Evaluate the following limit: Find
$$\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left[\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right]$$
MY TRY:
$$
\begin{align}
\lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}} 
+ \frac{1}{\sqrt{4}+\sqrt{6}} 
+ \cdots + \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr] \\
&= \lim_{n \to \infty} \frac{\sqrt{n}}{n} \biggl[ 
\frac{1}{\sqrt{2}+\sqrt{4}} 
+ \frac{1}{\sqrt{4}+\sqrt{6}} + \cdots 
+ \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr]
\end{align}
$$
Now using Cauchy first thm of limits
$$
a_n = \frac{\sqrt{n}}{\sqrt{2n}+\sqrt{2n+1}}
$$
The answer should be $\frac{1}{2\sqrt{2}}$.
But the answer is $1/\sqrt{2}$.
 A: $$\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=\frac{\sqrt{2k+2}-\sqrt{2k}}{2}$$
The sum telescopes to $\displaystyle \frac{1}{\sqrt n} \frac {\sqrt{2n+2} - \sqrt 2 }{2}$ which converges to $1/\sqrt{2}$

If you miss this, you can still use integrals to get bounds on $\sum_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$
A: If one attempts to use Cauchy's First Limit Theorem, then we start with
$$\begin{align}
\lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}&=\lim_{n\to \infty}\sqrt{n}\frac1n\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}} \tag 1\\\\
\end{align}$$
The Theorem reveals that since $\lim_{n\to \infty}\frac{1}{\sqrt{2n}+\sqrt{2n+2}}=0$, then $\lim_{n\to \infty}\frac1n\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=0$ 

Therefore, the limit on the right-hand side of $(1)$ is of indeterminate form (i.e., $\infty \times 0$) and we need to pursue its evaluation judiciously.

We revert to the left-hand side of $(1)$ and note that both $\sqrt{n}$ and $\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$ are monotonically increasing, divergent sequences.  We can invoke, therefore, the Stolz-Cesaro Theorem and write
$$\begin{align}
\lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}&=\lim_{n\to \infty}\frac{\sum_{k=1}^{n+1}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}-\sum_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}}{\sqrt{n+1}-\sqrt{n}} \\\\
&=\lim_{n\to \infty}\frac{\frac{1}{\sqrt{2n+2}+\sqrt{2n+4}}}{\sqrt{n+1}-\sqrt{n}}\\\\
&=\lim_{n\to \infty}\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{2n+2}+\sqrt{2n+4}}\\\\
&=\frac{1}{\sqrt{2}}
\end{align}$$
as was to be shown!
A: This is a classic partial sums of integrals problem. Your expression is:
$$\frac{1}{\sqrt{n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$$
write as 
$$\frac{1}{n}\sum\limits_{k=1}^n \frac{1}{\sqrt{2\frac{k}{n}}+\sqrt{2\frac{k}{n}+2}}$$
then the limit is the same as the integral
$$\int_0^1\frac{1}{\sqrt{2x}+\sqrt{2x+2}}dx=
\int_0^1\frac{\sqrt{2x}-\sqrt{2x+2}}{-2}dx
$$
