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

Cabiolab 2021-10-12 Answered

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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Expert Answer

aprovard
Answered 2021-10-13 Author has 11117 answers

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

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