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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!

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

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*}

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