Closed form for the series: $\sum_{n=1}^{\infty}x^{n^2}$

You are not going to get an elementary closed form for this sum. It is a linear transformation of the third Jacobi theta function: $$\sum_{n=1}^\infty x^{n^2}=\frac{\vartheta_3(0,x)-1}2$$ (I an using the notation used by Mathematica and mpmath. The theta functions have tons of different notations.)

There is no known closed form of $$ S(q):=\sum_{n=1}^{\infty}q^{n^2}, \qquad|q|<1. $$ in terms of elementary functions. A related function, called the Jacobi theta function, $$ \begin{align} \vartheta(z; \tau) &= \sum_{n=-\infty}^\infty \exp \left(\pi i n^2 \tau + 2 \pi i n z\right) \\ &= 1 + 2 \sum_{n=1}^\infty \left(e^{\pi i\tau}\right)^{n^2} \cos(2\pi n z) \\ &= \sum_{n=-\infty}^\infty q^{n^2}\eta^n \end{align} $$ has been studied showing very interesting properties. By setting $q=e^{-x \pi}$ and \begin{align} \varphi(e^{-\pi x}) = \vartheta(0; ix) = \sum_{n=-\infty}^\infty e^{-x \pi n^2}=2S(q)+1 \end{align} one may prove that \begin{align} \varphi\left(e^{-\pi} \right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)}, \\\\ \varphi\left(e^{-2\pi}\right) &= \frac{\sqrt[4]{6\pi+4\sqrt2\pi}}{2\Gamma\left(\frac34\right)}, \\\\ \varphi\left(e^{-3\pi}\right) &= \frac{\sqrt[4]{27\pi+18\sqrt3\pi}}{3\Gamma\left(\frac34\right)}, \\\\ \varphi\left(e^{-4\pi}\right) &= \frac{\sqrt[4]{8\pi}+2\sqrt[4]{\pi}}{4\Gamma\left(\frac34\right)}, \\\\ \varphi\left(e^{-5\pi}\right) &= \frac{\sqrt[4]{225\pi+ 100\sqrt5 \pi}}{5\Gamma\left(\frac34\right)}, \\\\ \varphi\left(e^{-6\pi}\right) &= \frac{\sqrt[3]{3\sqrt{2}+3\sqrt[4]{3}+2\sqrt{3}-\sqrt[4]{27}+\sqrt[4]{1728}-4}\cdot\sqrt[8]{243{\pi}^2}}{6\sqrt[6]{1+\sqrt6-\sqrt2-\sqrt3}{\Gamma\left(\frac34\right)}}. \end{align}