<s>[INST] The limit of sums of the form $ \frac{1}{\sqrt{2n}}- \frac{1}{\sqrt{2n+1}}+\frac{1}{\sqrt{2n+2}}-\dotsb-\frac{1}{\sqrt{4n}}$ I need to calculate limit:
$$
\lim\limits_{n \to \infty} \left ( \frac{1}{\sqrt{2n}}- \frac{1}{\sqrt{2n+1}}+\frac{1}{\sqrt{2n+2}}-\dotsb+\frac{1}{\sqrt{4n}}\right )
$$
Any hints how to do that would be appreciated.
 [/INST] The quantity inside the limit is between $\frac{1}{\sqrt{2n}}$ and $\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}$, hence the limit is zero by squeezing. To notice it, it is sufficient to couple consecutive terms: let $A_n=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$. 
Then $A_n>0$ and:
$$ \frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}+\ldots+\frac{1}{\sqrt{4n}}=\frac{1}{\sqrt{2n}}-\sum_{k=n}^{2n-1}A_{2k+1}<\frac{1}{\sqrt{2n}} $$
as well as:
$$ \frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}+\ldots+\frac{1}{\sqrt{4n}}=A_{2n}+\sum_{k=n+1}^{2n-1}A_{2k}+\frac{1}{\sqrt{4n}}>\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}.$$
</s>