Geometric Progression and Counting Partitions

Question

Related with The number of partitions of $n$ into distinct parts equals the number of partitions of $n$ into odd parts

I know that ${(x^n-1)\over (x-1)}=1+x+x^2+x^3+x^4+...$ as a Geometric Progress, but I do not see how $${1\over (1-x)(1-x^3)(1-x^5)\cdot\dots}=(1+x+x^{1+1}+\dots)(1+x^3+x^{3+3}+\dots)(1+x^5+\dots).$$ can anyone please explain?

Answer 1

Your question is why $1+x+x^2+x^3+x^4+...=\frac{1}{1-x}$

You know that ${x^n-1\over x-1}=1+x+x^2+x^3+x^4+...+x^{n-1}$

${(x^n-1)\over (x-1)}=\frac{x^n}{x-1}-\frac{1}{x-1}$

Let $n\to \infty$ and $-1<x<1$

$$\lim_{n\to \infty} \frac{x^n}{x-1}-\lim_{n\to \infty}\frac{1}{x-1}=0-\frac{1}{x-1}=\frac{1}{1-x}$$

If you substitute $x$ by $u^3$ or $u^5$ you´ll get similar results.