Strange series that converges I'm trying to evaluate this series
$$S= \sum_{n=2}^{\infty} a_n \frac{\ln (n)}{n}$$
But I have some conditions on $a_n$ making the problem hard. Namely, $a_n=3$ for $n = 2 \mod 4$ and $a_n = -1$ otherwise.
Albeit 2 mod 4 =2 is just a number, a friend in the comments suggested that the 2[mod 4] takes values 2,6,10,... which of course makes sense. Meaning that $a_n$ will go like $(3 -1 -1 -1 +3 -1 -1 -1 +3 + ...)$ for $n=2,3,4 ...$.
I don't really know how to do convergence test when I have constraints on a summation constant like this. I have tried to rewrite the sum to a simpler form, tried to separate is at two sums for the two different $a_n$'s but I don't really know how to do it and there is where I'm stuck...

As I pointed out in the comments, this is Question B-4 from the 2017 William Lowell Putnam competition.  For convenience, I am providing the first of two solutions provided from the link in my comment.  I take no credit--only minor changes to phrasing were made.
The key insight is to define an auxiliary telescoping series with terms $$a_k = \frac{\log k}{k} - \frac{\log (k+1)}{k+1},$$ for which we trivially have $$\sum_{k=1}^\infty a_k = 0. \tag{1}$$  Since $a_k > 0$ for $k \ge 3$, we also see that $(1)$ is absolutely convergent.
Having constructed such a series permitting rearrangement of its terms, we next observe $$3a_{4k+2} + 2a_{4k+3} + a_{4k+4} = (a_{4k+2} + a_{4k+4}) + 2(a_{4k+2} + a_{4k+3}),$$ hence
$$\begin{align}
S &= \sum_{k=0}^\infty \left(3 \frac{\log(4k+2)}{4k+2} - \frac{\log(4k+3)}{4k+3} - \frac{\log(4k+4)}{4k+4} - \frac{\log(4k+5)}{4k+5}\right) \\
&= \sum_{k=0}^\infty (3a_{4k+2} + 2a_{4k+3} + a_{4k+4}) \\
&= \sum_{k=1}^\infty a_{2k} + \sum_{k=0}^\infty 2(a_{4k+2} + a_{4k+3}). \tag{2}
\end{align} $$
We next observe $$2(a_{4k+2} + a_{4k+3}) = \frac{\log(4k+2)}{2k+1} - \frac{\log(4k+4)}{2k+2} = a_{2k+1} + \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right)\log 2.$$  Hence
$$\sum_{k=0}^\infty 2(a_{4k+2} + a_{4k+3}) = \sum_{k=0}^\infty a_{2k+1} + \log 2 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = \sum_{k=0}^\infty a_{2k+1} + \log^2 2. \tag{3}$$
Finally, putting $(1)$, $(2)$, and $(3)$ together yields the desired result:
$$S = \log^2 2 + \sum_{k=1}^\infty a_{2k} + \sum_{k=0}^\infty a_{2k+1} = \log^2 2 + \sum_{k=1}^\infty a_k = \log^2 2.$$
