Starting with the geometric series ∑n∞=0xn , find the sum of the series ∑n∞=1nxn−1 |x|<1

The geometric series ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}}$ with |x|<1, because it's the basic power series. The series they want the sum for is the derivative.
${\displaystyle \frac{d}{dx}\left(\sum _{n=0}^{\mathrm{\infty }}{x}^n\right)=\sum _{n=1}^{\mathrm{\infty }}n{x}^{n-1}}$
${\displaystyle \frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{0\cdot (1-x)-1\cdot (-1)}{{(1-x)}^2}=\frac{1}{{(1-x)}^2}}$
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}n{x}^{n-1}=\frac{1}{{(1-x)}^2},\left|x\right|<1}$
The derivative of the series representation of a function is equal to the derivative of the function.
The radius of convergence is the same as the original series, which is why there is a |x|<1.
Result:
${\displaystyle \frac{1}{{(1-x)}^2},\left|x\right|<1}$