Take the series
$$\sum_{n=1}^\infty \frac{cos(n\pi)}{(n+1)\ln(n+1)}$$
I want to show if the series converges absolutely, conditionally, or diverges.
$i)$ Let $a_n=\frac{cos(n\pi)}{(n+1)\ln(n+1)}$. See that $cos(n\pi)=1$ for $n=2k, k\in\mathbb{Z}$ and $cos(n\pi)=-1$ for $n=2k-1, k\in\mathbb{Z}$.
From this follows that $a_n=\frac{cos(n\pi)}{(n+1)\ln(n+1)}=\frac{(-1)^n}{(n+1)\ln(n+1)}$, for $cos(n\pi)=(-1)^n $ $ $ $\forall n \in \mathbb{N}$.
$ii$) Let's take
$$\sum_{n=1}^\infty |a_n|=\sum_{n=1}^\infty \frac{1}{(n+1)\ln(n+1)}$$
We can show this series diverges taking
$$\int_1 ^\infty \frac{1}{(x+1)\ln(x+1)}dx=\int_2 ^\infty \frac{1}{u\ln(u)}du=\int_{\ln(2)} ^\infty \frac{1}{v}dv$$
with $u=x+1$ and $v=ln(u)$.
$$\int_{\ln(2)} ^\infty \frac{1}{v}dv=\lim_{t\to\infty} \int_{\ln(2)}^t \frac{1}{v}dv=\lim_{t\to\infty}(\ln(t)-\ln(\ln(2))=\infty$$
and the integral diverges. Because the integral diverges, according to the integral test, we can conclude that the series $\sum_{n=1}^\infty |a_n|$ diverges.
$iii)$ Let's take
$$\sum_{n=1}^\infty \frac{cos(n\pi)}{(n+1)\ln(n+1)}=\sum_{n=1}^\infty \frac{(-1)^n}{(n+1)\ln(n+1)}=\sum_{n=1}^\infty (-1)^n*\frac{1}{(n+1)\ln(n+1)}$$
Let $b_n=\frac{1}{(n+1)\ln(n+1)}$. It is clear that $b_n>b_{n+1}$ because
$$(n+1)\ln(n+1)<(n+2)\ln(n+2) \implies \frac{1}{(n+1)\ln(n+1)}>\frac{1}{(n+2)\ln(n+2)}$$
It is also clear that $\lim_{n\to\infty} b_n=0$.
Therefore, because of the alternating series test, $\sum_{n=1}^\infty \frac{cos(n\pi)}{(n+1)\ln(n+1)}$ converges, and because of $ii)$ it converges conditionally.