Find nature of the following series: $$\sum_{n=1}^{\infty} {\frac{\cos{\frac{πn}{2}}}{n}}$$
It should be convergent, but I don’t know how to get to this result. It’s an exercise in a calculus 1 testbook.
I’d like a full answer which explains which test or theorem we use to prove the series rigorously, not a one-sentence-answer, because I need to understand how to prove it, and if I’m missing something that I could use to prove other series.
METHOD 1: FAILED
I have tried using the absolute convergence test, since it is not a series with non negative terms, and then using the ratio test, the root test, the comparison test or using asymptotic equivalences, but I wasn’t able to get to a conclusion.
METHOD 2: DOES THIS LEAD ANYWHERE?
I also tried re-writing the nth partial sums of the series $$ \sum_{n=1}^{\infty} {\left\vert \frac{\cos{\frac{πn}{2}}}{n}\right\vert} = \sum_{n=1}^{\infty} { \frac{\left\vert\cos{\frac{πn}{2}}\right\vert}{n}} $$ and, by ignoring all the terms for which $\cos{\frac{πn}{2}} = 0$, I got:
$$\sum_{n=1}^{m} {\frac{\left\vert\cos{\frac{πn}{2}}\right\vert}{n}} = \frac{0}{1} + \frac{1}{2} + \frac{0}{3} + \frac{1}{4} + … + \frac{\left\vert\cos{\frac{πn}{2}}\right\vert}{n} = \sum_{n=1}^{\frac{m}{2}} {\frac{1}{2n}}$$
if m is even, otherwise (if m is odd) just substitute $\frac{m-1}{2}$ instead of $\frac{m}{2}$ on the upper bound of the summation.
But even this doesn’t get me anywhere, since I don’t know how to say if the sequence of the nth partial sums of the series has a supremum.
NOTE
By calculating the limits of the ratio test and the root test i got, respectively, a non existing limit and l=1, for which the root test doesn’t give useful answers.