Sum of series which is not Arithmetic Progression or Geometric Progression What is the sum of following infinite series ?
I am not able to find it as modification of $\sin x$ or $\log(1+x)$ series
$$\frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dots$$
 A: tl;dr:

for $x\notin[-1,1]$, $$
-\ln\left(1-\frac{1}{x}\right)
 = \sum_{n=1}^\infty \frac{1}{nx^n}.
$$

How to get to the result:
Start with the series for $x\mapsto \ln(1+x)$: For $x\in(-1,1)$,
$$
\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}
$$
so, considering $-x$,
$$
\ln(1-x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}(-1)^nx^n}{n}
 = -\sum_{n=1}^\infty \frac{x^n}{n}
$$
and therefore, for $x\in(-1,1)$,
$$
-\ln(1-x)
 = \sum_{n=1}^\infty \frac{x^n}{n}.
$$
Now, this implies that for $x\notin[-1,1]$,
$$
-\ln\left(1-\frac{1}{x}\right)
 = \sum_{n=1}^\infty \frac{1}{nx^n}.
$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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With $\ds{\verts{x} > 1}$:

\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty}\,{1 \over nx^{n}}} & =
\sum_{n = 1}^{\infty}\,{1 \over x^{n}}\int_{0}^{1}y^{n - 1}\,\dd y =
{1 \over x}\int_{0}^{1}\sum_{n = 1}^{\infty}\pars{y \over x}^{n - 1}\,\dd y =
{1 \over x}\int_{0}^{1}{1 \over 1 - y/x}\,\dd y
\\[5mm] & =
\left.\vphantom{\Large A}-\ln\pars{\verts{y - x}}
\right\vert_{\ y\ =\ 0}^{\ y\ =\ 1}\,\,\,\,\,\, =\
-\ln\pars{\verts{1 - x \over 0 - x}} =
\color{#f00}{-\ln\pars{1 - {1 \over x}}}
\end{align}
A: Your series is $\frac 1x -\frac 1x \int{dx} \left(  \frac 1{x^2}+\frac 1{x^3}+\frac 1{x^4}\dots\right)$  where the first term is special because the constant in the denominator is $1$, which avoids problems with integrating to the log.  Bring the sum inside.
