9

I was reading this old question and fascinated by the second infinite sum $$\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n!}}.$$ This is clearly convergent (by comparison or ratio test) and, we can obtain some crude approximations of this using inequalities like $n^4\le n!\le (n!)^2,$ here the first inequality holds for all $n\ge7.$ But, I wonder if we can find the exact value of this series using (familiar) special functions/constants. How would you attack to this series?

Bumblebee
  • 18,220
  • 5
  • 47
  • 87
  • This is A248761 in OEIS, but no useful information is given. – Jair Taylor Apr 21 '20 at 19:26
  • 2
    The vast majority of convergent real series don't have 'nice' closed-form expressions. I've been wrong before, but this looks like it falls into that group. – Integrand Apr 21 '20 at 19:47
  • 2
    Just throwing something out: denote $T(k)=\sum\limits_{n\ge1}(n!)^{-1/2k}$. Then the desired value is $T(1)$ which satisfies $$T(1)=\lim_{x\to0}\frac x{2\pi i}\mathcal M^{-1}\left(x^{-s}T(s)\sin\frac{\pi s}2\Gamma(s-1)\right)$$ where $\mathcal M^{-1}$ is the inverse Mellin transform. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Apr 21 '20 at 20:04
  • 1
    Also an integral form: $$\sum_{n\ge1}\frac1{\sqrt{n!}}=\frac12\int_1^\infty\frac{\lfloor x\rfloor\psi^{(0)}(x+1)}{\sqrt{\Gamma(x+1)}},dx=1+\int_1^\infty\frac{dx}{\sqrt{\Gamma(x+1)}}-\frac12\int_1^\infty\frac{{x}\psi^{(0)}(x+1)}{\sqrt{\Gamma(x+1)}},dx$$ where $\lfloor\cdot\rfloor$ is the floor function, ${\cdot}$ denotes the fractional part and $\psi^{(0)}$ is the digamma function. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Apr 23 '20 at 10:34

0 Answers0