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
$$\int_{0}^\infty e^{-t^2}dt=\dfrac{\sqrt{\pi}}{2}.$$
Asked
Active
Viewed 596 times
1 Answers
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.
Mark Viola
- 179,405
Simply Beautiful Art
- 74,685
-
...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. – Jean Marie Jan 10 '17 at 02:06
-
Never mind, Wolfram was being weird – Simply Beautiful Art Jan 10 '17 at 02:08
-
Thank you. Do you know any lower/upper bounds on the series? – Zir Jan 10 '17 at 03:04
-
@Zir you might find such by taking the Riemann sum approximation of the integral you wrote. – Simply Beautiful Art Jan 10 '17 at 11:56
-
@SimplyBeautifulArt How does one prove that $$\sum_{n=0}^\infty e^{-\pi n^2}=\frac{\pi^{1/4}}{\Gamma(3/4)}\text{?}$$ – Franklin Pezzuti Dyer Jan 06 '18 at 22:00
-
@Nilknarf Paramanand Singh would almost certainly know better than me. – Simply Beautiful Art Jan 07 '18 at 13:20