The geometric series \(\displaystyle{\sum_{{{n}={0}}}^{{\infty}}}{x}^{{{n}}}={\frac{{{1}}}{{{1}-{x}}}}\) with |x|<1, because it's the basic power series. The series they want the sum for is the derivative.

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sum_{{{n}={0}}}^{{\infty}}}{x}^{{{n}}}\right)}={\sum_{{{n}={1}}}^{{\infty}}}{n}{x}^{{{n}-{1}}}\)

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{1}}}{{{1}-{x}}}}\right)}={\frac{{{0}\cdot{\left({1}-{x}\right)}-{1}\cdot{\left(-{1}\right)}}}{{{\left({1}-{x}\right)}^{{{2}}}}}}={\frac{{{1}}}{{{\left({1}-{x}\right)}^{{{2}}}}}}\)

\(\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{n}{x}^{{{n}-{1}}}={\frac{{{1}}}{{{\left({1}-{x}\right)}^{{{2}}}}}},{\left|{x}\right|}{<}{1}\)

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:

\(\displaystyle{\frac{{{1}}}{{{\left({1}-{x}\right)}^{{{2}}}}}},{\left|{x}\right|}{<}{1}\)