I am trying questions from Apostol Introduction to ANT of Chapter partitions and need help in deducing this identity.
Problem is question 6(a) which will use some information from 2 and 5(b).
Attempt : To use 5(b) I need to change index of summation so I changed $\sum_{-\infty}^{0} +\sum_{1}^{\infty} $ but the problem is as $\sum_{1}^{\infty} x^{m(m+1)} /2 = \sum_{1}^{\infty} x^{(m(m-1) /2 )} x^2 $. Now m=0 in $x^{m(m+1) /2} =1$ and $\sum_{-\infty}^{-1} x^{(m(m+1)/2} =\sum_{1}^{\infty} x^{(m(m+1) /2}$ so, I get $ (1-x^{2n+2}) \sum_{n=1}^{\infty}[ 1+x^{2n-1}]$ which is different from what I needed to do.
So, can you please tell how to approach the problem.
Substituting the identity in Exercise $2$ into the one in Exercise $5$(b) yields the identity
$$\sum_{m=1}^\infty x^{m(m-1)/2}=\prod_{n=1}^\infty(1+x^n)(1-x^{2n})\,.$$
Examining the exponents, we see that
$$\sum_{m=1}^\infty x^{m(m-1)/2}=\sum_{m=-\infty}^0x^{m(m-1)/2}\,,$$
so
$$\begin{align*} \sum_{m=-\infty}^\infty x^{m(m-1)/2}&=2\sum_{m=1}^\infty x^{m(m-1)/2}\\ &=2\prod_{n=1}^\infty(1+x^n)(1-x^{2n})\\ &=(1+x^0)\prod_{n=1}^\infty(1+x^n)(1-x^{2n})\\ &=\prod_{n=1}^\infty(1+x^{n-1})(1-x^{2n})\,. \end{align*}$$