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I am trying questions from Apostol Introduction to ANT of Chapter partitions and need help in deducing this identity.

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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.

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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

$$\begin{align*} \sum_{m=-\infty}^\infty x^{m(m-1)/2}&=2\sum_{m=1}^\infty x^{m(m-1)/2}\\ &=2\prod_{n=1}^\infty(1+x^n)(1-x^{2n})\\ &=(1+x^0)\prod_{n=1}^\infty(1+x^n)(1-x^{2n})\\ &=\prod_{n=1}^\infty(1+x^{n-1})(1-x^{2n})\,. \end{align*}$$

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  • $\begingroup$ I am unable to prove $\sum_{m=1}^\infty x^{m(m-1)/2}=\sum_{m=-\infty}^0x^{m(m-1)/2}$ , Rest of the proof is clear , can you please tell how to do it? $\endgroup$
    – user775699
    Commented Dec 2, 2020 at 12:52
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    $\begingroup$ @User: If $m\in\Bbb Z^+$, then $m(m-1)=(-m)(-m+1)$, so $$\prod_{m\ge 1}x^{m(m-1)/2}=\prod_{m\ge 1}x^{(-m+1)(-m)/2}=\prod_{k\le 0}x^{k(k-1)/2}\,,$$ where $k=-m+1$. $\endgroup$
    – Brian M. Scott
    Commented Dec 3, 2020 at 22:31

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