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.
 A: 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 $$\frac{-\ln(\frac{1}{2})}{\frac{1}{2}} = 1 + \sum_{i=1}^\infty \frac{1}{i+1}\frac{1}{2^i}$$
A: If we consider the more general case of $$S_n(x)=\sum_{k=1}^n \frac{k}{k+1}{x^k}=\sum_{k=1}^n x^k\biggl(1-\frac{1}{k+1}\biggl)=\sum_{k=1}^n x^k-\sum_{k=1}^n \frac{x^k}{(k+1)}$$ the problem looks quite simple when the summation is done up to $\infty$.
For finite $n$, at least to me, the problem seems to be difficult since we have $$\sum_{k=1}^n x^k=\frac{x \left(1-x^n\right)}{1-x}$$
$$\sum_{k=1}^n \frac{x^k}{(k+1)}=-1-\frac{\log (1-x)}{x}-x^{n+1} \Phi (x,1,n+2)$$ where appears the Lerch transcendent function. This makes 
$$S_n(x)=1+\frac{x \left(1-x^n\right)}{1-x}+\frac{\log (1-x)}{x}+x^{n+1} \Phi (x,1,n+2)$$
$$S_n\left(\frac 12\right)=2-2\log(2)+\frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}$$ As shown below, the last term decreases very fast
$$\left(
\begin{array}{cc}
 n & \frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}\\
 1 & 0.136294 \\
 2 & 0.052961 \\
 3 & 0.021711 \\
 4 & 0.00921103 \\
 5 & 0.00400269 \\
 6 & 0.00177055 \\
 7 & 0.000793989 \\
 8 & 0.000359961 \\
 9 & 0.000164649 \\
 10 & 0.0000758704
\end{array}
\right)$$ and a quick and dirty nonlinear regression (done for $1\leq n \leq 20$) seems to show a quite good approximation by $$\frac{\Phi \left(\frac 12,1,n+2\right)}{2^{n+1}}\approx 0.414102\, e^{-1.11129\, n^{0.888026}}$$
