I was trying to figure out how a limit was calculated and got stuck when trying to understand one of the proposed solutions: (note that this is just a small part of the solution, but the one that got me in trouble)

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

In my opinion, whether $$n$$ is odd or even has an impact on the sum. Plugging a few random $$n$$-s doesn't help to prove the validity of the formula for me.

I guess this is one of the cases when I am puzzled and can't see something obvious. If someone could clarify it for me, that would be great. Thanks!

• Prove that the effect of $n$ odd/even disappears in the limit. – user10354138 Jun 25 '19 at 17:47
• I mean that without considering the limit, the transition from one sum to the other is not correct for some finite $n$, right? – Don Draper Jun 25 '19 at 17:48
• When $n = 2m$, we get: $$\lim\limits_{m \to \infty} {\frac{1}{\sqrt{2m}} \sum \limits_{k=1}^{m}{\frac{1}{\sqrt{2k}+\sqrt{2k-1}}}}$$ And for odd $n$: $$\lim\limits_{m \to \infty} {\frac{1}{\sqrt{2m+1}} |\sum \limits_{k=1}^{m}{\frac{1}{\sqrt{2k}+\sqrt{2k-1}}-\sqrt{2m+1}|}}$$ – Don Draper Jun 25 '19 at 18:38
• Honestly, it's not obvious to me that the effect of $n$ being odd/even disappears – Don Draper Jun 25 '19 at 18:56

In order to show the validity of the claim it follows from your comment \begin{align*} \lim_{m\to \infty}&\frac{1}{\sqrt{2m}}\sum_{k=1}^m\frac{1}{\sqrt{2k}+\sqrt{2k-1}}\\ \lim_{m\to \infty}&\frac{1}{\sqrt{2m+1}}\left|\sum_{k=1}^m\frac{1}{\sqrt{2k}+\sqrt{2k-1}}-\sqrt{2m+1}\right|\\ &=\lim_{m\to \infty}\left|\frac{1}{\sqrt{2m+1}}\sum_{k=1}^m\frac{1}{\sqrt{2k}+\sqrt{2k-1}}-1\right|\\ \end{align*} that
\begin{align*} \lim_{m\to\infty}\frac{1}{\sqrt{m}}\sum\limits_{k=1}^{m} (-1)^k\sqrt{k} &=\lim_{m\to \infty}\frac{1}{\sqrt{2m}}\sum_{k=1}^m\frac{1}{\sqrt{2k}+\sqrt{2k-1}}\\ &=\lim_{m\to \infty}\frac{1}{\sqrt{2m+1}}\sum_{k=1}^m\frac{1}{\sqrt{2k}+\sqrt{2k-1}}\\ &=\frac{1}{2} \end{align*}