# Math Help - Multiplication of series

1. ## Multiplication of series

1. Multiply the series for (1-x)^-1 (valid when |x|<1) by itself and so obtain the expansion

(1-x)^-2 = sigma n=0 to infinity (n+1)x^n, |x|<1.

2. Derive the expansion

(1-x)^-3 = sigma n=0 to infinity ((n+1)(n+2)*x^n)/2 (|x|<1)

by multiplication of series. Use the result of Exercise 1.

2. Multiplying series is extra work in this case.

Notice that $\displaystyle \frac{d}{dx}\left(\frac{1}{1 - x}\right) = \frac{1}{(1 - x)^2}$ and $\displaystyle \frac{d}{dx}\left[\frac{1}{(1 - x)^2}\right] = \frac{2}{(1-x)^3}$...

3. $(1-x)^{-1}=\sum_{n=0}^{+\infty}x^n$ is absolutely convergent for $|x|<1$ . Use a well known result about Cauchy product of series.

Fernando Revilla