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.
 A: $\sum_\limits{n=1}^\infty \frac {1}{\sqrt{n^2 + 2}}$ can be compared to the harmonic series.  The harmonic series diverges... slowly.
The series is not absolutely convergent.  But, it still could be conditionally covergent.
The alternating series test doesn't help as $(-1)^n\cos 3n$ is not necessarily alternating.
That leaves Dirichlet's test.
If $\{a_n\}$ is a sequence of real numbers and $\{b_n\}$ is a sequence of complex numbers, such that:
$\{a_n\}$ is monotonic and $\lim_\limits{n\to\infty} a_n = 0$
And
$\forall N\in \mathbb N : \left|\sum_\limits{n=1}^N b_n \right|< M$ for some constant $M.$
$\sum_\limits{n=1}^\infty a_nb_n$ converges.
The first parts of the test, are easy enough and I will leave to you.
$(-1)^n\cos 3n = Re(e^{(3+\pi)in})$
And
$\left|\sum_\limits{n=1}^N Re(e^{(3+\pi)ni})\right| < \left|\sum_\limits{n=1}^N e^{(3+\pi)ni}\right|$
We still need to show that $\left|\sum_\limits{n=1}^N e^{(3+\pi)ni}\right|$ is bounded for all $N.$
$\sum_\limits{n=1}^N e^{(3+\pi)ni}$ is a geometric series.
$\sum_\limits{n=1}^N e^{(3+\pi)ni} =  e^{(3+\pi)i}\frac {1-e^{(3+\pi)Ni}}{1-e^{(3+\pi)i}}\\
\left| e^{(3+\pi)i}\frac {1-e^{(3+\pi)Ni}}{1-e^{(3+\pi)i}}\right| < 
\frac {2}{1+\cos 3}$
A: 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
