real analysis - Prove $limlimits_{nto infty}frac{1}{sqrt n}left|sumlimits_{k=1}^n (-1)^ksqrt kright|= frac{1}{2}$

I'm trying to show that $$\lim_{n\to \infty} x_n=\lim_{n\to \infty}\frac{1}{\sqrt n}\left|\sum_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}.$$
Assuming $\lim\limits_{n\to\infty} x_n=x$ exists, we have

$$x_{2n}=\frac{\sqrt{2n-1}(-x_{2n-1})+\sqrt {2n}}{\sqrt {2n}}$$

Letting $n\to \infty$,

$$\quad \quad x=-x+1$$
$$x=\frac{1}{2}$$

But I'm stuck on proving the existence of $\lim x_n$. Any idea?

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Update: I just solved the problem using sandwich theorem + integral test. Still, I would like to see a continuation of my initial idea, i.e. proving

$\{x_{2n}\}$ is monotonically increasing (similarly, $\{x_{2n+1}\}$ is monotonically decreasing)

Answer

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

$\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.

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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 $\,a>-x$