How to prove the identity $\prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1 $ for $|q|<1$?

Eulers product identity is as follows

\begin{align} \prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1 \end{align}

How one can explicitly prove this identity?

Note here $q$ deonotes a complex number satisfying $|q|<1$

Solution 1:

We have \begin{align}\prod_{n = 1}^\infty (1 - q^{2n-1})(1 + q^n) &= \prod_{n = 1}^\infty \frac{(1 - q^{2n-1})(1 - q^{2n})}{1 - q^n}\\ &= \frac{\prod_{n = 1}^\infty (1 - q^{2n-1})(1 - q^{2n})}{\prod_{n = 1}^\infty (1 - q^n)}\\ &= \frac{\prod_{n = 1}^\infty (1 - q^{2n-1})(1 - q^{2n})}{\prod_{n = 1}^\infty (1 -q^{2n})(1 - q^{2n-1})}\\ &= 1\end{align}

Solution 2:

For $\lvert q\rvert < 1$, define

$$f(q) = \prod_{n=1}^\infty (1-q^{2n-1})(1+q^n).$$

The product is absolutely and locally uniformly convergent on the open unit disk, hence $f$ is holomorphic there, and the product can be reordered and regrouped as desired. It is clear that $f(0) = 1$, and by reordering and regrouping, we have

$$\begin{aligned} f(q) &= \prod_{n=1}^\infty (1-q^{2n-1})(1+q^n)\\ &= \prod_{n=1}^\infty(1-q^{2n-1})(1+q^{2n-1})\cdot \prod_{n=1}^\infty (1+q^{2n})\\ &= \prod_{n=1}^\infty \bigl(1-(q^2)^{2n-1}\bigr)\bigl(1+(q^2)^n\bigr)\\ &= f(q^2). \end{aligned}$$

Since for $0 < \lvert q \rvert < 1$ the set $\{ q^{2^k} : k \in \mathbb{N}\}$ has an accumulation point in the unit disk (namely $0$), it follows that $f$ is constant.

Solution 3:

Another approach, similar to Daniel's, without an appeal to properties of holomorphic functions.

$$(1-q)(1+q)(1+q^2)\cdots (1+q^{2^n})\cdots = 1$$

since the partial product is $1-q^{2^{n+1}}$ .

Substituting $q^{2k-1}$, you get:

$$(1-q^{2k-1})(1+q^{2k-1})(1+q^{(2k-1)\cdot 2})\cdots\left(1+q^{(2k-1)\cdot 2^n}\right)\cdots = 1$$

So $$1=\prod_{k=1}^\infty \left((1-q^{2k-1})\prod_{n=0}^\infty \left(1+q^{(2k-1)2^n}\right)\right)$$

Now, ever positive integer $N$ can be written in exactly one way as $(2k-1)2^n$. So we can rearrange.