<s>[INST] 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 $g(x)$.
 [/INST] $$\lim _{ n\to \infty  } \frac { 1 }{ \sqrt { n }  } \left( \frac { 1 }{ \sqrt { 2 } +\sqrt { 4 }  } +\frac { 1 }{ \sqrt { 4 } +\sqrt { 6 }  } +.........+\frac { 1 }{ \sqrt { 2n } +\sqrt { 2n+2 }  }  \right) =\\ =\lim _{ n\to \infty  } \frac { 1 }{ \sqrt { n }  } \left( \frac { \sqrt { 2 } -\sqrt { 4 }  }{ -2 } +\frac { \sqrt { 4 } -\sqrt { 6 }  }{ -2 } +.........+\frac { \sqrt { 2n } -\sqrt { 2n+2 }  }{ -2 }  \right) =\\ =\lim _{ n\to \infty  } -\frac { 1 }{ 2\sqrt { n }  } \left( \sqrt { 2 } -\sqrt { 4 } +\sqrt { 4 } -\sqrt { 6 } +.........+\sqrt { 2n } -\sqrt { 2n+2 }  \right) =\\ =\lim _{ n\to \infty  } -\frac { 1 }{ 2\sqrt { n }  } \left( \sqrt { 2 } -\sqrt { 2n+2 }  \right) =\lim _{ n\to \infty  } \left( \frac { \sqrt { 2n+2 }  }{ 2\sqrt { n }  } -\frac { \sqrt { 2 }  }{ 2\sqrt { n }  }  \right) =\frac { \sqrt { 2 }  }{ 2 } $$
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