limit $\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right)$ 
$$\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt{2n} +\sqrt{2n+2}}\right)$$

To find the limit I think I can use the sandwich theorem: if $f,g$ and $h$ are real functions such that $f(x)\le g(x) \le h(x)$ for all $x$ and if $\lim_{x\to 0} f(x)=L= \lim_{x\to 0} h(x)$, then $\lim_{x\to 0} g(x)= L$.
I don't know that which series I can take for $f(x)$ and $h(x)$ to compare with $g(x)$.
 A: Note that
$$\frac{1}{\sqrt{k}+\sqrt{k+2}}=\frac{\sqrt{k+2}-\sqrt{k}}{2}$$
So we have 
\begin{align}
&\;\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt2n+\sqrt{2n+2}}\right)\\
=&\;\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{\sqrt 4-\sqrt2}{2}+\frac{\sqrt6-\sqrt4}{2}+\cdots+\frac{\sqrt{2n+2}-\sqrt{2n}}{2}\right)\\
=&\;\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{\sqrt{2n+2}-\sqrt{2}}{2}\right)\\
=&\;\lim_{n\to\infty}\left(\frac{\sqrt{2+\frac{2}{n}}-\sqrt{\frac{2}{n}}}{2}\right)\\
=&\;\frac{\sqrt{2}}{2}
\end{align}
A: HINT:$$\frac{1}{\sqrt{2n}+\sqrt{2n+2}}=\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\cdot\frac{\sqrt{2n+2}-\sqrt{2n}}{\sqrt{2n+2}-\sqrt{2n}}=\frac{\sqrt{2n+2}-\sqrt{2n}}{2}$$
A: Hint: write $$\frac{1}{\sqrt{2n}+\sqrt{2n+2}}=\frac{\sqrt{2n}-\sqrt{2n+2}}{-2}$$ etc
A: $$\lim _{ n\to \infty  } \frac { 1 }{ \sqrt { n }  } \left( \frac { 1 }{ \sqrt { 2 } +\sqrt { 4 }  } +\frac { 1 }{ \sqrt { 4 } +\sqrt { 6 }  } +.........+\frac { 1 }{ \sqrt { 2n } +\sqrt { 2n+2 }  }  \right) =\\ =\lim _{ n\to \infty  } \frac { 1 }{ \sqrt { n }  } \left( \frac { \sqrt { 2 } -\sqrt { 4 }  }{ -2 } +\frac { \sqrt { 4 } -\sqrt { 6 }  }{ -2 } +.........+\frac { \sqrt { 2n } -\sqrt { 2n+2 }  }{ -2 }  \right) =\\ =\lim _{ n\to \infty  } -\frac { 1 }{ 2\sqrt { n }  } \left( \sqrt { 2 } -\sqrt { 4 } +\sqrt { 4 } -\sqrt { 6 } +.........+\sqrt { 2n } -\sqrt { 2n+2 }  \right) =\\ =\lim _{ n\to \infty  } -\frac { 1 }{ 2\sqrt { n }  } \left( \sqrt { 2 } -\sqrt { 2n+2 }  \right) =\lim _{ n\to \infty  } \left( \frac { \sqrt { 2n+2 }  }{ 2\sqrt { n }  } -\frac { \sqrt { 2 }  }{ 2\sqrt { n }  }  \right) =\frac { \sqrt { 2 }  }{ 2 } $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\lim_{n \to \infty}\pars{{1 \over \root{n}}
\sum_{k = 1}^{n}{1 \over \root{2k} + \root{2k + 2}}}
\\[5mm] = &\
\lim_{n \to \infty}\pars{{1 \over \root{n + 1} - \root{n}}
{1 \over \root{2n + 2} + \root{2n + 4}}}\qquad
\pars{~Stolz-Ces\grave{a}ro\ Theorem~}
\\[5mm] = &\
{\root{2} \over 2}\lim_{n \to \infty}{
{\root{n + 1} + \root{n} \over \root{n + 1} + \root{n + 2}}} =
{\root{2} \over 2}\lim_{n \to \infty}{
{\root{1 + 1/n} + 1 \over \root{1 + 1/n} + \root{1 + 2/n}}}
\\[5mm] = &\
{\root{2} \over 2}\,{1 + 1 \over 1 + 1} = \bbx{\root{2} \over 2}
\end{align}
