Analyze if the series is convergent: $\sum_{n=0}^{\infty}(-1)^{n}\frac{\sqrt{n}}{n+1}$

For this series, I have used alternating series test because the series is alternating :P

  1. Show that $a_{n}=\frac{\sqrt{n}}{n+1}$ converges to zero for $n\rightarrow\infty$
  2. Show that $a_{n}$ is monotonic decreasing.

1) $\lim_{n\rightarrow \infty}\frac{\sqrt{n}}{n+1}$

Using L'Hôpital's rule, we get: $\lim_{n\rightarrow \infty}\frac{1}{2\sqrt{n}}=0$

2) $a_{n}\geq a_{n+1}$

$\Leftrightarrow \frac{\sqrt{n}}{n+1} \geq \frac{\sqrt{n+1}}{n+2}$ |$\cdot(n+2)$

$\Leftrightarrow \frac{\sqrt{n}(n+2)}{n+1}\geq \sqrt{n+1}$ |$\cdot(n+1)$

$\Leftrightarrow \sqrt{n}(n+2) \geq \sqrt{n+1}\cdot (n+1)$ |$()^{2}$

$\Leftrightarrow n(n+2)^{2} \geq (n+1)(n+1)^{2}$

$\Leftrightarrow n(n^{2}+4n+4) \geq (n+1)(n^{2}+2n+1)$

$\Leftrightarrow n^{3}+4n^{2}+4n \geq n^{3}+2n^{2}+n+n^{2}+2n+1$

$\Leftrightarrow n^{3}+4n^{2}+4n \geq n^{3}+3n^{2}+3n+1$

$\Leftrightarrow n^{2}+n \geq 1$

Which is true. Thus the series converges.

Is everything correct? Could it be shown easier/faster than this way?

Edit: Corrected mistake.

  • $\begingroup$ I think I found mistake: Did wrong at $()^{2}$...? $\endgroup$ – cnmesr Aug 4 '16 at 10:22
  • $\begingroup$ Corrected my mistake. Is it right now? $\endgroup$ – cnmesr Aug 4 '16 at 10:31
  • $\begingroup$ Just a note: The series is Conditionally convergent, not Absolutely convergent. Do you recognize why? $\endgroup$ – imranfat Nov 14 '16 at 4:13

You made a mistake in the step where you squared both sides of the inequality. You squared only the square roots, while you should also have squared the other factors.

Note that you actually "proved" that $a_n < a_{n+1}$ for all $n$, which would make the series increasing; that's obviously not true, as you already saw that it converges to 0.

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  • $\begingroup$ Thank you I have just realized that mistake too! :) Let's say it would end up like that (as the mistake). Could I conclude that the series diverges then? $\endgroup$ – cnmesr Aug 4 '16 at 10:23
  • $\begingroup$ Corrected. Is it correct now? ^^ $\endgroup$ – cnmesr Aug 4 '16 at 10:30

Your approach looks fine. Maybe is easier this way:

Write $a_n=f(n)$, where $f(x)=\frac{\sqrt{x}}{x+1}$ for $x\geq 1$.

Since $$f'(x)=\frac{1-x}{2\sqrt{x}(1+x)^2}\leq 0$$ for $x\geq 1$,

function $f$ is decreasing.

Therefore, $$a_n=f(n)\geq f(n+1)=a_{n+1}.$$

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  • $\begingroup$ But this is no function, it's series and sequence. Is it really allowed to do that? $\endgroup$ – cnmesr Aug 4 '16 at 11:15
  • $\begingroup$ I don't quite understand your question? f is a function and it is decreasing on $[1,\infty)$, therefore $f(n)\geq f(n+1)$ and $a_n\geq a_{n+1}$ holds for every n. This approach is maybe new to you, but everything is fine with it, why shouldn't it be allowed? :) $\endgroup$ – alans Aug 4 '16 at 11:21

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