Value of $\sum\limits^{\infty}_{n=1}\frac{\ln n}{n^{1/2}\cdot 2^n}$

Question

Here is a series:

$$\displaystyle \sum^{\infty}_{n=1}\dfrac{\ln n}{n^{\frac12}\cdot 2^n}$$

It is convergent by d'Alembert's law. Can we find the sum of this series ?

Answer 1

Consider $$f(s):=\sum_{n=1}^\infty \frac {\left(\frac 12\right)^n}{n^s}=\operatorname{Li}_s\left(\frac 12\right)$$ with $\operatorname{Li}$ the polylogarithm then (since $\,n^{-s}=e^{-s\ln(n)}$) : $$f'(s)=\frac d{ds}\operatorname{Li}_s\left(\frac 12\right)=-\sum_{n=1}^\infty \frac {\ln(n)}{n^s}\left(\frac 12\right)^n$$ giving minus your answer for $s=\frac 12$.

You may use the integrals defining the polylogarithm to get alternative formulations but don't hope much simpler expressions...

Answer 2

$f(n)=\dfrac{\ln n}{n^{1/2}\cdot 2^n}>0~\forall~n>1\\e^nf(e^n)=e^{n/2}\dfrac{n}{2^n}$

$[e^nf(e^n)]^{1/n}=\dfrac{\sqrt e n^{1/n}}{2}\to\sqrt e/2<1$

$\displaystyle \sum^{\infty}_{n=1}e^nf(e^n)$ is convergent implies $\displaystyle \sum^{\infty}_{n=1}f(n)$ is convergent.