<s>[INST] computation $ \prod_{n=1}^{\infty}(1+q^{2n})(1+q^{2(n-1)}) = 2 \prod_{n=1}^{\infty} (1+q^{2n})^2 $ 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?
 [/INST] $\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.
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