frac{1}{2n^2+2n} - \frac{1}{2n^2} + O(\frac{1}{n^3}) \right]
$$
This turns out to be zero.
