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

Harlen Pritchard 2021-06-13 Answered

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

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Anonym
Answered 2021-06-14 Author has 24929 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 \(\sum_{n=1}^\infty nx^{n-1},\ |x|<1\)
\(\sum_{n=1}^\infty nx^{n-1}=1+2x+3x^2+4x^3+...\)
\(=(1+x+x^2+x^3+...)+(x+2x^2+3x^2+4x^3+...)\)
\(=\frac{1}{1-x}+x(1+2x+3x^2+4x^3+...)\)
\(=\frac{1}{1-x}+x\sum_{n=1}^\infty nx^{n-1}\)
\(\Rightarrow\sum_{n=1}^\infty nx^{n-1}-x\sum_{n=1}^\infty nx^{n-1}=\frac{1}{1-x}\)
\((1-x)\sum_{n=1}^\infty nx^{n-1}=\frac{1}{1-x}\)
\(\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}\)

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