5

Is there a special function for the following series?

$$\sum_{m=0}^{\infty} {x^m \over (m!)^s}$$

Here, $s$ is a positive real number.

When, $s$ is an integer, $s=n \in \mathbb{Z}$, this series can be written in terms of the generalized hypergeometric function:

$$ \sum_{m=0}^{\infty} {x^m \over (m!)^n} = {}_0F_{n-1}(1,\cdots,1|x)~.$$

What is the analogue of the above for more general values of $s$, e.g. $s=1/2$?

TheTwistedSector
  • 153
  • 3
    even for $x=1$ the sum $\sum_{m=0}^\infty1/\sqrt{m!}=3.46950631\cdots$ does not seem to be any known special function. – Carlo Beenakker Oct 07 '21 at 21:41
  • 3
    Even for integral $n\ge3$, the expression ${}0F{n-1}(1,\ldots,1|x)$ is just tautological and thus hardly of any use. – Iosif Pinelis Oct 07 '21 at 22:01
  • 1
    @IosifPinelis Fair enough. The point of my question is whether an analytic continuation beyond integers would make sense. – TheTwistedSector Oct 07 '21 at 22:08
  • 5
    Check out the ancient question linked below and its related questions for some interesting observations: https://mathoverflow.net/questions/84958/is-sum-limits-n-0-infty-xn-sqrtn-positive – Suvrit Oct 07 '21 at 23:53
  • 1
    @Suvrit Thanks! Browsing through the linked questions, I found this https://mathoverflow.net/questions/353251/integral-expressions-for-bessel-like-power-series?noredirect=1&lq=1 The second answer there is interesting. For large $x$, there is an asymptotic expression for the above sum for $0\leq m \leq 4$. – TheTwistedSector Oct 08 '21 at 11:49

0 Answers0