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


$$ (1+x^2+x^3+\cdots)=\frac{1}{1-x}= \sum_{n=0}^\infty x^n $$ $$ (1-x^2+x^3-\cdots)=\frac{1}{1+x}= \sum_{n=0}^\infty (-1)^nx^n $$ 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

  • $\begingroup$ 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. $\endgroup$
    – Hamed
    Nov 2 '15 at 21:08
  • $\begingroup$ The sum $1+x^2+x^4 + \cdots $ converges to $\dfrac 1 {1-x^2}$, not $\dfrac 1 {1+x^2}$. ${}\qquad{}$ $\endgroup$ Nov 2 '15 at 21:18
  • $\begingroup$ @Tom . In Equation (2) the superscript k should be n. Then apply the next line to it. $\endgroup$ Nov 2 '15 at 23:13

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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