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Let $p$ be prime, and let $0\le j\le p-1$. I would like to find a closed expression for the infinite sum $$f(p,j)=\sum_{i=0}^{\infty}\frac{p^i}{(pi+j)!}$$ Initial computations show that $f(2,0)=\cosh\sqrt2-1$ and $f(2,1)=\frac{\sinh\sqrt 2}{\sqrt 2}-1$. Any ideas?

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up vote 1 down vote accepted

Recallin the Mittag-Leffler function

$$E_{\alpha,\beta}(z)= \sum_{k=0}^{\infty} \frac{z^k}{\Gamma(\alpha k+\beta)}. $$

Your series will be readily expressed in terms of the above function as

$$ f(p,j)=\sum_{i=0}^{\infty}\frac{p^i}{(pi+j)!}= \sum_{i=0}^{\infty}\frac{p^i}{\Gamma(pi+j+1)} = E_{p,j+1}(p). $$.

Note that, $n!=\Gamma(n+1)$.

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Making this connection is very helpful. Thank you. Is anything known about the special value $E_{p,j+1}(p)$? Or do we know how they multiply, i.e., do we know the value of $E_{p,j+1}(p)\cdot E_{q,k+1}(q)$ in terms of other values of Mittag-Leffler functions? – Jared Mar 28 '13 at 21:49
@Jared: In general, multiplying two Mittag-Leffler functions does not belong to the same class of functions. – Mhenni Benghorbal Mar 28 '13 at 22:00

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