TITLE: Does $\sum_{k=1}^{\infty}k(p^{\frac{(k-1)k}{2}}-p^{\frac{(k+1)k}{2}})$ converge?
QUESTION [5 upvotes]: Does the sum:
$$\sum_{k=1}^{\infty}k(p^{\frac{(k-1)k}{2}}-p^{\frac{(k+1)k}{2}})$$
$$ p\in\mathbb{R}|0{\leq}p<1$$
converse, and if so, to what function?

REPLY [4 votes]: Let's simplify your expression to get Carl Najafi's expression :
\begin{align}
\sum_{k=1}^{\infty}k\left(p^{\frac{(k-1)k}{2}}-p^{\frac{(k+1)k}{2}}\right)&=\sum_{k=1}^{\infty}k\;p^{\frac{(k-1/2)^2}2-\frac 18}-\sum_{k=1}^{\infty}k\;p^{\frac{(k+1/2)^2}2-\frac 18}\\
&=\sum_{k=0}^{\infty}(k+1)\;p^{\frac{(k+1/2)^2}2-\frac 18}-\sum_{k=1}^{\infty}k\;p^{\frac{(k+1/2)^2}2-\frac 18}\\
&=\sum_{k=0}^{\infty}\;p^{\frac{(k+1/2)^2}2-\frac 18}\\
&=\sum_{k=0}^{\infty}\;p^{\frac{k(k+1)}2}\\
\end{align}
Proving Carl's claim.
After that you'll simply have to use the definition of the second theta function
$$\theta_2(0,\sqrt{p})=2\sum_{k=0}^{\infty}\;\sqrt{p}^{(k+1/2)^2}$$
to get the Alpha result  (since $\sqrt{p}^{1/4}=\sqrt[8]{p}$) :
$$\frac {\theta_2(0,\sqrt{p})}{2\;\sqrt[8]{p}}\quad\text{for}\ 0<p<1$$