Series expansion of $\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$? How would I find the series expansion $\displaystyle\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}$ so that it will turn into an infinite power series again?? 
 A: If you set
$$f(x)=\prod_{n=0}^\infty (1-x^n)^{-1}=\sum_{n=0}^\infty p(n)x^n$$
we see that yours is just
$$f(x^2)=\sum_{n=0}^\infty p(n)x^{2n}.$$
A: $$\frac{1}{1-x^k}=1+x^k+x^{2k}+x^{3k}\cdots$$
$$\frac{1}{1+x^k}=1-x^k+x^{2k}-x^{3k}\cdots$$
A: I suppose that you can make the expresion shorter using $(1+a)(1-a)=1-a^2$ so $$A=\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}=\frac{1}{(1−x^2)(1−x^4)(1-x^6)\cdots}$$ and now use the fact that $$\frac{1}{1-y}=\sum_{i=0}^{\infty}y^i$$ and replace successively $y$ by $x^2$, $x^4\cdots$,$x^{2n}$ before computing the overall product.
Doing so, for a large number of terms, you should arrive to $$A=1+x^2+2 x^4+3 x^6+5 x^8+7 x^{10}+11 x^{12}+15 x^{14}+22 x^{16}+30 x^{18}+42 x^{20}+56
   x^{22}+77 x^{24}+101 x^{26}+135 x^{28}+176 x^{30}+O\left(x^{31}\right)$$ and, as mentioned earlier, the coefficients correspond to sequence $\text{A00041}$ of $\text{EOIS}$ where $a_n$ is the number of partitions of $n$ (the partition numbers). 
Probably off-topic, for an infinite number of terms, $$A=\frac{1}{\left(x^2;x^2\right){}_{\infty }}$$ where appears the q-Pochhammer symbol and which, for sure, leads to the same expansion.
