Proving Absolute Convergence of $\sum_{n=1}^{\infty} (-1)^n\ln \left [ 1+\sin \left (\frac{\pi}{n\sqrt{n}} \right) \right ]$

I am trying with no success to prove the Absolute/Conditional Convergence / Divergence of the following series:

$$\sum_{n=1}^{\infty} (-1)^n\ln \left [ 1+\sin \left (\frac{\pi}{n\sqrt{n}} \right) \right ]$$

And if it is conditionally divergent, how can I prove that it is not absolutely convergent?

Thanks!

Solution

I managed to solve this with @Mr.Gandalf Sauron hint. Heres the solution:

This sereis is absolutly convergent:

$$\sum_{n=1}^{\infty} \left | (-1)^n\ln \left [ 1+\sin \left (\frac{\pi}{n\sqrt{n}} \right ) \right ]\right| = \sum_{n=1}^{\infty} \left | \ln \left [ 1+\sin \left (\frac{\pi}{n\sqrt{n}} \right ) \right ] \right |$$

I will now use the limit comparison test for positive series, with the series $$\sum_{n=1}^{\infty} \left | \sin \left ( \frac{\pi}{n\sqrt{n}} \right ) \right |$$ :

$$\lim_{n\to\infty} \left| \frac{\ln \left (1+\sin \left (\frac{\pi}{n\sqrt{n}} \right ) \right )}{\sin \left (\frac{\pi}{n\sqrt{n}} \right )} \right|$$

Now let $$t=\sin \left (\frac{\pi}{n\sqrt{n}} \right )$$, so $$n \to \infty \implies t \to 0$$

We get

$$\lim_{t \to 0} \left | \frac{\ln (1 + t)}{t} \right | = \left \{\frac{0}{0} \right \} \underset{L'Hôpital's}{=} \lim_{t \to 0} \left | \frac{\frac{1}{1+t}}{1}\right | = \lim_{t \to 0} \left | \frac{1}{1+t} \right | = \left | \frac{1}{1+0} \right | = 1$$

so by the limit comparison test both series have the same behavior.

I will now show that $$\sum_{n=1}^{\infty} \sin \left (\frac{\pi}{n\sqrt{n}} \right)$$ converges:

I will use the inequality: $$\forall x \in \mathbb{R}^{+}: |\sin(x)| \leq x$$, and the comparison test:

$$\sum_{n=1}^{\infty} \left | \sin \left (\frac{\pi}{n\sqrt{n}} \right ) \right | \leq \sum_{n=1}^{\infty} \frac{\pi}{n\sqrt{n}} = \pi \sum_{n=1}^{\infty}\frac{1}{n^{1.5}}$$

and that is convergent (Generlized harmonic series $$\sum_{n=1}^{\infty} \frac{1}{n^\alpha}$$ is convergent for $$\alpha > 1$$).

And therefore because of the comparison test, $$\sum_{n=1}^{\infty} \left | \sin \left (\frac{\pi}{n\sqrt{n}} \right ) \right |$$ is also convergent, and because of the limit comparison test we get that $$\sum_{n=1}^{\infty} \left| (-1)^n\ln \left [ 1+\sin \left(\frac{\pi}{n\sqrt{n}} \right ) \right ]\right|$$ is also convergent and therfore:

$$\sum_{n=1}^{\infty} (-1)^n\ln \left [1+\sin \left (\frac{\pi}{n\sqrt{n}} \right) \right ]$$

Is absolutely convergent

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Commented Mar 15, 2023 at 9:09

Hint:-

Do you know that $$\lim_{h\to 0^{+}}|\frac{\ln(1+h)}{h}|\to 1$$ . Then what can you say about $$|\frac{\ln(1+\sin(\frac{\pi}{n\sqrt{n}})}{\sin(\frac{\pi}{n\sqrt{n}})}|$$ by Limit comparison test?

Is it equivalent to the series $$\sum_{n}|\sin(\frac{\pi}{n\sqrt{n}})|$$ converging or diverging?. If yes then what can you say by the fact that $$|\sin(x)|\leq x\,,x\geq 0$$ ?

• Thank you very much!! I managed to solve it now Commented Mar 15, 2023 at 9:36
• @YahavB Besides accepting the answer, you can also upvote it :) Commented Mar 15, 2023 at 17:14
• @SineoftheTime One needs atleast 20 reputation, I believe , to upvote. Commented Mar 15, 2023 at 18:03
• @Mr.GandalfSauron that makes sense, my bad Commented Mar 15, 2023 at 18:23

$$\sum_{n=1}^{+\infty} (-1)^n\log \left ( 1+\sin \left (\frac{\pi}{n\sqrt{n}} \right) \right )$$ Recalling that $$\log(1+t) \sim t$$ as $$t\to 0$$ and $$\sin t \sim t$$ as $$t \to 0$$, you have: $$\sum_{n=1}^{+\infty}\left| (-1)^n\log \left ( 1+\sin \left (\frac{\pi}{n\sqrt{n}} \right) \right )\right|\sim\sum_{n=1}^{+\infty}\frac{1}{n\sqrt n}=\sum_{n=1}^{+\infty}\frac{1}{n^{3/2}}$$ which converges since $$3/2>1$$.

• Thank you!, When I have enough reputation I’ll upvote you :) Commented Mar 16, 2023 at 9:21
• @YahavB upvote it only if it's useful :) Commented Mar 16, 2023 at 13:53