How to solve this sum limit? $\lim_{n \to \infty } \left( \frac{1}{\sqrt{n^2+1}}+\cdots+\frac{1}{\sqrt{n^2+n}} \right)$ How do I solve this limit?
$$\lim_{n \to \infty } \left( \frac{1}{\sqrt{n^2+1}}+\cdots+\frac{1}{\sqrt{n^2+n}} \right)$$
Thanks for the help!
 A: $$ \frac n{\sqrt{n^2+n}}\le  \frac1{\sqrt{n^2+1}}+\cdots+\frac{1}{\sqrt{n^2+n}}\le \frac n{\sqrt{n^2+1}}$$
Apply  Squeeze theorem
Observe that for finite $a,b$ $$\lim_{n\to\infty^+}\frac n{\sqrt{n^2+an+b}}=\lim_{n\to\infty^+}\sqrt{\frac{n^2}{n^2+an+b}}=\lim_{n\to\infty^+}\sqrt{\frac1{1+a\cdot\frac1n+b\cdot\frac1{n^2}}}=1$$
A: Extract n from the radical. So, you have to sum n terms which are close to 1 and you divide the sum by n. Are you able to continue with this ?
A: First of all, you need to prove that $\left\{s_n\right\}_{n \in \mathbb{N}} = \left\{ \frac{1}{\sqrt{n^2+1}}+\cdots+\frac{1}{\sqrt{n^2+n}} \right\}_{n \in \mathbb{N}} $ converges. In order to do that, we will find $\left\{a_n\right\}_{n \in \mathbb{N}} $ and $\left\{b_n\right\}_{n \in \mathbb{N}}$ such that:


*

*$a_n \leq s_n \leq b_n$ for all $n \in \mathbb{N}$

*$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = l$


And in this situation, we will be able to conclude that $\lim_{n \to \infty} s_n = l$.
Since $\frac{1}{\sqrt{n^2+n}} \leq \frac{1}{\sqrt{n^2+i}} \leq \frac{1}{\sqrt{n^2+1}}$ for all $i\in\left\{1, \ldots, n \right\}$, we have that $\frac{n}{\sqrt{n^2+n}} \leq s_n \leq \frac{n}{\sqrt{n^2+1}}$.
Moreover, $\lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}} = \lim_{n \to \infty} \frac{n}{\sqrt{n^2+1}} = 1$. That can be easily seen dividing numerator and denominator by $n$ in both sequences. 
Hence, we can conclude that $$ \lim_{n \to \infty} \frac{1}{\sqrt{n^2+1}}+\cdots+\frac{1}{\sqrt{n^2+n}} = 1$$
