# Cauchy Product - Power Series Representation

Question

Use Cauchy product to find a power series representation of $$(1+x^2+x^3+\cdots)(1-x^2+x^3-\cdots)$$

Solution

$$(1+x^2+x^3+\cdots)=\frac{1}{1-x}= \sum_{n=0}^\infty x^n$$ $$(1-x^2+x^3-\cdots)=\frac{1}{1+x}= \sum_{n=0}^\infty (-1)^nx^n$$ The Cauchy Product states: $$\left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty b_n x^n\right) =\sum_{n=0}^\infty\left(\sum_{j=0}^n a_jb_{n-j}\right)x^n\tag{2}$$ and $$\sum_{j=0}^n a_jb_{n-j}=\sum_{j=0}^n (-1)^{n-j}= \begin{cases} 0 & \text{even} \\ 1 & \text{odd} \end{cases}$$

But now I am stuck and not sure quite what to do even though I know the answer is just $$\frac{1}{1+x}\cdot\frac{1}{1-x}=\frac{1}{1-x^2} = 1+x^2+x^4+\cdots=\sum_{n=0}^\infty x^{2n}$$

• You did a little mistake. Sine the sum runs through $j=0$ to $n$, if $n=2k+1$, then there are $2k+2$ terms which are alternating. So the Caucy coefficient is 0 when $n$ odd and $1$ when $n$ even. Nov 2 '15 at 21:08
• The sum $1+x^2+x^4 + \cdots$ converges to $\dfrac 1 {1-x^2}$, not $\dfrac 1 {1+x^2}$. ${}\qquad{}$ Nov 2 '15 at 21:18
Hint. Since $$\frac{1+(-1)^n}2= \left\{\begin{array}{ll}1 &\quad n = 2k\\ 0 &\quad n=2k+1, \end{array} \right.$$ why not just write $$\frac{1}{1+x}\times\frac{1}{1-x}=\frac{1}{1-x^2} =\sum_{n=0}^{\infty}{x^{2n}}=\sum_{n=0}^{\infty}\frac{1+(-1)^n}2x^{n}, \quad |x|<1,$$ the latter series being a power series representation?