I would give a slightly more direct explanation than WhatsUp. The Poisson summation formula states that for $f$ a smooth rapidly decreasing function on $\mathbb R$, $\hat{f}$ the Fourier transform, and $y>0$ that

$$ y\sum_{n \in \mathbb Z} f(yn) = \sum_{n \in \mathbb Z} \hat{f} (y^{-1} n)$$

where $\hat{f} = \int_{-\infty}^\infty f(t)dt$ is the integral to which $y\sum_{n \in \mathbb Z} f(yn)$ is a Riemann sum. Hence $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is the error in the Riemann sum approximation to this integral.

For $y$ small, $y^{-1}$ is large, so $y^{-1} n$ is large for all nonzero $n$. Thus the more rapidly decreasing $\hat{f}$ is, the smaller the error term $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is, and the more rapidly this Riemann sum converges.

The Fourier transform of $e^{-x^2}$ is another function of the form $C e^{ - B x^2}$ and thus is pretty rapidly decreasing.

The Poisson summation formula is exactly how the functional equation of the theta function is proved, showing the relation to WhatsUp's answer.

I would give a slightly more direct explanation than WhatsUp. The Poisson summation formula states that for $f$ a smooth rapidly decreasing function on $\mathbb R$, $\hat{f}$ the Fourier transform, and $y>0$ that

$$ y\sum_{n \in \mathbb Z} f(yn) = \sum_{n \in \mathbb Z} \hat{f} (y^{-1} n)$$

where $\hat{f}(0) = \int_{-\infty}^\infty f(t)dt$ is the integral to which $y\sum_{n \in \mathbb Z} f(yn)$ is a Riemann sum. Hence $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is the error in the Riemann sum approximation to this integral.

For $y$ small, $y^{-1}$ is large, so $y^{-1} n$ is large for all nonzero $n$. Thus the more rapidly decreasing $\hat{f}$ is, the smaller the error term $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is, and the more rapidly this Riemann sum converges.

The Fourier transform of $e^{-x^2}$ is another function of the form $C e^{ - B x^2}$ and thus is pretty rapidly decreasing.

The Poisson summation formula is exactly how the functional equation of the theta function is proved, showing the relation to WhatsUp's answer.

I would give a slightly more direct explanation than WhatsUp. The Poisson summation formula states that for $f$ a smooth rapidly decreasing function on $\mathbb R$, $\hat{f}$ the Fourier transform, and $y>0$ that

$$ y\sum_{n \in \mathbb Z} f(yn) = \sum_{n \in \mathbb Z} \hat{f} (y^{-1} n)$$

where $\hat{f} = \int_{-\infty}^\infty f(t)dt$ is the integral to which $y\sum_{n \in \mathbb Z} f(yn)$ is a Riemann sum. Hence $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is the error in the Riemann sum approximation to this integral.

For $y$ small, $y^{-1}$ is large, so $y^{-1} n$ is large for all nonzero $n$. Thus the more rapidly decreasing $\hat{f}$ is, teh smaller the error term $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is, and the more rapidly this Riemann sum converges.

The Fourier transform of $e^{-x^2}$ is another function of the form $C e^{ - B x^2}$ and thus is pretty rapidly decreasing.

The Poisson summation formula is exactly how the functional equation of the theta function is proved, showing the relation to WhatsUp's answer.

I would give a slightly more direct explanation than WhatsUp. The Poisson summation formula states that for $f$ a smooth rapidly decreasing function on $\mathbb R$, $\hat{f}$ the Fourier transform, and $y>0$ that

$$ y\sum_{n \in \mathbb Z} f(yn) = \sum_{n \in \mathbb Z} \hat{f} (y^{-1} n)$$

where $\hat{f} = \int_{-\infty}^\infty f(t)dt$ is the integral to which $y\sum_{n \in \mathbb Z} f(yn)$ is a Riemann sum. Hence $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is the error in the Riemann sum approximation to this integral.

For $y$ small, $y^{-1}$ is large, so $y^{-1} n$ is large for all nonzero $n$. Thus the more rapidly decreasing $\hat{f}$ is, the smaller the error term $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is, and the more rapidly this Riemann sum converges.

The Fourier transform of $e^{-x^2}$ is another function of the form $C e^{ - B x^2}$ and thus is pretty rapidly decreasing.

The Poisson summation formula is exactly how the functional equation of the theta function is proved, showing the relation to WhatsUp's answer.