# Decide if the next infinite series are convergent or divergent.

I need help with this problem:

Decide if the next infinite series are convergent or divergent. The tools that you need to use are the Leibniz Theorem, the Comparison Test, Quotient Test and the Integral Test.

1. $$\sum_{n=1}^\infty {{\sin\ (n\theta)}\over{n^2}}$$
2. $$\sum_{n=1}^\infty (-1)^n {{\log\ n}\over{n}}$$

I don't know how to start. Can you please help me and explain me how to decide when to use one of the tests? Thanks.

$$1$$. Note that, Every absolutely convergent series is convergent.
Now, $$\displaystyle\sum_{n=1}^\infty |\frac{\sin(n\theta)}{n^2}|\le\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$. So, the series $$\sum_{n=1}^\infty\frac{\sin (n\theta)}{n}$$ is absolutely convergent and hence convergent.
$$2$$.For the second one we use alternating series test. Set $$a_n=(-1)^n\frac{\log n}{n}$$ then, $$|a_n|=\frac{\log n}{n}$$ decreases monotonically as $$n>\log n$$ for all $$n>1$$ with $$\displaystyle\lim_{n\to \infty}a_n=0$$. Hence the series $$\sum_{n=1}^\infty (-1)^n\frac{\log n}{n}$$ converges.
• How did you found that $\displaystyle\sum_{n=1}^\infty |\frac{\sin(n\theta)}{n^2}|\le\sum_{n=1}^\infty \frac{1}{n^2}$? Why did you choose $\sum_{n=1}^\infty \frac{1}{n^2}$? Sorry I find this topic pretty difficult. – davidllerenav Jan 28 at 4:48
• Oh, I see. Recall that $|\sin x|\le 1$ for all $x\in\mathbb{R}$ so each term of the $n$-th partial sum of the above series i.e. $\displaystyle\sum_{n=1}^N |\frac{\sin(n\theta)}{n^2}|$ is less than $\sum_{n=1}^N \frac{1}{n^2}$ for all $N=1,2,...$ – Sujit Bhattacharyya Jan 28 at 4:51
• Or, simply $|\frac{\sin(n\theta)}{n^2}|\le\frac{1}{n^2}$ and take finite sum both sides then take the sum to infinite sum. – Sujit Bhattacharyya Jan 28 at 4:53
• So $|sin(n\theta)|$ is always $\leq 1$? And then the inequality will remain when we divide by $n^2$, right? – davidllerenav Jan 28 at 5:00