# Factorial series $j(D)=\sum_{n=1}^\infty \frac{1}{(n^D)!}$ and hypergeometric functions

For positive integer $$D$$, define $$j(D)=\sum_{n=1}^\infty \frac{1}{(n^D)!}$$.

For $$D \le 6$$, sage finds closed form in terms of hypergeometric functions at algrebraic arguments and fails to find closed form for $$D>6$$.

According to sage $$j(2)= \,_1F_4\left(\begin{matrix} 1 \\ -i + 1,i + 1,-i \, \sqrt{2} + 1,i \, \sqrt{2} + 1 \end{matrix} ; 1 \right)$$

The length of the latex sage's closed form of $$j(6)$$ is 2534 characters long.

Sage's results in machine readable form are available

Wolfram Alpha can't solve $$D>1$$ and chatGPT "believes" there isn't simple closed form for $$j(2)$$.

Q1 Are sage's results correct?

Q2 Is there closed form for $$D>6$$?

• All these results cannot be true since $(n^2)!$ grows too fast to be expressible by a finite product of pochammer symbols, so no hypergeometric evaluation is possible. In addition, since all the series converge extremely fast, you can immediately check that the result for $D=2$ is numerically wrong. Feb 12 at 11:10
• for $d=2$ your hypergeometric function evaluates to 1.17227, while the correct answer is 1.04167 Feb 12 at 11:13
• The bug comes from Maxima's simplify_sum command used by Sage, see groups.google.com/g/sage-devel/c/E-JooEu5QTo/m/uR3RCXP8AQAJ Feb 14 at 19:42