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. 
$$\lim _{n\to \infty }\left(\frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}\right)$$
 A: 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.
A: 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.
A: 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...
A: 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 $\lim_{n\to \infty}a_n = 0$.
$$\left(  \lim_{n\to \infty}b_n =  \lim_{n\to \infty}c_n = 0 \wedge \left(\forall n \in \mathbb{N}\right)\left( n \geq 1 \Rightarrow b_n \leq a_n \leq c_n \right)\right) \Longrightarrow \lim_{n\to\infty}a_n = 0$$
