<s>[INST] Does the series $\sum\frac{(-1)^n cos(3n)}{n^2+n}$ converge absolutely? $\sum|\frac{(-1)^n \cos(3n)}{n^2+n}| \le \sum\frac{1}{n^2+n}$
since $-1 \le cos(3n) \le 1$
$\sum\frac{1}{n^2+n} = \sum\frac{1}{n(n+1)} = \sum(\frac{1}{n} - \frac{1}{n+1})$
$\int(\frac{1}{x} - \frac{1}{x+1}) dx = \log|x| + \log|x+1| + C$
which diverges.
Therefore by the integral test, the series diverges.
However, the answer is that the series converges absolutely. Where have I gone wrong?
 [/INST] In fact, it can be simple. If $n \geq 1$ then
$$
\bigg| \frac{(-1)^{n}\cos 3n}{n^{2}+n} \bigg| \leq \frac{1}{n^{2}+n} < \frac{1}{n^{2}}.
$$
But $\sum_{n} \frac{1}{n^{2}}$ converges.
</s>