Absolute and conditional convergence of a series with $\sin$ $$\sum_{n=2}^{\infty}\frac{(-1)^n \sin^4{n}}{\sqrt{n\ln{(n)} }}$$
I can't imagine how to approach to this problem. Could you give any tips?
 A: You can't apply the alternating series test or Dirichlet test directly because $\sin^4 n / \sqrt{n \ln n}$ is not monotonically decreasing.
First, note that
$$\sin^4 n = \frac{3}{8} - \frac{\cos 2n}{2} + \frac{\cos 4n}{8}.$$
Then consider the sum of three separate series.
$$\sum_{n=2}^\infty \frac{(-1)^n\sin^4 n}{\sqrt{n \ln n}}= \frac{3}{8}\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n \ln n}} -   \frac{1}{2}\sum_{n=2}^\infty \frac{(-1)^n \cos 2n}{\sqrt{n \ln n}} + \frac{1}{8}\sum_{n=2}^\infty \frac{(-1)^n \cos 4n}{\sqrt{n \ln n}}. $$
Each of the three series on the RHS converge by the Dirichlet test and, hence, the original series converges. This follows since $1/\sqrt{n \ln n}$ converges monotonically to $0$ and we can show that the numerators have bounded partial sums.
This is obvious for the first series, since
$$\left|\sum_{n=2}^{m}(-1)^n\right| \leqslant 1$$
More work is needed for the remaining two series.
For example, $(-1)^n \cos 2n = \cos \pi n \cos 2n = \cos (\pi -2)n, $ and
$$\sum_{n=1}^m \cos nx =\frac{\sin(mx/2)\cos[(m+1)x/2]}{\sin(x/2)}.$$
The series fails to converge absolutely.  Note that 
$$\sqrt{n \ln n} \leqslant n \implies \frac{1}{\sqrt{n \ln n}} \geqslant \frac{1}{n},$$
and
$$\sum_{n=2}^\infty \frac{\sin^4 n}{\sqrt{n \ln n}} \geqslant \frac{3}{8}\sum_{n=2}^\infty \frac{1}{n} -   \frac{1}{2}\sum_{n=2}^\infty \frac{\cos 2n}{n} + \frac{1}{8}\sum_{n=2}^\infty \frac{ \cos 4n}{n}. $$
The first series on the RHS is a divergent harmonic series and the other two converge by the Dirichlet test. Hence the series on the LHS diverges and the original series converges conditionally.
