Find $\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$ I don't know how to find the sum of $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$. After rationalization we have
$\left(\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right){/}\left((\sqrt{2}+\sqrt{4})(\sqrt{4}+\sqrt{6})...(\sqrt{2n}+\sqrt{2n+2})\right)=$
$(\sqrt{4}+\sqrt{6})...(\sqrt{2n}+\sqrt{2n+2})+(\sqrt{2}+\sqrt{4})...(\sqrt{2n}+\sqrt{2n+2})+...+(\sqrt{2}+\sqrt{4})...(\sqrt{2n-2}+\sqrt{2n})$
How to reduce this sum to general form? 
 A: $$\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k+2}+\sqrt{2k}}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{(\sqrt{2k+2}+\sqrt{2k})(\sqrt{2k+2}-\sqrt{2k})}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{2k+2-2k}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{2}\right)$$
$$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})\right)$$
$$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\cdot (\sqrt{n+1}-1)\right)$$
$$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left( \sqrt{1+\frac{1}{n}}-\frac{1}{\sqrt{n}}\right)$$
$$=\frac{1}{\sqrt{2}}$$
A: Hint rationlize you will get a telescoping series for eg after rationalizing first two terms we get $\frac{\sqrt{4}-\sqrt{2}+\sqrt{6}-\sqrt{4}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2}$ on rationalizing till $n$ see which terms remain then you will get your steps
