eukrasicx

2022-09-30

Series involving a Logarithm

Consider the series

$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty}}[\frac{n}{a}\mathrm{ln}(1+\frac{a}{n})-1+\frac{a}{2n}].\end{array}$

Is there a closed form solution to this series and what is the value when a=1,2 ?

Consider the series

$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty}}[\frac{n}{a}\mathrm{ln}(1+\frac{a}{n})-1+\frac{a}{2n}].\end{array}$

Is there a closed form solution to this series and what is the value when a=1,2 ?

Leslie Braun

Beginner2022-10-01Added 7 answers

For small $|a|$, $\mathrm{ln}(1+a/n)=\sum _{j=1}^{\mathrm{\infty}}(-1{)}^{j-1}(a/n{)}^{j}/j$, so your series is

$\sum _{n=1}^{\mathrm{\infty}}\sum _{j=2}^{\mathrm{\infty}}(-1{)}^{j}{\displaystyle \frac{{a}^{j}}{(j+1)\phantom{\rule{thickmathspace}{0ex}}{n}^{j}}}=\sum _{j=2}^{\mathrm{\infty}}{\displaystyle \frac{(-a{)}^{j}}{j+1}}\zeta (j)$

I don't think this has a closed form.

$\sum _{n=1}^{\mathrm{\infty}}\sum _{j=2}^{\mathrm{\infty}}(-1{)}^{j}{\displaystyle \frac{{a}^{j}}{(j+1)\phantom{\rule{thickmathspace}{0ex}}{n}^{j}}}=\sum _{j=2}^{\mathrm{\infty}}{\displaystyle \frac{(-a{)}^{j}}{j+1}}\zeta (j)$

I don't think this has a closed form.

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