Proving the following formula of $\ln2$
$$\ln2=\frac{1}{2}\left(\frac{3}{2}\right)-\frac{1}{4}\left(\frac{3}{4}\right)+\frac{1}{6}\left(\frac{9}{8}\right)-\frac{1}{8}\left(\frac{15}{16}\right)+\frac{1}{10}\left(\frac{33}{32}\right)-\frac{1}{12}\left(\frac{63}{64}\right)\cdots$$
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 $\,\displaystyle -\log(1-x)=\sum_{n=1}^\infty \frac{x^n}n\;$)
Hint: What is the series expansion for $$ \frac{1}{1-x} $$ where does it converge? Notice that $$ \log(1-x) = - \int \frac{dx}{1-x} $$
