5
$\begingroup$

Has anyone studied a function of the form $$\eta(s) = \sum_{n=1}^{\infty} \frac{1}{\Gamma(n)^{s}} = \sum_{n=0}^{\infty}\frac{1}{k!^s}$$ This series is appearing in my research on the volumetric properties of infinite families of polytopes and I am not sure how to evaluate the sum, or if this is a well-known function other than the fact that $\eta(2) = I_{0}(2)$, where $I_{\nu}(x)$ is the modified Bessel function of the first kind as seen here. I am interested in evaluating this series for any $s \in (1,\infty)$. For integer values of $s$ is it equal to a generalized hypergeometric function?

EDIT: It would be sufficient for my purposes to prove that $\eta(s)$ is finite for any $s \in (0,\infty)$, any ideas?

$\endgroup$
  • 5
    $\begingroup$ You might as well write the sum as $\sum_{k=0}^\infty 1/k!^s$. I doubt that there's any kind of closed form of this, even allowing hypergeometric fuctions and the like, for $s \notin {\bf Z}$. Last year there was some MO discussion of the power series $\sum_{k=0}^\infty x^k / k!^s$ for $0<s<1$ and real but negative $x$ (it turns out that it's always positive, see the answers at mathoverflow.net/questions/84958 and mathoverflow.net/questions/85013 ). $\endgroup$ – Noam D. Elkies Jun 14 '13 at 2:35
  • 3
    $\begingroup$ $\eta(s) < \infty$ for $s>0$ is clear by comparison with a geometric series (e.g. show that $k! \geq 2^{k-1}$ for all $k$). $\endgroup$ – Noam D. Elkies Jun 15 '13 at 4:32
  • $\begingroup$ Thanks! I can't believe I didn't see that, too much geometry dulls the mind I suppose. $\endgroup$ – Samuel Reid Jun 15 '13 at 5:10
  • $\begingroup$ I was always interesting to connect this function with some known say just for $s=1/2$. Its convolution square equals exp, if consider $\sum \frac{x^k}{(\sqrt{k!})}$. May somebody know something nontrivial for this function??? $\endgroup$ – Sergei Sep 29 '14 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.