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 $1/\sqrt{2}$.

$$\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=\frac{\sqrt{2k+2}-\sqrt{2k}}{2}$$
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 $\sum_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$
