The closed form of $\sum_{k=1}^{\infty}\left(\frac{2}{3}\right)^{k^2}\frac{3^k}{2^k-3^k}$ Are you kind to show me the way? I want to find its closed form.
$$\sum_{k=1}^{\infty}\left(\frac{2}{3}\right)^{k^2}\frac{3^k}{2^k-3^k}$$
 A: Let $S=\sum_{k=1}^{\infty}\left(\frac{2}{3}\right)^{k^2}\frac{3^k}{2^k-3^k}$. We calculate:
\begin{eqnarray}
S &=& -\sum_{k=1}^{\infty}\frac{(2/3)^{k^2}}{1-(2/3)^k} \\
&=& -\sum_{k=1}^{\infty}\left((2/3)^{k^2}+(2/3)^{k^2+k}+(2/3)^{k^2+2k}+\cdots\right) \\
&=& -\sum_{i=1}^{\infty} a_i(2/3)^i
\end{eqnarray}
Where $a_i$ is the number of pairs of integers $(k,\ell)$ with $k > 0$, $\ell \geq 0$ such that $i = k^2+k\ell=k(k+\ell)$. Equivalently, letting $m=k$ and $n=k+\ell$, $d_i$ is the number of pairs of integers $(m,n)$ with $m,n > 0$ and $m \leq n$ such that $i = mn$. Therefore, $a_i=\left \lceil{d_i/2}\right \rceil $, where $d_i$ is the number-of-divisors function $d_i = \sum_{k|i}1$. Since $d_i$ is only odd when $i$ is square, we have $a_i = \begin{cases}(d_i+1)/2 & i\text{ square} \\ d_i/2 & i\text{ not square}\end{cases}$. Therefore:
\begin{eqnarray}
S &=& -\frac{1}{2}\left(\sum_{i=1}^{\infty} d_i(2/3)^i + \sum_{i=1}^\infty (2/3)^{i^2}\right)\\
&=& \frac{1}{4}\Big(1-2L(2/3) -\theta_3(0,2/3)\Big)
\end{eqnarray}
Where $L(x)=\sum_{i=1}^\infty d_i x^i=\sum_{i=1}^\infty x^i/(1-x^i)$ is the Lambert Series which is the generating function of the $d_i$, and $\theta_3(0,x)=\sum_{i=-\infty}^\infty x^{i^2}$ is a Jacobi theta function. This may not count as a closed form expression, but it's better than nothing. Finding a closed form would be equivalent to showing that $2L(2/3)+\theta_3(0,2/3)$ can be expressed in closed form, which may or may not be possible.
