<s>[INST] Harmonic series with sign alternates every $n$ terms. Let $A(1)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$
Let $A(2)=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\dots$
Let $A(3)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}+\dots$
In general, for $n \in \mathbb{N}$,  let $A(n)$ be the same series as the above with sign alternates every $n$ terms.
I know that $A(1)=\log(2)$, and $A(2)=\frac{1}{4}(\pi+\log(4))$.
How can we evaluate $A(n)$ for $n \in \mathbb{N_{\ge3}}$?
 [/INST] $A(3) = \lim_{m\to\infty} K_m$ where
$$
K_m = \sum_{k=1}^m\frac{1}{6k-5}
+\sum_{k=1}^m\frac{1}{6k-4}
+\sum_{k=1}^m\frac{1}{6k-3}
-\sum_{k=1}^m\frac{1}{6k-2}
-\sum_{k=1}^m\frac{1}{6k-1}
-\sum_{k=1}^m\frac{1}{6k}
$$
Then use estimates like
$$
\sum_{k=1}^m\frac{1}{6k-a} =
\frac{1}{6}\left(\log m-\psi\left(1-\frac{a}{6}\right)\right)+O(1/m)
$$
with the Gauss digamma theorem for $\psi$ of rational number.  Result:
$$
A(3) = \frac{\log 2}{3}+\frac{2\pi}{3\sqrt{3}} .
$$
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