By geometric series, we have, for − 1 < x < 1 {\displaystyle -1<x<1} ,
1 1 − x = ∑ n = 0 ∞ x n . {\displaystyle {\begin{aligned}{\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}.\end{aligned}}}
Multiplying both sides by x 3 {\displaystyle x^{3}} yields
x 3 1 − x = x 3 ∑ n = 0 ∞ x n = ∑ n = 0 ∞ x n + 3 {\displaystyle {\begin{aligned}{\frac {x^{3}}{1-x}}=x^{3}\sum _{n=0}^{\infty }x^{n}=\sum _{n=0}^{\infty }x^{n+3}\end{aligned}}}
which is the power series we are looking for.
,
yields
