6
$\begingroup$

I am curious about the value of

$$\sum_{n=0}^\infty e^{-n^2}.$$

It comes to my mind by observing Gauss integral that it is equal to

$$\int_{0}^\infty e^{-t^2}dt=\dfrac{\sqrt{\pi}}{2}.$$

1 Answers 1

8

You've actually stumbled upon an Jacobi theta function:

$$\vartheta_3(z,q)=\sum_{n=-\infty}^{+\infty}q^{n^2}e^{2\pi iz}$$

With $z=0$ and $q=e^{-1}$ and taking advantage that the sum is symmetric, we end up with

$$\frac12\left(\vartheta_3(0,e^{-1})+1\right)=\sum_{n=0}^\infty e^{-n^2}$$

Though if I may interest you a bit, there are some closed forms for similar series:

$$\frac{\pi^{1/4}}{\Gamma(3/4)}=\frac12\left(\vartheta_3(0,e^{-\pi})+1\right)=\sum_{n=0}^\infty e^{-\pi n^2}$$

Such identities begin at $(45)$ on the given link.

  • 0
    ...which can as well be written $\psi(1/\pi)+1$, where $\psi$ is a function that satisfies functional equation $\dfrac{1+2 \psi(x)}{1+2 \psi(1/x)}=\dfrac{1}{\sqrt{x}}$ on the same link. – 2017-01-10
  • 0
    Never mind, Wolfram was being weird – 2017-01-10
  • 0
    Thank you. Do you know any lower/upper bounds on the series? – 2017-01-10
  • 0
    @Zir you might find such by taking the Riemann sum approximation of the integral you wrote. – 2017-01-10
  • 0
    @SimplyBeautifulArt How does one prove that $$\sum_{n=0}^\infty e^{-\pi n^2}=\frac{\pi^{1/4}}{\Gamma(3/4)}\text{?}$$ – 2018-01-06
  • 0
    @Nilknarf [Paramanand Singh](https://math.stackexchange.com/a/1798992) would almost certainly know better than me. – 2018-01-07