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

Consider the geometric series, ∑n=0∞xn=1+x+x2+x3+... =11−x Find the sum of the series ∑n=1∞nxn−1, |x|&lt;1 ∑n=1∞nxn−1=1+2x+3x2+4x3+... =(1+x+x2+x3+...)+(x+2x2+3x2+4x3+...) =11−x+x(1+2x+3x2+4x3+...) =11−x+x∑n=1∞nxn−1 ⇒∑n=1∞nxn−1−x∑n=1∞nxn−1=11−x (1−x)∑n=1∞nxn−1=11−x ∑n=1∞nxn−1=1(1−x)2

"Starting with the geometric series ∑n=0∞xn, find the..." solved and explained

Harlen Pritchard

Answered question

2021-06-13

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

Answer & Explanation

Anonym

Skilled2021-06-14Added 108 answers

Consider the geometric series,
$\sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + . . .$
$= \frac{1}{1 - x}$
Find the sum of the series $\sum_{n = 1}^{\infty} n x^{n - 1}, | x | < 1$
$\sum_{n = 1}^{\infty} n x^{n - 1} = 1 + 2 x + 3 x^{2} + 4 x^{3} + . . .$
$= (1 + x + x^{2} + x^{3} + . . .) + (x + 2 x^{2} + 3 x^{2} + 4 x^{3} + . . .)$
$= \frac{1}{1 - x} + x (1 + 2 x + 3 x^{2} + 4 x^{3} + . . .)$
$= \frac{1}{1 - x} + x \sum_{n = 1}^{\infty} n x^{n - 1}$
$\Rightarrow \sum_{n = 1}^{\infty} n x^{n - 1} - x \sum_{n = 1}^{\infty} n x^{n - 1} = \frac{1}{1 - x}$
$(1 - x) \sum_{n = 1}^{\infty} n x^{n - 1} = \frac{1}{1 - x}$
$\sum_{n = 1}^{\infty} n x^{n - 1} = \frac{1}{(1 - x)^{2}}$

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