I want to find the sum function of the series
$$ \sum_{n=0}^{\infty}\frac{n}{n+1}x^{n} $$
and I would like to get your assessment of my work cause I'm really not sure about its validity?
I define the sum $$s(x) = \sum_{n=0}^{\infty}\frac{n}{n+1}x^{n}$$ and multipy with x
$$ s(x)\cdot x = \sum_{n=0}^{\infty}\frac{n}{n+1}x^{n+1} $$
then differentiate
$$ (s(x)\cdot x)' = \sum_{n=0}^{\infty}nx^{n} $$
divide by x
$$ \frac{(s(x)\cdot x)'}{x} = \sum_{n=0}^{\infty}nx^{n-1} $$
then integrate and find the sum of a regular geometric series
$$ \int\frac{(s(x)\cdot x)'}{x} = \sum_{n=0}^{\infty}x^{n} = \frac{1}{1-x} $$
differentiate again
$$ (\int\frac{(s(x)\cdot x)'}{x})' = (\frac{1}{1-x})' \Rightarrow \frac{(s(x)\cdot x)'}{x} = \frac{1}{(1-x)^{2}} $$
multiply by x and take the integral
$$ s(x)\cdot x = \int_{0}^{x}\frac{t}{(1-t)^{2}}dt = \ln(|x-1|) - \frac{1}{x-1} + C $$
To find the constant C let $x=0$ and find that $C=-1$. Lastly divide by x to get the final result
$$ s(x) = \frac{\ln(|x-1|)}{x} - \frac{1}{x(x-1)} - \frac{1}{x} $$
Can anyone verify that this is correct? Or is there another way to get the sum?
Thanks!
Yes, it is correct. But there is a simpler way. Note that\begin{align}\sum_{n=0}^\infty\frac n{n+1}x^n&=\sum_{n=0}^\infty x^n-\sum_{n=0}^\infty\frac{x^n}{n+1}\\&=\frac1{1-x}-\sum_{n=0}^\infty\frac{x^n}{n+1}.\end{align}In order to compute$$\sum_{n=0}^\infty\frac{x^n}{n+1},$$if $s(x)$ is its sum, then$$\bigl(xs(x)\bigr)'=\sum_{n=0}^\infty x^n=\frac1{1-x}$$and therefore $xs(x)=-\log|x-1|$. So, the sum of your series is$$\frac1{1-x}+\frac{\log|x-1|}x.$$