Does $\sum_{n=1}^{\infty} \frac{1}{2^n}\left(\frac11+\frac12+\frac 13+...+\frac1 n\right)$ have a closed form solution?

Question

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.

Answer 1

We are interested in $$ S=\sum_{n=1}^\infty \frac{1}{2^n}\sum_{k=1}^n\frac{1}{k}. $$ By changing the order of summation, we have $$S=\sum_{k=1}^\infty\frac{1}{k}\sum_{n=k}^\infty\frac{1}{2^n}=\sum_{k=1}^\infty\frac{1}{k\cdot2^{k-1}}\sum_{n=1}^\infty\frac{1}{2^n}=\sum_{k=1}^\infty\frac{1}{k\cdot2^{k-1}}=2\sum_{k=0}^\infty\frac{1}{k\cdot2^k},$$ where we use $$\sum_{n=1}^\infty \frac{1}{2^n}=1.$$ Now $$ \sum_{k=1}^\infty\frac{x^k}{k}=-\ln(1-x), $$ so our sum is $$S=-2\cdot\ln(1-\tfrac{1}{2})=\ln(4)$$

Answer 2

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 $2\log 2$.

Answer 3

Write this as a double sum, then interchange order of summation: $$ \sum_{n=1}^\infty\frac1{2^n}\sum_{k=1}^n\frac1k= \sum_{k=1}^\infty\sum_{n=k}^\infty\frac1{2^n}\frac1k =\sum_{k=1}^\infty\frac1k\frac{(\frac12)^k}{1-\frac12} =2\sum_{k=1}^\infty \frac{(\frac12)^k}k $$ This last sum can be evaluated using the power series for $\log(1-x)$: $$ \log(1-x)=-\sum_{k=1}^\infty\frac{x^k}k $$ to obtain the value $-2\log (1/2)=\log 4$.