<s>[INST] Stupid question about $1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$ I have
$$1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$$
Partial sum $S_{3n}$ of the above is:
$$(1 - \frac{1}{2}-\frac{1}{4})+(\frac{1}{3}- \frac{1}{6}-\frac{1}{8})+(\frac{1}{5}-\dots$$
But what is $S_{3n-1}$ and $S_{3n-2}$ ?
 [/INST] By guessing the pattern hidden in the ellipsis, it seems that you consider the series
$\sum_{k=1}^\infty a_k  $
where $a_{3n-2}=\frac1{2n-1}$, $a_{3n-1}=-\frac1{4n-2}$, $a_{3n}=-\frac1{4n}$ for $n\ge 1$.
Thus $a_{3n-2}+a_{3n-1}+a_{3n}=\frac1{2n-1}-\frac1{4n-2}-\frac1{4n}=\frac1{4n(2n-1)}$ and we have the partial sums
$$S_{3n}=\sum_{k=1}^{3n} a_k  =\sum_{k=1}^n\frac1{4n(2n-1)}$$
$$S_{3n-1}=\sum_{k=1}^{3n-1} a_k  =\sum_{k=1}^n\frac1{4n(2n-1)}+\frac1{4n}$$
$$S_{3n-2}=\sum_{k=1}^{3n-2} a_k  =\sum_{k=1}^n\frac1{4n(2n-1)}+\frac1{4n}+\frac1{4n-2}$$
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