Does the series $\sum_{n\ge1}\left(-1\right)^n\frac{\cos\left(\alpha n\right)} {\sqrt n}$ converge? Does the series $\sum_{n\ge1}\left(-1\right)^n\frac{\cos\left(\alpha n\right)} {\sqrt n}$
converge ?
I tried to deal with this problem this way.
Let $S_k$ be a sequence of partial sums of the given series.
Than 
$S_{2k}=\sum_{k=2}^{2n} \frac{\cos\left(\alpha n\right)} {\sqrt n}$,
series $\sum_{k=2}^\infty \frac{\cos\left(\alpha n\right)} {\sqrt n}$ converges by Dirichlet Convergence Test, therefore $S_{2k}$ converges.
And $S_{2k+1}=-\sum_{k=1}^{2n+1} \frac{\cos\left(\alpha n\right)} {\sqrt n}$.
If $S_{2k}\to 0$ than also $S_{2k+1}\to 0$ and the given series converges,
but I do not think that $S_{2k}$ must converge to zero.
Are there another approach to this problem ?
Thanks.
 A: Answer. The series converges iff $a\ne (2k+1)\pi$. 
This can be proved using Abel's summation method,
since it is a series of the form
$$
\sum_{n=1}^\infty a_nb_n
$$
with $a_n=\frac{1}{\sqrt{n}}$ decreasing and tending to zero, and $b_n=(-1)^n\cos(an)$ which has bounded partial sums.
A: We know that $\cos(\alpha n$) oscillates between $-1$ and $1$. So, for now, we look at the following series, 
$$\sum_{n=1}^{\infty} (-1)^n \frac{a}{\sqrt{n}}$$
The series shown above converges because of the alternating series test. You must prove:
$$(1) \lim \limits_{x \to \infty} \frac{a}{\sqrt{n}}=0$$
$$(2)\frac{a}{\sqrt{n}} \text{is decreasing}$$
$$(3) \frac{a}{\sqrt{n}}>0$$
All three are obviously true in this case. Hence, the series is convergent. Notice that its conditionally convergent since,
$$\sum_{n=1}^{\infty}  \frac{a}{\sqrt{n}} \text{is divergent}$$
So, now we come back to $\cos(\alpha n)$. The function $\cos(\alpha n)$ if $\alpha=(2b+1)\pi$ for $n,b \in \mathbb{Z}$ and $n>1$ is equivalent to $(-1)^n$. Hence, if if $\alpha=(2b+1)\pi$ when $n,b \in \mathbb{Z}$ then the series is divergent:
$$\sum_{n=1}^{\infty} (-1)^n \frac{(-1)^n}{\sqrt{n}}=\sum_{n=1}^{\infty}  \frac{1}{\sqrt{n}}$$
The next question to answer is what if $\alpha \neq (2b+1)\pi$
It would most definitely converge when $\alpha=0$
This series is the real part of $\sum\limits_{n\geqslant1}n^{-a}z^n$. Dirichlet's test would be appropriate to prove that it is convergent.
