How to find the summation of the following series? .

Question

If $$S=\sum_{i=1}^{n} \frac{1}{i2^{i}},$$ Then how can I find the summation of the above series up to $n^{th}$ terms? I can't solve this question because I don't know whether this summation is a series expansion of any function or not. I know the series expansions of $\cos(x)$, $\sin(x)$, $\tan(x)$, $e^{x}$, $\tan^{-1}x$, $\ln(1+x)$, $\ln(1-x)$, etc. But I am confused that the above summation is the series expansion of which function. Please help me out.

Answer 1

Let us define a function for $|x| < 1$ and $n \in \mathcal{N}$, $$S(x,n) = \sum_{i=1}^n \frac{x^i}{i}$$

Differentiating this function w.r.t $x$, $$\frac{d}{dx} S(x,n) = \sum_{i=1}^n x^{i-1} = \frac{1-x^n}{1-x}$$

Integrating this function from $0$ t0 $x$, $$S(x,n) = \int_{u=0}^x \frac{1-u^n}{1-u} du$$

Your requirement is the value $S(\frac{1}{2}, n)$