<s>[INST] Prove specially rearranged alternating harmonic series converges to $\frac 12 \ln{\frac{4p}{q}}$ By Leibnitz's test the alternating series is convergent.
$\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac 1n 
=\frac 12 \ln{\frac{4×1}{1}}$
\begin{align}
& \left(1-\frac 12- \frac 14\right)+\left(\frac 13-\frac 16-\frac 18\right)+...\\
& =\frac 12 \ln {\frac{4\cdot 1}{2}}
\end{align}
Let $\sum \sigma$ is a rearrangement of the alternating series.$\sum t$ is obtained by grouping it's terms in such  a manner that each group contains $p$ positive terms and $q$ negative terms. Prove that $\sum t=\frac 12\ln {\frac{4p}{q}}$
 [/INST] $t_{(p+q)n}$
\begin{align}
& =\left(1+\frac 13+...+\frac{1}{2p-1}-\frac 12-\frac 14-...-\frac{1}{2q}\right)+\left(\frac{1}{2p+1}+...+\frac{1}{4p-1}-\frac{1}{2q+2}-...-\frac{1}{4q}\right)+...+\left(\frac{1}{2p(n-1)+1}+...+\frac{1}{2pn-1}-\frac{1}{2q(n-1)+2}-...-\frac{1}{2qn}\right)\\
&=\left(1+\frac 13+...+\frac{1}{2pn-1}\right)-\frac 12\left(1+\frac 12+\frac 13+...+\frac{1}{qn}\right)\\ 
&=\left(1+\frac 12+\frac 13+...+\frac{1}{2pn}\right)-\left(\frac 12+\frac 14+...+\frac{1}{2pn}\right)-\frac 12(\gamma_{qn}+\ln{qn})\\
 &=(\gamma_{2pn}+\ln{2pn})-\frac 12(\gamma_{pn}+\ln{pn})-\frac 12(\gamma_{qn}+\ln{qn}) \\
& =\left(\frac 12\ln 2^2+\frac 12\ln\frac pq\right)+\left(\gamma_{2pn}-\frac 12\gamma_{pn}-\frac 12\gamma_{q n}\right)
 \end{align}
$\therefore \displaystyle \lim_{n \to \infty}t_{(p+q)n}=\frac 12\ln{\frac{4p}{q}} $
</s>