Determining convergence of a series $\sum_n (-1)^n \sin a_n $ I need to determine if the following series is convergent:
$$\sum_{n=2}^\infty (-1)^n\sin\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n\ln^2(n)}\right).$$
I've tried to use alternating series test but got stuck.
 A: The series converges:
Using $\sin x = x - \frac{x^3}{6} + o(x^4)$ around $0$, we get that when $n\to \infty$
$$
\sin\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n\ln^2(n)}\right)
 = 
\underbrace{\frac{\sin(3n)}{\sqrt{n}}}_{a_n} + \underbrace{\frac{1}{n\ln^2(n)} + O\left(\frac{1}{n^{3/2}}\right)}_{b_n}.
$$
The series $\sum_{n=2}^\infty (-1)^n b_n$ converges absolutely by theorems of comparison (as the sum of a Bertrand series and something negligible in front of it), so our original series will be convergent if, and only if, the series $\sum_{n=2}^\infty  (-1)^n a_n$ converges.
We will show it is indeed convergent (conditionally). We cannot apply the alternating series test, since it does not look like the assumptions (monotonicity and positivity of $a_n$) hold at all. But we can apply Dirichlet's test:


*

*The sequence $(\alpha_n)_{n\geq 1} = (\frac{1}{\sqrt{n}})_{n\geq 1}$ is positive and decreasing to $0$;

*The partial sums of the sequence $(\beta)_{n\geq 1} = ((-1)^n \sin 3n)_{n\geq 1}$ are bounded: for all $N\geq 2$,
$$
\left\lvert \sum_{n=2}^N \beta_n\right\rvert
=
\left\lvert \operatorname{Im} \sum_{n=2}^N (-1)^ne^{i3n}\right\rvert
=
\left\lvert \operatorname{Im} \sum_{n=2}^N (e^{(3+\pi)i})^n\right\rvert
=
\left\lvert \operatorname{Im}\left( e^{2(3+\pi)i}\frac{1-e^{(3+\pi)i(N-1)}}{1-e^{(3+\pi)i}}\right)\right\rvert \leq M
$$
for some absolute constant $M>0$ ($\dagger$). By Dirichlet's test, we therefore have (conditional) convergence of the series
$$
\sum_{n=2}^\infty \alpha_n\beta_n = \sum_{n=2}^\infty (-1)^n \frac{\sin 3n}{\sqrt{n}}
$$
which by the above implies (conditional) convergence of the original series
$$
\sum_{n=2}^\infty  (-1)^n \sin\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n\ln^2(n)}\right).
$$
($\dagger$) This step would require more detail, in a full proof.
