closed form sum: $\sum_{i=1}^n \frac{2^{-i}}{i}$

Question

I came across this gnarly sum and was wondering if anyone knew any tricks to get a closed form for it:

$$\sum_{i=1}^n \frac{2^{-i}}{i}$$

Answer 1

As said by Batman in his comment, this summation involves Lerch transcendent function and simply write $$\sum_{i=1}^n \frac{2^{-i}}{i}=\log (2)-2^{-(n+1)} \Phi \left(\frac{1}{2},1,n+1\right)$$ For $n$ going to infinity, the limit is $\log (2)$.

Also as mentioned by Apurv, for large values of $n$ this summation looks like the infinite Taylor series of $$-\log (1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} +\frac{x^8}{8}+O\left(x^9\right)$$ If you plug $x=\frac{1}{2}$, you end with a limit equal to $-\log \frac{1}{2}= \log (2)$.

Answer 2

A closed form of the sum does not exist. See this.

However, you can find the limit of the sum as n tends to infinity, using Taylor series for $\log (1+x)$ and then putting $x=\dfrac{1}{2}$.