Convergence to $0$ of Jacobi theta function

UPD: the previous version contained a square which shouldn't be there.

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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 $$f(y\rightarrow0)\sim 2\sqrt{\frac{\pi}{y}} \exp\left\{-\frac{\pi^2}{4y}\right\}.$$