Closed form of the sum $\sum\limits_{n=0}^\infty \exp(-n^3)$

Question

I am trying to calculate the sum of the series

$$\sum_{n=0}^\infty \exp(-n^3)$$

Can it be expressed in terms of known mathematical functions?

Answer 1

No, such expression is not known in the world of commonly used special functions. Its closest analog is the sum $\sum\nolimits_{n=0}^{\infty}e^{-n^2}$, which is already quite nontrivial: it is expressed in terms of elliptic theta functions.

One way to convince yourself that the sum is "too exotic" is to consider a generalization $$f(z)=\sum\limits_{n=0}^{\infty}z^{n^3}.$$ This series converges inside the unit circle $|z|=1$ but the circle itself is a natural boundary of $f(z)$ - that is, a dense set of singularities across which the function cannot be analytically continued. This is to be contrasted with "normal" special functions which usually have a finite or countable number of isolated singular points in the complex plane.

Answer 2

I guess the answer ("it is hard") have been given.

The numerical computation gives

$\sum_{n=0}^\infty{\mathrm e}^{-n^3}=1.368\dots$

I want to add that a natural first order approximation given by the Euler–Maclaurin formula, is

$$\sum_{n=0}^\infty{\mathrm e}^{-n^a}\approx\frac{1}{2}+\Gamma\left(1+\frac{1}{a}\right)\approx 1.5-\frac{\gamma}{a},$$

where $\gamma=0.577\dots$ is the Euler–Mascheroni constant.

The approximation is best at about $a\approx 4.5$. For you with $a=3$ we find

$\frac{1}{2}+\Gamma\left(1+\frac{1}{3}\right)=1.392\dots$