Prove $\lim\limits_{n\to \infty}\frac{1}{\sqrt n}\left|\sum\limits_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}$

Solution 1:

You can also apply Stolz–Cesàro, for $$z_{2n}=\frac{\sum\limits_{i=1}^{n}\left(\sqrt{2i}-\sqrt{2i-1}\right)}{\sqrt{2n}}=\frac{a_n}{b_n}$$ where $b_n=\sqrt{2n}$ is strictly monoton and divergent. Then $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}= \frac{\sqrt{2(n+1)}-\sqrt{2n+1}}{\sqrt{2(n+1)}-\sqrt{2n}}=\\ \frac{1}{2}\cdot \frac{\sqrt{2(n+1)}+\sqrt{2n}}{\sqrt{2(n+1)}+\sqrt{2n+1}}\to \frac{1}{2}, n\to\infty$$ and finally $$z_{2n+1}=\left|z_{2n}\cdot\sqrt{\frac{2n}{2n+1}}-1\right|\to\left|\frac{1}{2}-1\right|=\frac{1}{2}, n\to\infty$$

Solution 2:

$\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n}}\left|\sum\limits_{k=1}^n (-1)^k\sqrt{k}\right| = \lim_{n\to\infty}\frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k-1}+\sqrt{2k}} $

$\displaystyle \frac{1}{2\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{ \sqrt{2k} } < \frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{\sqrt{2k-1} +\sqrt{2k} } $$\displaystyle < \frac{1}{2\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{\sqrt{2k-1} } < \frac{1}{2\sqrt{2n}}\left(1+\sum\limits_{k=1}^{n-1} \frac{1}{ \sqrt{2k} }\right)$

With Riemann we get:

$\displaystyle \frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{ \sqrt{2k} } = \frac{1}{2n}\sum\limits_{k=1}^n\frac{1}{\sqrt{k/n}} \to \frac{1}{2}\int\limits_0^1\frac{dx}{\sqrt{x}} = \sqrt{x}|_0^1 =1$

And therefore the confirmation of the claim.

About the monotonie.

$\displaystyle s_n := \sum\limits_{k=1}^n \frac{1}{\sqrt{2k-1}+\sqrt{2k}}$

Assumption about the monotonie: $\enspace \displaystyle \frac{s_n}{\sqrt{2n}} \enspace$ is strictly increasing.

$\displaystyle \frac{s_n}{\sqrt{2n}} < \frac{s_{n+1}}{\sqrt{2{n+2}}} \,$ leads to

$\displaystyle \left(\frac{1}{\sqrt{2n}} - \frac{1}{\sqrt{2n+2}}\right)s_n < \frac{1}{\sqrt{2n+2}}\frac{1}{ \sqrt{2n+1} + \sqrt{2n+2} }\,$ and therefore to

$\displaystyle s_n < a_n:=\frac{\sqrt{2n}}{2}\frac{ \sqrt{2n+2}+\sqrt{2n} }{\sqrt{2n+2} +\sqrt{2n+1} } $

For $n=1$ it’s o.k. . Assume it’s correct for $n$ . $(*)$

Then we have to show that it's also correct for $n \to n+1$ .

$\displaystyle s_{n+1} < \frac{\sqrt{2n}}{2}\frac{ \sqrt{2n+2}+\sqrt{2n} }{\sqrt{2n+2} +\sqrt{2n+1} } + \frac{1}{\sqrt{2n+1} +\sqrt{2n+2} } < a_{n+1}$

The first inequation follows from the assumption $(*)$ and the second inequation can be proved by some transformations, $2n$ replaced by $x$ and $x\geq 0$:

$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+4}+\sqrt{x+3}) <$

$<\sqrt{x+2}(\sqrt{x+4}+ \sqrt{x+2})( \sqrt{x+2}+\sqrt{x+1})$

Transformations:

$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+4}+\sqrt{x+2}) $$+ (\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+3}-\sqrt{x+2}) <$ $<\sqrt{x+2}(\sqrt{x+4}+ \sqrt{x+2})( \sqrt{x+2}+\sqrt{x+1})$

$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+3}-\sqrt{x+2}) $$< (\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+1})-(\sqrt{x}(\sqrt{x+2}+\sqrt{x})+2))(\sqrt{x+4}+\sqrt{x+2})$

$(\sqrt{x}\sqrt{x+2} +x+2)(\sqrt{x+3}-\sqrt{x+2}) $$< (\sqrt{x+2}\sqrt{x+4} +x+2)(\sqrt{x+1}-\sqrt{x})$

This is true because of:

$\sqrt{x}\sqrt{x+2} +x+2< \sqrt{x+2}\sqrt{x+4} +x+2$

$\sqrt{x+3}-\sqrt{x+2} < \sqrt{x+1}-\sqrt{x}$

Note: $\enspace\sqrt{x+1+a}-\sqrt{x+a}~$ is decreasing by growing $\,a>-x$