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Define the $q$-analog $(a;q)_n = \prod_{k=0}^n \left(1 - aq^k\right)$.

I want to prove the identity $\frac{(q^2;q^2)_\infty}{(q;q)_\infty}=\frac{1}{(q;q^2)_\infty}$.

I viewed the LHS this way:

$\frac{1 - q^2}{1 - q} \frac{1 - q^4}{1 - q^2} \frac{1 - q^6}{1 - q^3} \frac{1 - q^8}{1 - q^4} \frac{1 - q^{10}}{1 - q^5} \frac{1 - q^{11}}{1 - q^6} \cdots$

and you can imagine getting the RHS by canceling the $k$th term on the top with the $2k+1$th term on the bottom.

Can we really cancel that way though? Since the gap between $k$ and $2k+1$ keeps getting bigger it might not be valid.

1 Answers 1

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Take the logarithm of the quotient of infinite products to get a difference of infinite sums. Both sums converge absolutely, and hence you can rearrange their terms at will and let them cancel.