Check for convergence $$\sum_{n = 2}^\infty (-1)^n \sin\left( \frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n  (\ln n)^2}\right)$$
I tried to use Maclaurin series, but failed to evaluate little-o.
 A: Write $\sin x=x-x^3/6+x^3\varepsilon(x)$, where $\varepsilon$ is such that $\lim_{x\to 0}\varepsilon(x)=0$. Consequently, 
$$ (-1)^n \sin\left( \frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n  (\ln n)^2}\right)=(-1)^n\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n  (\ln n)^2}\right)-\frac{(-1)^n}6\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n  (\ln n)^2}\right)^3\\
+\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n  (\ln n)^2}\right)^3\delta_n,$$
where the sequence $\left(\delta_n\right)_{n\geqslant 1}$ converges to $0$ as $n$ goes to infinity. Since 
$$\left|\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n  (\ln n)^2}\right)^3\right|\leqslant cn^{-3/2}$$
for some $c$ independent of $n$ and the series $\sum_{n\geqslant 2}1/(n(\ln n)^2)$ is convergent, the problem reduces to the simpler one: determine the convergence of the series $\sum_{n\geqslant 2}(-1)^n\sin(3n)/\sqrt n$.
A: First, note that since $\sin x - x \le Cx^3$ for $|x|\le1$ and some constant $C$, we have
\begin{multline*}
\bigg| \sum_{n=2}^\infty \bigg\{ (-1)^n \sin \bigg( \frac{\sin3n}{\sqrt n} + \frac1{n\log^2n} \bigg) - (-1)^n \bigg( \frac{\sin3n}{\sqrt n} + \frac1{n\log^2n} \bigg) \bigg\} \bigg| \\ \le C \sum_{n=2}^\infty \bigg| \frac{\sin3n}{\sqrt n} + \frac1{n\log^2n} \bigg|^3,
\end{multline*}
and the latter series converges by comparison with $\sum n^{-3/2}$. So it suffices to investigate the series
$$
\sum_{n=2}^\infty (-1)^n \bigg( \frac{\sin3n}{\sqrt n} + \frac1{n\log^2n} \bigg) = \sum_{n=2}^\infty (-1)^n \frac{\sin3n}{\sqrt n} + \sum_{n=2}^\infty \frac{(-1)^n}{n\log^2n}.
$$
But this second series converges absolutely (or by the alternating series test), so it boils down to the series $\sum_{n=2}^\infty (-1)^n \frac{\sin3n}{\sqrt n}$. This series doesn't converge absolutely, nor does the alternating series test apply because the signs aren't truly alternating (and the unsigned summand isn't eventually decreasing). However, the oscillations in $\sin 3n$ should be enough to make the series converge.
There are multiple ways to make manifest this philosophy, but my go-to method is summation by parts:
$$
\sum_{n=2}^N a_nb_n = b_{N+1}A_N + \sum_{n=2}^N A(n)(b_n - b_{n+1}),
$$
where $A_n = \sum_{k=2}^n a_k$. Here, take $a_n = (-1)^n\sin 3n = -\sin((3+\pi)n)$ and $b_n = \frac1{\sqrt n}$. It is known (and can be deduced from geometric series and Euler's formula for $e^{ix}$, for example) that
$$
\sum_{k=2}^n \sin \alpha n = \frac{\cos \left(\frac{3 \alpha
   }{2}\right)-\cos \left(\frac{(2 n+1)\alpha}{2}\right)}{2\sin\frac\alpha2}
$$
for any constant $\alpha$ that's not a multiple of $2\pi$; in particular, $A(n)$ is bounded, and
$$
\sum_{n=2}^N \frac{\sin(3+\pi)n}{\sqrt n} = \frac{A(N)}{\sqrt{N+1}} + \sum_{n=2}^N A(n)\bigg( \frac1{\sqrt n}-\frac1{\sqrt{n+1}} \bigg).
$$
The first term vanishes as $N\to\infty$; in the second term, the difference $\frac1{\sqrt n}-\frac1{\sqrt{n+1}}$ is bounded above by a constant times $n^{-3/2}$ by the mean value theorem, and so the resulting series converges absolutely. This shows that the original series converges.
(Note that the second half of this proof is working through, in detail, the proof of Dirichlet's test.)
