I'm trying to prove that a function $$f(y) = \sum_{k=-\infty}^{+\infty}{(-1)^ke^{-k^2y}}$$ is $O(y)$ while $y$ tends to $+0$. I have observed that $f(y) = \vartheta(0.5,\frac{iy}{\pi})$ where $\vartheta$ is a Jacobi theta function. It seems that these functions are very well studied but I am not too familiar with this area.
Any useful links or suggestions are very appreciated.
UPD: the previous version contained a square which shouldn't be there.
Actually, your function is even more simply expressed in terms of $\vartheta_4$-function. Also, I prefer this notation in which $$f(y)=\vartheta_4(0,e^{-y})=\vartheta_4\Bigl(0\Bigr|\Bigl.\frac{iy}{\pi}\Bigr).$$ I.e. I use the convention $\vartheta_k(z,q)=\vartheta_k(z|\tau)$.
Then, to obtain the asymptotics as $y\rightarrow 0^+$, we need two things:
Jacobi's imaginary transformation, after which the transformed nome and half-period behave as $q'\rightarrow0$, $\tau'\rightarrow i\infty$ (instead of $q\rightarrow1$, $\tau\rightarrow0$): $$\vartheta_4\Bigl(0\Bigr|\Bigl.\frac{iy}{\pi}\Bigr)=\sqrt{\frac{\pi}{y}}\vartheta_2\Bigl(0\Bigr|\Bigl.\frac{i\pi}{y}\Bigr).$$
Series representations for theta functions (e.g. the formula (8) by the first link), which implies that $$\vartheta_2(0,q')\sim 2(q')^{\frac14}$$ as $q'\rightarrow 0$. Note that you can also obtain an arbitrary number of terms in the asymptotic expansion if you want.
Taking into account the two things above, we obtain that the leading asymptotic term is given by $$f(y\rightarrow0)\sim 2\sqrt{\frac{\pi}{y}} \exp\left\{-\frac{\pi^2}{4y}\right\}.$$
This does not appear to be correct. If $y>0$ then each summand is greater than 1 in absolute value, so there is no change the series will converge. In particular, one cannot claim that it is $O(y)$ in any obvious sense.
