Upper limit for a real gaussian finite sum. Prove $\sum_{i=0}^{N} (2\pi\sigma^2)^{-1/2}e^{-(i+0.5)^2/(2\sigma^2)} < 0.5$?

Question

Let

$$ S = \sum_{i=0}^{N} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(i+0.5)^2/(2\sigma^2)} $$

where $N$ is a non-negative integer and $\sigma>0$ real. It seems (from simulations) that $S<0.5$. Any ideas?

Answer 1

Hint: $\int_{-.5}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(t+0.5)^2/(2\sigma^2)} dt = .5$ because this is the integral of a Gaussian density with mean $-.5$ and variance $\sigma^2$. Which means $\int_{0}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(t+0.5)^2/(2\sigma^2)} dt < .5$.

Answer 2

OK this is my solution:

$$ S < \sum_{i=0}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{-(i+0.5)^2/(2\sigma^2)}= \frac{\vartheta_2(q)}{2\sqrt{2\pi}\sigma} $$

where $\vartheta_2$ is the Jacobi Theta Function and $q = e^{-1/(2\sigma^2)}=e^{-\pi K^\prime/K}$. Here, $K(k)$ is the Complete Elliptic Integral of the First Kind, $k$ is the elliptic modulus, and $K^\prime(k)=K(\sqrt{1-k^2})$. Using this, together with $\vartheta_2(q)=\sqrt{2kK/\pi}$, I find

$$ S < \frac{1}{2} \sqrt{\frac{2kK^\prime}{\pi}} $$

The function $f(k)=2kK^\prime/\pi$ is a strictly increasing function of $k$ and maps the interval $k \in (0, 1)$ (corresponding to $\sigma \in (0, \infty)$ since $K^\prime/K$ is a strictly decreasing function of $k$ mapping $(0, 1)$ to $(0, \infty)$) to the interval $f(k) \in (0, 1)$. Thus $f(k)<1$ in this interval so $S<0.5$