Why Does This Infinite Product Identity Hold for Integer Partitions?

[futurebird]

Homework Statement

I am reading about integer partitions. I'm learning a proof and I don't understand what would seem to be a simple step... as the book presents it without comment:

\prod_{n=1}^{\infty} \frac{1-q^{2n}}{1-q^{n}}=\prod_{n=1}^{\infty} \frac{1}{1-q^{2n-1}}

The fractions presented are not algebraically equivalent outside of the infinite product... So, it's something about the product that makes this possible. I also know that |q|<1... What is going on?

 

[LCKurtz]

Look at a partial product:

\left(\frac{1-q^2}{1-q}\right)\left(\frac{1-q^4}{1-q^2}\right)\left(\frac{1-q^6}{1-q^3}\right)\left(\frac{1-q^8}{1-q^4}\right)...

At least intuitively, each factor in the numerator is canceled by the corresponding factor in the denominator with q to that same even exponent. So in the limit you are left with 1 on top and odd powers of q in the factors on the bottom.

 

[futurebird]

Thank you so much... I wish I'd seen that, it makes so much sense.