TITLE: Prove $\lim\limits_{n\to \infty}\frac{1}{\sqrt n}\left|\sum\limits_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}$
QUESTION [8 upvotes]: 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)

REPLY [5 votes]: 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
$$
\lim_{n\to\infty}\frac1{\sqrt{n}}\left|\,\sum_{k=1}^n(-1)^k\sqrt{k}\,\right|=\frac12\tag7
$$