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.
 A: The limit you are asking for appears to be:
$$\lim_{n\to\infty} \sum_{m=2n}^{4n} \frac{(-1)^m}{\sqrt{m}}.$$
Note that the series $$\sum_{n=0}^\infty \frac{(-1)^m}{\sqrt{m}}$$ converges by the alternating series test. A series converges if for every $\epsilon > 0$ there exists an integer $N$ for which given any $n_2>n_1 > N$ we have
$$\left| \sum_{m=n_1}^{n_2} \frac{(-1)^m}{\sqrt{m}} \right| < \epsilon.$$
We can choose $n_1 = 2n$ and $n_2=4n$. Thus for any $\epsilon>0$, $$\left| \sum_{m=2n}^{4n} \frac{(-1)^m}{\sqrt{m}} \right| < \epsilon$$ for sufficiently large $n$. Since $\epsilon$ can be as small as we want, the limit must be zero.
A: 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:
$$ \frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}+\ldots+\frac{1}{\sqrt{4n}}=A_{2n}+\sum_{k=n+1}^{2n-1}A_{2k}+\frac{1}{\sqrt{4n}}>\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}.$$
