Infinite product equality $\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$

Question

Prove the following equation ($|x|<1$) $$\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$$

I made this question and I have the following answer but I think it may be incomplete. If anyone can point out a flaw in my proof or give a better proof then it would be appreciated.

My solution:

$$\begin{align} &f(x)=\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right)\\ &N(p,q)=\{n\in\mathbb N \mid n\ne (2m-1)2^{k-1};m,k\in\mathbb N,2m-1\leq p,k\leq q\}\\ &f(x)(1+x+x^2)=(1+x+x^2)(1-x+x^2)\prod_{n\in N(1,1)}^{\infty} \left(1-x^n+x^{2n}\right)\\ &=(1+x^2+x^4)\prod_{n\in N(1,1)}^{\infty} \left(1-x^n+x^{2n}\right)=(1+x^2+x^4)(1-x^2+x^4)\prod_{n\in N(1,2)}^{\infty} \left(1-x^n+x^{2n}\right)\\ &=(1+x^4+x^8)\prod_{n\in N(1,2)}^{\infty} \left(1-x^n+x^{2n}\right)=(1+x^4+x^8)(1-x^4+x^8)\prod_{n\in N(1,3)}^{\infty} \left(1-x^n+x^{2n}\right)\\ &=(1+x^8+x^{16})\prod_{n\in N(1,3)}^{\infty} \left(1-x^n+x^{2n}\right)=\cdots\\ &=\lim_{k\to\infty}(1+x^{2^k}+x^{2^{k+1}})\prod_{n\in N(1,k)}^{} \left(1-x^n+x^{2n}\right)=\prod_{n\in N(1,\infty)}^{} \left(1-x^n+x^{2n}\right)\\ &\text{Similarly,}\\ &f(x)(1+x+x^2)(1+x^3+x^6)=\prod_{n\in N(3,\infty)}^{} \left(1-x^n+x^{2n}\right)\\ &f(x)(1+x+x^2)(1+x^3+x^6)(1+x^5+x^{10})=\prod_{n\in N(5,\infty)}^{} \left(1-x^n+x^{2n}\right)\\ &\cdots\\ &f(x)\prod_{m=1}^{\infty} \left(1+x^{2m-1}+x^{2(2m-1)}\right)=\lim_{p\to\infty} \prod_{n\in N(p,\infty)}^{} \left(1-x^n+x^{2n}\right)=1 \end{align}$$

(*) Is it obvious that $\{(2m-1)\cdot2^{k-1}\mid m,k\in \mathbb N\}$ is equivalent to $\mathbb N$, or should I also prove it?

Thanks.

Answer 1

We have

\begin{align} \prod_{n = 1}^\infty (1 - x^n + x^{2n}) &= \prod_{n = 1}^\infty \frac{1+x^{3n}}{1+x^n} \\ &= \prod_{n = 1}^\infty \frac{1 - x^{6n}}{1-x^{3n}}\prod_{n=1}^\infty\frac{1-x^{n}}{1-x^{2n}}\\ & = \prod_{n = 1}^\infty \frac{1}{1-x^{3(2n-1)}}\prod_{n = 1}^\infty (1 - x^{2n-1})\\ &= \prod_{n = 1}^\infty \frac{1-x^{2n-1}}{1-(x^{2n-1})^3}\\ &= \prod_{n = 1}^\infty \frac{1}{1 + x^{2n-1} + x^{4n-2}}. \end{align}

Answer 2

Here is a different/more streamlined/possibly less formal approach. To prove that $$\prod_{n=1}^{\infty}(1-x^n+x^{2n})(1+x^{2n-1}+x^{4n-2})=1$$ Separate the first type of factors into factors coming from even $n$ and odd $n$: $$\begin{align} &\prod_{n=1}^{\infty}(1-x^{2n}+x^{4n})(1-x^{2n-1}+x^{4n-2})(1+x^{2n-1}+x^{4n-2})\\ &=\prod_{n=1}^{\infty}(1-x^{2n}+x^{4n})(1+x^{4n-2}+x^{8n-4}) \end{align}$$ Note this is the same as the product from the first line, with $x$ replaced by $x^2$. You can repeat inductively so that the product has higher and higher powers of $x$. If it is convergent at all, it must converge to $1$.