Find $\lim _{n\to \infty }\left(\frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}\right)$

We have $\dfrac1{n^2+k} < \dfrac1{n^2}$. Hence, $$\sum_{k=2}^n \dfrac1{n^2+k} < \sum_{k=2}^n \dfrac1{n^2} < \dfrac{n}{n^2} = \dfrac1n$$ Now conclude what the limit should be.

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.

Assuming the sum starts with $\dfrac1{n^2+1}$ ....
We have $0 < \sum_{k=1}^n \dfrac1{n^2+k} < \sum_{k=1}^n\dfrac1{n^2+1}= \frac{n}{n^2+1}$ since each of the summands is smaller than $\dfrac1{n^2+1}$.

Now use the squeeze theorem...

I assume that it's $\lim_{n\to \infty}{\frac{1}{n^2+1} + \frac{1}{n^2+2}+ \cdots + \frac{1}{n^2+n}}$, otherwise it is meaningless for me.

Easy way to solve this problem is use squeeze theorem.

$$\left(\lim_{n\to\infty} b_n = \lim_{n\to\infty}c_n = m \wedge \left(\exists \delta \in \mathbb{R}\right)\left(\forall n \in \mathbb{N}\right)\left( n \geq \delta \Rightarrow b_n \leq a_n \leq c_n \right)\right) \Rightarrow \lim_{n\to\infty}a_n = m$$

We have $\left(a_n\right) \wedge a_n = \sum_{i=1}^{n}{\frac{1}{n^2+i}}$, so we have to select the appropriate $\left(b_n\right), \left(c_n\right)$. Let $b_n = \sum_{i=1}^{n}{\frac{1}{n^2+n}} \wedge c_n =\sum_{i=1}^{n}{\frac{1}{n^2}}$. It's not hard to see, that $\left(\forall n \in \mathbb{N}^{+}\right)\left(b_n \leq a_n \leq c_n\right)$.

Now we should compute $\lim_{n \to \infty}b_n, lim_{n \to \infty}c_n$

$$\begin{align*}\begin{split} \lim_{n\to \infty}b_n &= \sum_{i=1}^{n}{\frac{1}{n^2+n}} = \lim_{n \to \infty} \frac{n}{n^2+n} = \lim_{n \to \infty}\frac{1}{n+1} = 0 \\ \lim_{n\to \infty}c_n &= \lim_{n\to \infty}\sum_{i=1}^{n}{\frac{1}{n^2}} = \lim_{n\to \infty} \frac{n}{n^2} = \lim_{n\to \infty}\frac{1}{n} = 0 \end{split}\end{align*}$$

So, by virtue of squeeze theorem, we obtain $\lim_{n\to \infty}a_n = 0$.

$$\left( \lim_{n\to \infty}b_n = \lim_{n\to \infty}c_n = 0 \wedge \left(\forall n \in \mathbb{N}\right)\left( n \geq 1 \Rightarrow b_n \leq a_n \leq c_n \right)\right) \Longrightarrow \lim_{n\to\infty}a_n = 0$$