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

Question

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

Answer 1

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 $|\sin(x)|\leq x\,,x\geq 0$ ?

Answer 2

$$\sum_{n=1}^{+\infty} (-1)^n\log \left ( 1+\sin \left (\frac{\pi}{n\sqrt{n}} \right) \right ) $$ 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 $3/2>1$.