Sum of series which is not Arithmetic Progression or Geometric Progression

Question

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

$$\frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dots$$

Answer 1

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

Answer 2

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.

Answer 3

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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}