Summation of a series-combination of a GP and another series Find the sum of the series:
$$\frac{1}{2}\biggl(\frac{1}{2}\biggl)+\frac{2}{3}\biggl({\frac{1}{2}\biggl)}^2+\frac{3}{4}\biggl({\frac{1}{2}\biggl)}^3+\cdots+
n^{th}terms$$
I calculated the $n^{th}$ term to be $\frac{n}{n+1}\frac{1}{2^n}$
$$\sum_{k=1}^n \frac{k}{k+1}\frac{1}{2^k}=\sum_{k=1}^n \frac{1}{2^k}\biggl(1-\frac{1}{k+1}\biggl)=\sum_{k=1}^n \frac{1}{2^k}-\sum_{k=1}^n \frac{1}{2^k(k+1)} $$
The first summation is a GP. But then I could not solve the second summation as it was not telescoping. Please help.

You have $\frac{1}{1-x} = \sum_{i=0}^\infty x^i$
So $-\ln(1-x) = \sum_{i=0}^\infty \frac{1}{i+1}x^{i+1}$
Then $$\frac{-\ln(1-x)}{x} = \sum_{i=0}^\infty \frac{1}{i+1}x^{i}$$
So, you have $$\frac{-\ln(\frac{1}{2})}{\frac{1}{2}} = 1 + \sum_{i=1}^\infty \frac{1}{i+1}\frac{1}{2^i}$$
