Dealing with $\sin$ and $\cos$ in a series

Question

Decide if the series converge or diverge: $\sum_{n=1}^{\infty} \frac {\ln(n)+\cos(n)}{n^2}$

I don't know how to approach this.. or well series which involves more functions in general with sum between them such $\ln(n) + \cos(n)$ or $\sin(n) + \cos(n)$ since $\sin$ and $\cos$ have a weird behaviour when plugged in integres, I tried ratio test, thought about some limit comparison with $\frac 1n$ or $\frac 1{n^2}$ but doesn't seem to get anywhere, what test should I use?

Answer 1

In cases such as this, you can usually bound the $\cos$ part of the series, since $$-1<\cos(n)<1$$

In particular, you have $$\frac{|\ln n + \cos n|}{n^2}\leq \frac{\ln n + 1}{n^2}$$

and you can now use any number of methods to show that this series converges.

One would be to simply further rewrite it as

$$\frac{\ln n + 1}{n^2} = \frac{\ln n + 1}{\sqrt{n}}\cdot \frac{1}{n^\frac32}$$ and note that the first term in the expression is bounded (because its limit is $0$).

Answer 2

HINT: note that $\frac{\ln n+\cos n}{n^2}\le\frac{\ln n}{n^2}+\frac1{n^2}$.

Answer 3

Since the series $\displaystyle\sum_{n=1}^\infty\frac{\cos n}{n^2}$ converges (since $(\forall n\in\mathbb{N}):-1\leqslant\cos n\leqslant 1$), what remains to be determined is whether the series $\displaystyle\sum_{n=1}^\infty\frac{\ln n}{n^2}$ converges or not. For that use the fact that $\displaystyle\lim_{n\to\infty}\frac{\ln n}{\sqrt n}=0$ and that $\displaystyle\sum_{n=1}^\infty\frac1{n^{\frac32}}$ converges.