12
$\begingroup$

There seem to be rather nice values for low integer values for $x$, as follows:

$$\begin{array}{|c|c|} \hline x & \displaystyle\sum_{n=0}^\infty {1\over (xn)!} \\[1ex] \hline 1 & e \\[2ex] 2 & \cosh(1) \\[2ex] 3 & \displaystyle{e^{3/2} + 2\cos\frac{\sqrt 3}{2} \over 3 \sqrt e} \\[2ex] 4 & \displaystyle\frac12(\cos1+\cosh1)\\ \hline \end{array}$$

(These were verified with WolframAlpha.)

Is there any closed form for this, or some other function that it can be expressed in?

This is just out of interest by the way; I have no dire need for an answer, although one would certainly be appreciated!

$\endgroup$

3 Answers 3

13
$\begingroup$

Let $k=1,2,3,\cdots$ be a fixed integer. One may consider the set of roots $$ R_k=\{\omega\in\mathbb C\mid \omega^k=1\} $$ then, for each integer $n$, one has $$\sum\limits_{\omega \in R_k}\omega^n = \begin{cases} k, & \text{if $n=0$ (mod $k$)} \\ 0, & \text{otherwise} \end{cases}$$ then $$ \sum_{\omega \in R_k} e^{\omega z}=\sum\limits_{\omega \in R_k} \sum\limits_{n\ge0}\frac{\omega^nz^n}{n!}=\sum\limits_{n\ge0}\frac{z^n}{n!}\cdot\sum_{\omega \in R_k}\omega^n=k\cdot\sum\limits_{n\ge0}\frac{z^{kn}}{(kn)!} $$ that is

$$ \sum_{n\ge0}\frac{z^{kn}}{(kn)!}=\frac1k\cdot\sum_{\omega \in R_k} e^{\omega z} $$

from which one deduces the considered cases with $z=1$.

$\endgroup$
3
$\begingroup$

In 1903, the Swedish mathematician Gösta Mittag-Leffler introduced the following special function: $$ E_\alpha(z)=\sum_{n=0}^{\infty}\frac{z^n}{\Gamma(\alpha n+1)}, $$ where $z$ is a complex number, $\Gamma$ is the gamma function and $\alpha \geq 0$. The Mittag-Leffler function is a direct generalization of the exponential function to which it reduces for $\alpha = 1$. You can find more details about the function in this survey article.

We can express your sum in terms of the Mittag-Leffler function as the following. For a positive integer $k$, we have

$$ \sum_{n=0}^\infty \frac{1}{(kn)!} = E_k(1). $$

Note that for a positive integer $n$, we have $\Gamma(n) = (n-1)!$

$\endgroup$
2
$\begingroup$

Assuming $x$ is a positive integer, $$ \sum_{n=0}^\infty\frac1{(xn)!}=(e^{1}+e^{\zeta}+e^{\zeta^2}+\dots+e^{\zeta^{x-1}})/x\tag1 $$ where $\zeta$ is a primitive $x^\text{th}$ root of unity, e.g. $\zeta=\exp(i2\pi/x)$.

To see this, rewrite the sum as $$ \sum_{n=0}^\infty \frac{[\,x\text{ divides n}\,]}{n!}\tag2 $$ where $[P]$ is the Iverson bracket, equal to $1$ if $P$ is true and $0$ otherwise.

I claim that $$ (1+(\zeta^n)+(\zeta^n)^2+\dots+(\zeta^n)^{x-1})/x=[x\text{ divides }n]\tag3 $$ Once this is proved, substituting (3) into (2), and splitting into several sums, we obtain several exponential series and obtain (1).

If $n$ is a multiple of $x$, then $\zeta^n=1$, so (3) holds because $$ 1+(\zeta^n)+(\zeta^n)^2+\dots+(\zeta^n)^{x-1}=1+1+\dots+1=x $$ On the other hand, if $n$ is not a multiple of $x$, then $\zeta^n\neq 1$, so we can use the geometric series formula: $$ 1+(\zeta^n)+(\zeta^n)^2+\dots+(\zeta^n)^{x-1}=\frac{(\zeta^n)^x-1}{\zeta^n-1}=\frac{(\zeta^x)^n-1}{\zeta^n-1}=\frac{1^n-1}{\zeta^n-1}=0 $$ This proves both cases of (3).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .