# Proving convergence or divergence using the comparison test.

The question is to determine if the series

$$\sum_{n=1}^\infty \frac{1+\cos(nx)}{n^{4}}$$

converges or diverges using the comparison test. I established that for $$x=\frac{\pi}{2}, \frac{3\pi}{2},$$ etc (first time using MSE so I don’t know the code properly sorry). Then I tried to split the series to

$$\sum_{n=1}^\infty \frac{1}{n^4} + \frac{cos(nx)}{n^4}$$ and tried to compare this to

$$\sum_{n=1}^\infty \frac{1}{n^4}$$ But here my mind has gone blank and I’m not too sure how to show if it diverges or converges as I’m guessing its depends for different values of x. Thank you for any help.

It suffices to observe that

$$0\le \frac{1+\cos(nx)}{n^{4}}\le \frac{2}{n^{4}}$$

to conclude by direct comparison test that the given series converges.

As an alternative by your first idea we have that

$$\sum_{n=1}^\infty \frac{1+\cos(nx)}{n^{4}}=\sum_{n=1}^\infty \frac{1}{n^{4}}+\sum_{n=1}^\infty \frac{\cos(nx)}{n^{4}}$$

and by absolute convergence the second series converges, indeed

$$\sum_{n=1}^\infty \left|\frac{\cos(nx)}{n^{4}}\right|\le \sum_{n=1}^\infty \frac{1}{n^{4}}$$

Note that $$cos(nx) \le 1$$ for every $$n\in \mathbb{N}$$ and $$x$$. Thus: $$\frac{1+\cos(nx)}{n^{4}}\le \frac{1}{n^{4}}+\frac{1}{n^{4}}= \frac{2}{n^{4}}$$