<s>[INST] Proving the following formula of $\ln(2)$ Proving the following formula of  $\ln2$
$$\ln2=\frac{1}{2}\left(\frac{3}{2}\right)-\frac{1}{4}\left(\frac{3}{4}\right)+\frac{1}{6}\left(\frac{9}{8}\right)-\frac{1}{8}\left(\frac{15}{16}\right)+\frac{1}{10}\left(\frac{33}{32}\right)-\frac{1}{12}\left(\frac{63}{64}\right)\cdots$$
 [/INST] Your formula is :
\begin{align}
S&:=-\sum_{n=1}^\infty \frac{(-1)^n}{2\,n}\frac{2^n-(-1)^n}{2^n}\\
&=-\sum_{n=1}^\infty \frac{(-1)^n}{2\,n}+\sum_{n=1}^\infty \frac{(-1)^{2n}}{2\,n}\frac 1{2^n}\\
&=-\frac 12\sum_{n=1}^\infty \frac{(-1)^n}{n}+\frac 12\sum_{n=1}^\infty \frac 1{n} \left(\frac 12\right)^n\\
&=\frac 12\log(1-(-1))-\frac 12\log(1-1/2)\\
&=\log 2\\
\end{align}
( using only $\,\displaystyle -\log(1-x)=\sum_{n=1}^\infty \frac{x^n}n\;$)
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