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...
 A: Hint:
$$\frac{n}{\sqrt{n^2+n}}\leq(\frac{1}{\sqrt{n^2+1}} + ... + \frac{1}{\sqrt{n^2+n}}) \leq \frac{n}{\sqrt{n^2+1}}$$
A: I want to propose a more powerful method which might be of your use in harder problems. That is use of the Riemann definition for the integral. For your question, we could write $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{1}{{\sqrt {{n^2} + i} }}}  = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{1}{{\sqrt {1 + \frac{i}{n}\frac{1}{n}} }}\frac{1}{n}} $$now, lets assume that this is a discritized version of a continuous integral, for that assume that $x = \frac{i}{n}$, clearly in this case $dx$ which is the difference between two "consecutive" values of $x$ is ${\frac{1}{n}}$ so that we get $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{1}{{\sqrt {{n^2} + i} }}}  = \int\limits_0^1 {\frac{{dx}}{{\sqrt {1 + xdx} }}}  = \int\limits_0^1 {dx} $$Okay, I get that this is like using a cannon for a fly, but things would have got more interesting if the $i$ in the denominator of your question was also squared. Hope it helps ;)
