The alternating harmonic series $a:=\sum_{n=1}^{\infty}{\frac{(-1)^{n-1}}n}$ converges by the Leibnitz-criteria. Now consider the following rearrangement of this series given by: \begin{align} S:=1+\frac{1}{3}-\frac{1}{2}-\frac{1}{4}+\frac{1}{5}+\frac{1}{7}-...+\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{4n-2}-\frac{1}{4n}+... \end{align} I want to see that this series converges aswell and has the same value $a$. I am stuck with this one. Does someone have a hint how to start this?
Edit: this not a duplicate to the question in the link below since I am asking on how to get the limit value of this series.
Hint: Let $S_n$ be the $n$th partial sum of the alternating harmonic series, and let $T_n$ be the $n$th partial sum of the given rearrangement. Then $S_{4n}= T_{4n}$ for all $n.$ How far from $T_{4n}$ can $T_{4n+1},T_{4n+2},T_{4n+3}$ be?