Let a series be given as $$\sum_{m=2}^{\infty}\frac{\cos(\pi m)}{\ln(\ln(m))}$$
Is it converges conditionally, converges absolutely or diverges.
Attempt
From computation I see that it does indeed converge (and I can show that it converges using alternating series test / Leibnitz). The thing that is left is to show that the $$\sum_{m=2}^{\infty}\left | \frac{\cos(\pi k)}{\log(\log(k))} \right |$$ diverges and thus that it does not converge absolutely but only conditionally. But trying to use that if $a_n$ does not converge to $0$ then the series diverges I get stock as I see that the sequence $a_n$ does converge to $0$.
For $m\in\mathbb{Z}$ we have $\cos(\pi m)=(-1)^m$, so we may write $$\sum_{m=2}^{\infty}\left | \frac{\cos(\pi m)}{\log(\log(m))} \right |=\sum_{m=2}^{\infty}\left | \frac{(-1)^m}{\log(\log(m))} \right|=\sum_{m=2}^{\infty}\frac{1}{\lvert\log(\log(m))\rvert}.$$ Also observe that $\log\log m <m$ for $m>1$. So a lower bound is the harmonic series.