<s>[INST] Sum of a rearranged alternating harmonic series, with three positive terms followed by one negative term The series is this:
$$ 1 + 1/3 + 1/5 - 1/2 + 1/7 + 1/9 + 1/11 - 1/4 + 1/13 + 1/15 + 1/17 - 1/6 ...$$
The hint is to consider partial sums to $4n$ terms. I did that, and got the summation
$$ \sum 1/(6x-5) + 1/(6x-3) + 1/(6x-1) - 1/(2x) $$
But then I got stuck. The terms don't seem to simplify to anything... Wolfram alpha gives the result of this summation as $\log(12)/2$, but I have no idea how to get there.
Any help greatly appreciated.
 [/INST] Hint:
Try writing the sum as
$$
\sum_{k=0}^\infty \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{3}{6k+6}\tag{1}
$$
Then use the fact that
$$
\begin{align}
\sum_{k=0}^n\frac{1}{2k+1}
&=\log(2n+1)+\gamma-\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\log(2)+\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{2}
\end{align}
$$
and that
$$
\begin{align}
\sum_{k=0}^n\frac{1}{2k+2}
&=\tfrac{1}{2}(\log(n+1)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{3}
\end{align}
$$
Addendum:
Both formulas $(2)$ and $(3)$ follow from the asymptotic expansion for the partial harmonic series:
$$
\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+O\left(\tfrac{1}{n}\right)
$$
where $\gamma$ is the Euler-Mascheroni Constant. Thus,
$$
\begin{align}
\sum_{k=0}^n\frac{1}{2k+1}
&=\sum_{k=1}^{2n+1}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{2k}\\
&=\log(2n+1)+\gamma-\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\log(2)+\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)
\end{align}
$$
and
$$
\begin{align}
\sum_{k=0}^n\frac{1}{2k+2}
&=\sum_{k=1}^{n+1}\frac{1}{2k}\\
&=\tfrac{1}{2}(\log(n+1)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)
\end{align}
$$
Using $(2)$, we get
$$
\begin{align}
\sum_{k=0}^n \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}
&=\log(2)+\tfrac{1}{2}(\log(3n+2)+\gamma)+O\left(\tfrac{1}{n}\right)\\
&=\tfrac{1}{2}(\log(n)+\log(12)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{4}
\end{align}
$$
and using $(3)$, we get
$$
\begin{align}
\sum_{k=0}^n \frac{3}{6k+6}
&=\tfrac{1}{2}(\log(n)+\gamma)+O\left(\tfrac{1}{n}\right)\tag{5}
\end{align}
$$
Combining $(4)$ and $(5)$, we get
$$
\sum_{k=0}^n \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{3}{6k+6}=\tfrac{1}{2}\log(12)+O\left(\tfrac{1}{n}\right)
$$
</s>