How to evaluate $\sum \frac{1}{2^{n-r}}\frac{1}{n}$?

Question

$$\sum_{n=r+1}^{\infty}\frac{1}{2^{n-r}}\frac{1}{n}$$

,where r is a positive integer.

First, I wanna know if this is convergent(I guess so) and WHY.

And if so, how to evaluate it?

Answer 1

\begin{align*} \sum_{n=r+1}^\infty \frac{1}{2^{n-r}}\cdot\frac{1}{n}=2\Phi\left(\frac{1}{2},1,r+1\right) \end{align*} where $\Phi$ is Lerch zeta function defined as \begin{align*} \Phi(z,s,\alpha)=\sum_{n=0}^\infty \frac{z^n}{(n+\alpha)^s} \end{align*} This series only converges for any real number $\alpha>0$ where $|z|<1$ or $\operatorname*{Re}(s)>1$ and $|z|=1$.

In this problem, using Lerch's integral representation, \begin{align*} \Phi(z,s,\alpha)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{t^{s-1}e^{-\alpha t}}{1-ze^{-t}}dt \end{align*} let $z=1/2, \ s=1, \ \alpha=r+1$, then \begin{align*} \Phi(1/2,1,r+1)=\frac{1}{\Gamma(1)}\int_0^\infty \frac{t^{1-1}e^{-(r+1)t}}{1-\frac{1}{2}e^{-t}}dt=\int_0^\infty\frac{2e^{-rt}}{2e^t-1}dt\overset{\triangle}{=}I \end{align*} set $u=e^t, \ du=e^tdt=udt$, so we get \begin{align*} I=\int_1^\infty \frac{2}{2u^{r+2}-u^{r+1}}du \end{align*} notice that $u^{r+2}<2u^{r+2}-u^{r+1}<2u^{r+2}$ when $u>1, \ r>1$ which implies \begin{align*} \frac{1}{r+1}=\int_{1}^\infty \frac{1}{u^{r+2}}du<I<\int_0^\infty \frac{2}{u^{r+2}}du=\frac{2}{r+1} \end{align*}