Cabiolab

2021-10-12

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

aprovard

The geometric series $\sum _{n=0}^{\mathrm{\infty }}{x}^{n}=\frac{1}{1-x}$ with |x|<1, because it's the basic power series. The series they want the sum for is the derivative.
$\frac{d}{dx}\left(\sum _{n=0}^{\mathrm{\infty }}{x}^{n}\right)=\sum _{n=1}^{\mathrm{\infty }}n{x}^{n-1}$
$\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{0\cdot \left(1-x\right)-1\cdot \left(-1\right)}{{\left(1-x\right)}^{2}}=\frac{1}{{\left(1-x\right)}^{2}}$
$\sum _{n=1}^{\mathrm{\infty }}n{x}^{n-1}=\frac{1}{{\left(1-x\right)}^{2}},|x|<1$
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:
$\frac{1}{{\left(1-x\right)}^{2}},|x|<1$

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