Does the sequence converge, alternating series fails? Does $\sum_{n=1}^{\infty} (-1)^n \frac{2}{n} \cos^2(\frac{\pi n}{2})$ converge?


The alternating series test fails, because I am unable to prove that $f'(n) &lt; 0$, I can get to $f'(n) = 0$, but that fails the test.
notice that $\cos^2(\frac{\pi n}{2})$ is $1$ if $n$ is odd and $0$ otherwise, so letting $n=2m$ and adding only over even terms our sum is equal to:

$\sum\limits_{m=1}^{\infty}(-1)^{2m} \frac{2}{2m}=\sum\limits_{m=1}^\infty \frac{1}{m}$
No, it doesn't... when $n$ is odd. $\cos^2(\frac {n\pi}{2}) = 0$

and when n is even $\cos^2(\frac {n\pi}{2}) = 1$

So we can re-write this as $\sum \frac 1{n}$ which does not converge.