<s>[INST] showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $ How to show that the following infinite series
$$
1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0?
$$
The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$, where $f$ is a periodic arithmetical function of period $4$, with the values $f(1)=f(3)=f(4)=1$ and $f(2)=-3$. Since $\sum_{1 \le i \le 4} f(i)=0$, it is assured that this series is convergent.
 [/INST] I'm a little wary of doing this, since it's not absolutely convergent, but can you rewrite each period as
$$
\left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) -
\left(\frac{2}{4k-2}-\frac{2}{4k}\right)
$$
which can in turn be rewritten as
$$
\left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) -
\left(\frac{1}{2k-1}-\frac{1}{2k}\right)
$$
and then you have two parallel alternating harmonic series, both of which sum to $\ln 2$.  Subtract one from the other and you get $0$.
But, as I say, I'm not sure that's kosher.
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