TITLE: Prove that $\sum_{n=0}^\infty\frac{x^{2n+1}}{1+x^{4n+2}}\neq0$ for complex $x$
QUESTION [5 upvotes]: For $x\in\mathbb C$ with $0<|x|<1$, does this function have any roots?
$$f(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{1+x^{4n+2}}$$
$$=\sum_{n=0}^\infty x^{2n+1}\sum_{m=0}^\infty(-x^{4n+2})^m$$
$$=\sum_{n=0}^\infty\sum_{m=0}^\infty(-1)^mx^{(2n+1)(2m+1)}$$
Finding an infinite product representation would help. I don't remember where I found this, but here's another form of the function:
$$f(x)\overset?=\sum_{n=0}^\infty\sum_{m=0}^\infty x^{((4n+1)^2+(4m+1)^2)/2}$$

REPLY [3 votes]: Just to mention that we are dealing with modular forms.
Using that $\frac{\pi}{\cosh(\pi t)}$ is its own Fourier transform and the Poisson summation formula we get that $$f(z)=\sum_{n=0}^\infty\frac{e^{2i\pi (2n+1) z}}{1+e^{2i\pi (4n+2) z}}$$ is a weight $1$ modular form for $\Gamma_1(8)$, 
then $$F(z)=\prod_{\gamma \in \Gamma_1(8)\setminus SL_2(Z)} f(\gamma(z))(cz+d)^{-1}$$ is a weight $[SL_2(Z): \Gamma_1(8)]=48$ cusp form for $SL_2(Z)$. We find that $F(z)=O(e^{8i\pi z})$ at $i\infty$, thus $F(z)/\Delta(z)^4$ is a weight $0$ modular form, it has to be constant. 
$\Delta(z)$ is given by an infinite product, thus $F$ doesn't vanish, $f$ doesn't vanish.