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

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)

 A: $\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}) $$< (\sqrt{x+2}\sqrt{x+4} +x+2)(\sqrt{x+1}-\sqrt{x})$
This is true because of: 
$\sqrt{x}\sqrt{x+2} +x+2< \sqrt{x+2}\sqrt{x+4} +x+2$
$\sqrt{x+3}-\sqrt{x+2} < \sqrt{x+1}-\sqrt{x}$
Note: $\enspace\sqrt{x+1+a}-\sqrt{x+a}~$ is decreasing by growing $\,a>-x$
A: Since
$$
\frac1{\sqrt{2k+1}+\sqrt{2k-1}}\le\frac1{\sqrt{2k}+\sqrt{2k-1}}\le\frac1{\sqrt{2k}+\sqrt{2k-2}}\tag1
$$
we have
$$
\tfrac12\left(\sqrt{2k+1}-\sqrt{2k-1}\right)\le\sqrt{2k}-\sqrt{2k-1}\le\tfrac12\left(\sqrt{2k}-\sqrt{2k-2}\right)\tag2
$$

Summing over an even number of terms gives
$$
\begin{align}
\hspace{-1cm}\frac1{\sqrt{2n}}\sum_{k=1}^{2n}(-1)^k\sqrt{k}
&=\frac1{\sqrt{2n}}\sum_{k=1}^n\left(\sqrt{2k}-\sqrt{2k-1}\right)\\
&=\frac1{\sqrt{2n}}\sum_{k=1}^n\left[\tfrac12\left(\sqrt{2k+1}-\sqrt{2k-1}\right),\tfrac12\left(\sqrt{2k}-\sqrt{2k-2}\right)\right]_\#\\
&=\frac1{\sqrt{2n}}\left[\,\tfrac12\left(\sqrt{2n+1}-1\right),\tfrac12\sqrt{2n}\,\right]_\#\tag3
\end{align}
$$
where $[a,b]_\#$ represents a number in $[a,b]$.
Therefore, by the Squeeze Theorem,
$$
\lim_{n\to\infty}\frac1{\sqrt{2n}}\sum_{k=1}^{2n}(-1)^k\sqrt{k}=\frac12\tag4
$$

Summing over an odd number of terms gives
$$
\begin{align}
\frac1{\sqrt{2n+1}}\sum_{k=1}^{2n+1}(-1)^k\sqrt{k}
&=\frac1{\sqrt{2n+1}}\left(\sum_{k=1}^n\left(\sqrt{2k}-\sqrt{2k-1}\right)-\sqrt{2n+1}\right)\\
&=\frac1{\sqrt{2n+1}}\left[\,\tfrac12\left(\sqrt{2n+1}-1\right),\tfrac12\sqrt{2n}\,\right]_\#-1\tag5
\end{align}
$$
Therefore, by the Squeeze Theorem,
$$
\lim_{n\to\infty}\frac1{\sqrt{2n+1}}\sum_{k=1}^{2n+1}(-1)^k\sqrt{k}=-\frac12\tag6
$$

Combining $(4)$ and $(6)$ yields
$$
\lim_{n\to\infty}\frac1{\sqrt{n}}\left|\,\sum_{k=1}^n(-1)^k\sqrt{k}\,\right|=\frac12\tag7
$$
A: Here are some steps.


*

*Let $a_n:=n^{-1/2}\sum_{k=1}^n(-1)^k\sqrt k$. Study separately the behavior of $(a_{2n})$ and $(a_{2n+1})$. 

*For $a_{2n}$: notice that this term is equal to $$(2n)^{-1/2}\sum_{j=1}^n\left(\sqrt{2j}-\sqrt{2j-1}\right)=(2n)^{-1/2}\sum_{j=1}^n\frac{1}{\sqrt{2j}+\sqrt{2j-1}}.$$

*Add and substract $(2n)^{-1/2}\sum_{j=1}^n\frac{1}{2\sqrt{2j} }$ and show that 
$$
(2n)^{-1/2}\sum_{j=1}^n\frac{1}{\sqrt{2j}+\sqrt{2j-1}} -(2n)^{-1/2}\sum_{j=1}^n\frac{1}{2\sqrt{2j} }
$$
goes to zero. 

*It remains to study the limit of $4^{-1}n^{-1/2}\sum_{k=1}^nk^{-1/2}$ for example comparing with an integral.

A: Sums like this can be solved in general by the method of analytic regularization i.e.: From $$z^{-\epsilon} = \frac{1}{\Gamma(\epsilon)} \int_0^\infty \frac{{\rm d}t}{t} \, t^{\epsilon} \, {\rm e}^{-zt}$$ for $z=\frac{k}{n}$ one obtains
$$\sum_{k=1}^{n} (-1)^k \left(\frac{k}{n}\right)^{-\epsilon} = \frac{1}{\Gamma\left(\epsilon\right)} \int_0^\infty \frac{{\rm d}t}{t} \, t^{\epsilon} \sum_{k=1}^{n} (-1)^k \,  {\rm e}^{-{\frac{k}{n}}\,t} = \frac{1}{\Gamma\left(\epsilon\right)} \int_0^\infty \frac{{\rm d}t}{t} \, t^{\epsilon} \, \frac{(-1)^n \, {\rm e}^{-t}-1}{{\rm e}^{{t}/{n}}+1} $$
This line is fully valid for all $\epsilon>0$, in which case however, when taking the limit $n\rightarrow \infty$, the LHS manifestly diverges. Nevertheless, when $\epsilon<0$ the RHS can be understood as a to be regularized integral, meaning if it acquires a value it must correspond to a value of the LHS for $\epsilon<0$. The last equality also shows that one has to differ between even and odd $n$. When making a choice for either, the limit $n \rightarrow \infty$ exists and the value can be recovered by analytic regularization. Hence:
$$\stackrel{n \rightarrow \infty}{=} \frac{1}{2\,\Gamma\left(\epsilon\right)} \int_0^\infty \frac{{\rm d}t}{t} \, t^{\epsilon} \left\{(-1)^n \, {\rm e}^{-t}-1 \right\} = \frac{(-1)^n\Gamma\left(\epsilon\right) - 0}{2\,\Gamma\left(\epsilon\right)} = \frac{(-1)^n}{2}$$
where the $(-1)^n$ stands symbolically for the choice of even or odd $n$. The second integral is zero in terms of analytic continuation, since it is a pure power (Veltman rule).

monotonicity: For even $n$ the above integral is convergent even for $-1<\epsilon<0$ in which case $\Gamma(\epsilon)<0$ and the integral then represents the continuous version of the discret LHS for even $n$. The derivative of the integrand with respect to $n$ (once the even branch is chosen: $n\rightarrow 2n$)
$$\frac{t^\epsilon \left(1-{\rm e}^{-t}\right) {\rm e}^{\frac{t}{2n}}}{-2n^2\,\Gamma(\epsilon) \left( {\rm e}^{\frac{t}{2n}} + 1 \right)^2}$$
is manifestly positive.
A: 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
$$z_{2n+1}=\left|z_{2n}\cdot\sqrt{\frac{2n}{2n+1}}-1\right|\to\left|\frac{1}{2}-1\right|=\frac{1}{2}, n\to\infty$$
