Loretta Cannon

## Answered question

2022-02-27

Is the series
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\left(\sum _{k=1}^{n}\frac{1}{k}{\left(\frac{1}{2}\right)}^{n-k}\right)$

### Answer & Explanation

cassidacc4

Beginner2022-02-28Added 7 answers

If we denote
$f\left(a,n\right)=\sum _{k=1}^{n}\frac{{e}^{ak}}{k}⇒{f}^{\prime }\left(x,n\right)=\frac{d}{dx}f\left(x,n\right)=\sum _{k=1}^{n}{e}^{xk}$ and
$f\left(a,n\right)={\int }_{-\mathrm{\infty }}^{a}{f}^{\prime }\left(x,n\right)dx$
Then the initial sum
$S=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\left(\sum _{k=1}^{n}\frac{1}{k}{\left(\frac{1}{2}\right)}^{n-k}\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\frac{1}{{2}^{n}}\sum _{k=1}^{n}\frac{{2}^{k}}{k}$
$=\sum _{\frac{1}{n}}\frac{1}{{2}^{n}}f\left(\mathrm{ln}2,n\right)$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\frac{1}{{2}^{n}}{\int }_{-\mathrm{\infty }}^{\mathrm{ln}2}{f}^{\prime }\left(x,n\right)dx$
Evaluating ${f}^{\prime }\left(x,n\right)$
${f}^{\prime }\left(x,n\right)=\sum _{k=1}^{n}{e}^{xk}=\frac{{e}^{x}\left(1-{e}^{xn}\right)}{1-{e}^{x}}$
Therefore, our sum can be presented in the form
$S=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\frac{1}{{2}^{n}}{\int }_{-\mathrm{\infty }}^{\mathrm{ln}2}\frac{{e}^{x}\left(1-{e}^{xn}\right)}{1-{e}^{x}}dx$