TITLE: Does $\sum_{n=1}^{\infty} \frac{1}{2^n}\left(\frac11+\frac12+\frac 13+...+\frac1 n\right)$ have a closed form solution?
QUESTION [5 upvotes]: I need to know whether
$$\sum_{n=1}^{\infty} \frac{1}{2^n}\left(\frac11+\frac12+\frac 13+...+\frac1 n\right)$$has a close form solution? 
It is easy to prove this series converge ,because $\frac11+\frac12+\frac 13+...+\frac1 n \sim \ln n$ so $$\sum_{n=1}^{\infty} \frac{1}{2^n}\left(\frac11+\frac12+\frac 13+...+\frac1 n\right)\sim\sum_{n=1}^{\infty} \frac{\ln n}{2^n}<\sum_{n=1}^{\infty} \frac{n}{2^n}=4$$ 
I tried to get the answer by a MATLAb program: suppose $$\quad{a_k=\sum_{n=1}^{k} \frac{1}{2^n}\left(\frac11+\frac12+\frac 13+...+\frac1 n\right)\\a_=0.5\\a_2=    0.8750   \\a_3= 1.1042 \\a_4=   1.2344\\a_5=    1.3057\\\vdots\\a_{20}=1.3863\\\vdots\\a_{100}= 1.3863\\\vdots\\a_{10000}= 1.3863\\\vdots\\a_{10^6}<1.5}$$ So I think It converges to $1.3863 \leq \lim_{k\to \infty}a_k \leq 1.5$  
My question is about an analytic solution, Does it exist?
  Thanks in advance for any Idea.

REPLY [7 votes]: Note that $$\sum_{n=1}^{\infty}\frac{x^{n}}{n}=-\log(1-x)$$ and $$\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$$ and multiplying these series using Cauchy product we get $$\sum_{n=1}^{\infty}\left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)x^{n}=-\frac{\log(1-x)}{1-x}$$ and putting $x=1/2$ we get the desired sum as $2\log 2$.