On the convergence of $\sum_{n=1}^{+\infty}\frac{1}{n}\,\cos\left(\frac{\pi n}{2}\right)$ Does the following series absolutely converge, conditionally converge or diverge? 
$$\sum_{n=1}^{+\infty}\frac{1}{n}\,\cos\left(\frac{\pi n}{2}\right)$$
My answer:
$$ 0<\frac{1}{\sqrt{n}}<\left|\frac{\cos(\pi n/2)}{n}\right| $$
and $\sum_{n\geq 1}\frac{1}{\sqrt{n}}$ diverges by p-series test so by comparison test, the original series must also diverge.
 A: This series is conditionally convergent. Its terms are:
$$\frac{1}{n}\cos\left(\frac{n\pi}{2}\right) =
\left\{
 \begin{array}{llll}
    \frac{1}{n} & \mbox{if } n=4k , k>0\\
   0 & \mbox{if } n=4k+1 , k\ge0\\
   -\frac{1}{n} & \mbox{if } n=4k+2 , k\ge0\\
   0& \mbox{if } n=4k+3, k\ge0\\
 \end{array}
\right. $$
By removing zero expression, and just considering $n=4k$ and $n=4k+2$, we could replace and change indices. By write some sentences for this series:
$$\sum_{n=1}^{\infty}\frac{1}{n}\cos\left(\frac{n\pi}{2}\right)=0-\frac{1}{2}+0+\frac{1}{4}+0-\frac{1}{6}+0+\frac{1}{8}+0+...=\\-\frac{1}{2}\Big(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...\Big)=-\frac{\ln{2}}{2}\approx-0.3466$$

It's conditionally convergences because 
$$\sum \left|\frac{\cos\left(\frac{n\pi}{2}\right)}{n}\right|=\frac{1}{2}\Big(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\Big)$$
doesn't converge.
References:
https://en.wikipedia.org/wiki/Alternating_series_test
A: Let $M$ any natural number, then
$$
2\sum^{M}_{n=1}\cos(n\pi/2)=-1+\cos(M\pi/2)+\sin(M\pi/2).
$$ 
Hence the partial sums of $\cos(n\pi/2)$ are bounded. Also $\frac{1}{n}$ is a null monotonic sequence. Hence from Dirichlet's test (see [1] pg. 315), the series
$$
\sum^{\infty}_{n=1}\frac{\cos(n\pi/2)}{n}
$$ 
converges to a number. Moreover the series
$$
\sum^{\infty}_{n=1}\frac{\cos(n x)}{n^a}\textrm{, }a>0
$$
converges in every interval of the form $\epsilon\leq x\leq 2\pi-\epsilon$, where $\epsilon$ any number in $(0,\pi)$ (see [1] pg.349).
References.
[1]: Konrad Knopp. ''Theory and application of infinite series''. Dover. (1990)
