Series convergence or divergence $\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}$ Can you give me some hint for the series
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}
\end{align*} I have tried Leibniz test. It didn't help me.

Hint: use Dirichlet's test. See more at Wikipedia
It's clear that $\left\{\dfrac 1 {\sqrt{n^2 + 2}}\right\}$ is monotonic and converges to 0
Now the task is to prove $-\cos 3 + \cos 6 - \cos 9 + \cos 12 + ...$ is bounded
Recognize that $\cos 6 - \cos 3 = -2\sin \frac 9 2 \sin \frac 3 2$
$\cos 12 - \cos 9 = -2\sin \frac {21} 2 \sin \frac 3 2$
$\cos 18 - \cos 15 = -2\sin \frac {33} 2 \sin \frac 3 2$
and $-2 \sin 3\left(\sin \frac 9 2 + \sin \frac {21} 2 + \sin \frac {33} 2 ... \right)= \cos \frac {15} 2 - \cos \frac 3 2 + \cos \frac {27} 2 - \cos \frac {15} 2 +...$
You can see that $\sum\limits_{i=1}^{2n} (-1)^k \cos {3k} = \dfrac{\sin \frac 3 2}{\sin 3} \left|\cos(2n + 1.5) - \cos 1.5 \right|$, which is always bounded
Therefore the series converges
