How to solve this limit $\lim_{n\to\infty} ((\frac{1}{\sqrt{n^2+1}}) + \cdots + (\frac{1}{\sqrt{n^2+n}}))$ How do I solve this limit $$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}} + \cdots + \frac{1}{\sqrt{n^2+n}}\right)\text{ ?}$$
(n goes to plus infinity.)
I tried putting in $n=1,2,3,4,\ldots$ to find some pattern but it's hard to see where it's going.
For example, $n=1$, limit is $\dfrac{1}{\sqrt{2}}$
For example, $n=2$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}}$ 
For example, $n=3$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{12}}$
For example, $n=4$, limit is $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{12}} + \dfrac{1}{\sqrt{20}}$
I'm not exactly sure what this limit is converging to...

Hint:
$$\frac{n}{\sqrt{n^2+n}}\leq(\frac{1}{\sqrt{n^2+1}} + ... + \frac{1}{\sqrt{n^2+n}}) \leq \frac{n}{\sqrt{n^2+1}}$$
