I'm interested in approximating the infinite sum $$ \sum_{i=1}^\infty Z\left(\frac{\alpha i\pm1}{\beta}\right) $$
where $\alpha,\beta$ are constant and $$ Z(a\pm b)=\frac{1}{2\pi}\int_{a-b}^{a+b}e^{-x^2/2}dx=\frac{\operatorname{erf}(a+b)-\operatorname{erf}(a-b)}{2} $$ is the standard normal distribution. Any useful asymptotics?
I am not sure this is much helpful, just some thoughts.
Let us denote $a=\alpha/\beta$, $b=1/\beta$, then differentiating the sum $$ Z(a,b)=\sum_{k=1}^{\infty}Z(ak\pm b)$$ with respect to $b$, we find $$\frac{\partial}{\partial{b}}Z(a,b)=\frac{1}{2\pi}\sum_{k=1}^{\infty}\left(e^{-(ak+b)^2/2}+e^{-(ak-b)^2/2}\right)=\frac{e^{-b^2/2}}{2\pi}\sum_{k=1}^{\infty}e^{-a^2k^2/2}\cosh (abk).$$ This can be expressed in terms of Jacobi theta functions, since for example $$\vartheta_3(z|\tau)-1=2\sum_{k=1}^{\infty}e^{i\pi\tau k^2}\cos 2kz.$$ So in this case the half-period ratio is pure imaginary, $\tau=\frac{ia^2}{2\pi}$, and $z=\frac{iab}{2}$. Integrating back with respect to $b$, one then obtains $$Z(a,b)=\frac{1}{4\pi}\int_0^be^{-b^2/2}\left[\vartheta_3\Bigl(\frac{iab}{2}\Bigl|\Bigr.\frac{ia^2}{2\pi}\Bigr)-1\right]db.$$ I haven't double-checked the calculations, and didn't verify that we can indeed exchange summation and integration, so this is just a formal result.