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

Question

I want to compute the following identity

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

Can anyone gives some explict procedure of this identity?

Answer 1

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

$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}]\cdot [\prod_{n=1}^\infty{(1+q^{2 \cdot (n - 1)})}] = $

$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}] \cdot [\prod_{n=0}^\infty{(1+q^{2 \cdot n})}] = $

$[\prod_{n=1}^\infty{(1+q^{2 \cdot n})}] \cdot (1+q^0) \cdot [\prod_{n=1}^\infty{(1+q^{2 \cdot n})}] = $

$2 \cdot \prod_{n=1}^\infty{(1+q^{2 \cdot n})^2}$

It's the same as simplifying a telescoping sequence like $\sum_{n=1}^\infty{\frac{1}{n \cdot (n+1)}}$ but with multiplication instead of addition.

Answer 2

Observe that the product can be written as $A\cdot B$ where $B=2A, A=\prod_{n=1}^\infty (1+q^{2n})$