# Starting with the geometric series\infty\sum n=0x^{n}, find the

Starting with the geometric series
$$\displaystyle\sum_{n}^\infty={0}{x}^{{{n}}}$$
, find the sum of the series
$$\displaystyle\sum_{n}^\infty={1}{n}{x}^{{{n}-{1}}}$$
$$|x|<1$$

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aprovard

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