<s>[INST] Find sum of the series $ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$ I have to find the sum of the given series $$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$$
My attempt
$$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$$
or,
$$ S=1 + \left( \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \right)+\left(\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}\right)+\left(\frac{1}{11}-\frac{1}{6}+\frac{1}{13}\right)+\left( \frac{1}{15}-\frac{1}{8}+\frac{1}{17}\right)+ \cdots$$
or,
$$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{4n-1}-\frac{1}{2n}+\frac{1}{4n+1}\right)$$
or,
$$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{4n-1}-\frac{1}{4n}-\frac{1}{4n}+\frac{1}{4n+1}\right)$$
or,
$$S= 1+  \sum_{n=1}^\infty \left( \frac{1}{(4n-1)(4n)}-\frac{1}{(4n)(4n+1)} \right)$$
Since rearrrangement is allowed now, we have
$$S= 1+\sum_{n=1}^\infty  \frac{1}{(4n-1)(4n)}-\sum_{n=1}^\infty\frac{1}{(4n)(4n+1)}$$
or,
$$ S = 1+ I_1 - I_2$$
But even after many trials, I was unable to find the value of either $I_1$ or $I_2$, any help regarding this will be much appreciated.
 [/INST] Let $s_n$ be the sum of the first $n$ terms of the series.  It is clear that if $s_{3n}$ converges, then the series converges to the same sum.  $$\begin{align}
s_{3n}&=\sum_{k=1}^{2n}\frac1{2k-1}-\sum_{k=1}^n\frac1{2k}\\
&=\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1{2k}-\sum_{k=1}^{n}\frac1{2k}\\
&=\sum_{k=1}^{4n}\frac1k-\frac12\sum_{k=1}^{2n}\frac1{k}-\frac12\sum_{k=1}^{n}\frac1{k}
\end{align}$$ 
Can you continue from here?  Use the standard approximation for $\sum\frac1k$.
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