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

The geometric series ∑n=0∞xn=11−x with |x|&lt;1, because it's the basic power series. The series they want the sum for is the derivative. ddx(∑n=0∞xn)=∑n=1∞nxn−1 ddx(11−x)=0⋅(1−x)−1⋅(−1)(1−x)2=1(1−x)2 ∑n=1∞nxn−1=1(1−x)2,|x|&lt;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|&lt;1. Result: 1(1−x)2,|x|&lt;1

Solved: "Starting with the geometric series ∑n∞=0xn , find..."

Cabiolab

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2021-10-12

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

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aprovard

Skilled2021-10-13Added 94 answers

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

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