<s>[INST] 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 
$$\frac{1}{1+x}\cdot\frac{1}{1-x}=\frac{1}{1-x^2} = 1+x^2+x^4+\cdots=\sum_{n=0}^\infty x^{2n}$$
Any advice would be helpful thank you in advance
 [/INST] 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?
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