what is the expansion for $\log \sum q^{n^2}$ ?
Using the Jacobi triple product $$ (q;q)_\infty(z;q)_\infty (q/z ;q)_\infty = \sum_{k \in \mathbb{Z}} z^k q^{\binom{k}{2}} $$ if we set $q \to q^2$ and $z \to q$ $$ (q^2;q^2)_\infty(q;q^2)_\infty (q ;q^2)_\infty = \sum_{n \in \mathbb{Z}} q^{n^2} \tag{1}$$ quite confusingly Wikipedia has $$ (q^2;q^2)_\infty(-q;q^2)_\infty (-q ;q^2)_\infty = \sum_{n \in \mathbb{Z}} q^{n^2} \tag{2}$$
In one paper I found: $$\theta(x; q) = (x;q)_\infty (qx^{-1}; q)_\infty = \exp \left( - \sum_{ m \neq 0} \frac{x^m}{m(1-q^m)} \right) \tag{$\ast$} $$ however there is a factor of $(q;q)$ missing. Then if we plug in $x = q^2$ and $ p = q$: $$\theta(q; q^2) = (q^2 ; q^2)(q;q^2)_\infty (q; q^2)_\infty = \exp \left( - \sum_{ m \neq 0} \frac{q^{2m}}{m(1-q^{m})} \right) $$ We we expand out the Pocchammer symbols: $$ \sum_{n \in \mathbb{Z}} q^{n^2} = \prod (1 - q^{2m+1})^2(1 - q^{2m})$$ Then I might take the log of both sides and get my answer: $$ \log \sum_{n \in \mathbb{Z}} q^{n^2} = \sum \Big[2\log (1 - q^{2m+1}) + \log(1 - q^{2m})\Big]$$
Then recall the formula for $\log$ $$ \log (1 -x ) = \sum \frac{x^m}{m}$$