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?

Of course, the absolute value of $(-1)^n\cos(3n)$ is $\leq 1$ and the series $\sum_{n\geq 1}\frac{1}{n(n+1)}$ is absolutely convergent (to $1$). Moreover, since:
$$ \sum_{n\geq 1}\frac{x^n}{n(n+1)}=\sum_{n\geq 1}\frac{x^n}{n}-\sum_{n\geq 1}\frac{x^n}{n+1}=1-\left(1-\frac{1}{x}\right)\log(1-x)\tag{1}$$
by taking $x=-e^{3i}$, then the real part, we have:
$$ \sum_{n\geq 1}\frac{(-1)^n \cos(3n)}{n(n+1)} = 1-\frac{3}{2}\sin(3)-2\cos^2\left(\frac{3}{2}\right)\log\left(2\cos\left(\frac{3}{2}\right)\right).\tag{2}$$
