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stiv31 [10]
2 years ago
15

(a) Starting with the geometric series [infinity] xn n = 0 , find the sum of the series [infinity] nxn − 1 n = 1 , |x| < 1.

Mathematics
1 answer:
Mila [183]2 years ago
5 0

Let <em>f(x)</em> be the sum of the geometric series,

f(x)=\displaystyle\frac1{1-x} = \sum_{n=0}^\infty x^n

for |<em>x</em>| < 1. Then taking the derivative gives the desired sum,

f'(x)=\displaystyle\boxed{\dfrac1{(1-x)^2}} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}

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Answer:

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