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Does a closed form exist for

$$\sum \limits_{n=0}^{\infty} \frac{1}{(kn)!}$$

in terms of $k$ and other functions? The best that I have been able to do is solve the case where $k=1$, since the sum is just the infinite series for $e$. I would guess that any closed form must involve the exponential function, but am at a loss to prove it.

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    $\begingroup$ For small $k$ there are fairly easy to find closed forms involving functions like cos, cosh. In the general case the closed form is a messy expression involving the generalised hypergeometric function. $\endgroup$ – almagest Mar 22 '16 at 16:24
  • $\begingroup$ sum diverges at $k=0$ and conveges to $\cos (ix)$ for $k=2$ $\endgroup$ – gt6989b Mar 22 '16 at 16:26
  • $\begingroup$ @gt6989b you could also write $\cosh(1)$ for $k=2$ $\endgroup$ – zz20s Mar 22 '16 at 16:27
  • $\begingroup$ If you could find a general function that has the property $f(x)=f^{(k)}(x)$, and $f^{(k)}(a)=1$ for all $k$ and fixed $a$, then you could make this the Taylor series of that? $\endgroup$ – Simply Beautiful Art Mar 23 '16 at 12:00
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    $\begingroup$ Possible duplicate of Closed-forms of infinite series with factorial in the denominator $\endgroup$ – Yanior Weg Nov 27 at 19:16
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If $(c_n)$ is any sequence with period $k$ (that is, $c_{n+k}=c_n$) then it's possible to evaluate $\sum c_n/n!$ using tricks involving $k$-th roots of unity.

Let $\omega=e^{2\pi i/k}$. Consider the $k$ sequences

$s_0:1,1,1\dots$

$s_1: 1, \omega,\omega^2,\omega^3,\dots$

$s_2: 1, \omega^2,\omega^4,\omega^6,\dots$

$s_3: 1, \omega^3,\omega^6,\omega^9,\dots$

...

$s_{k-1}: 1, \omega^{k-1},\omega^{2(k-1)},\dots$.

Using tricks analogous to finding Fourier coefficients you can find $a_0,\dots a_{k-1}$ so that $$(c_n)=a_0(s_0)+\dots+a_k(s_{k-1}).$$

Hence $$\sum\frac{c_n}{n!}=\sum_{j=0}^{k-1}a_j\sum_n\frac{\omega^{jn}}{n!} =\sum_{j=0}^{k-1}a_je^{\omega^j}.$$

If you do that for your sequence you get $$\sum_{n=0}^\infty\frac{1}{(kn)!}=\frac1k\sum_{j=0}^{k-1}e^{\omega^j}.$$

Edit: Thomas Andrews makes a comment that I should have included: Since the original sum is real, it follows that $$\sum_{n=0}^\infty\frac{1}{(kn)!}=\frac{1}{k}\sum_{j=0}^{k-1}e^{\cos 2\pi j/k}\cos(\sin2\pi j/k).$$

Edit: If one is familiar with "abstract harmonic analysis", here just harmonic analysis on compact abelian groups, one sees that those "tricks analogous to finding Fourier coefficients" are in fact finding Fourier coefficients for a certain function on the group $\Bbb Z/k\Bbb Z$.

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    $\begingroup$ And, since the imaginary part is necessarily zero, this also gives the formula:$$\frac{1}{k}\sum_{j=0}^{k-1}e^{\cos 2\pi j/k}\cos(\sin2\pi j/k)$$ $\endgroup$ – Thomas Andrews Mar 22 '16 at 16:56
  • $\begingroup$ @ThomasAndrews Indeed, I shoulda said that, thanks. $\endgroup$ – David C. Ullrich Mar 22 '16 at 17:03
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    $\begingroup$ Interestingly, this is a Riemann sum for $\int_{0}^{1} e^{\cos 2\pi x}\cos(\sin 2\pi x)\,dx$. $\endgroup$ – Thomas Andrews Mar 22 '16 at 17:49
  • $\begingroup$ @ThomasAndrews Yes, that's interesting. As is the fact that it follows that that integral equals $1$. $\endgroup$ – David C. Ullrich Mar 22 '16 at 17:56
  • $\begingroup$ Yeah, I was just trying to see if I could prove that directly. $\endgroup$ – Thomas Andrews Mar 22 '16 at 17:56
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This is a different approach to the idea in David Ullrich's answer.


As long as $\frac nk\not\in\mathbb{Z}$, $$ \begin{align} \sum_{j=0}^{k-1}e^{2\pi ij\frac nk} &=\frac{e^{2\pi in}-1}{e^{2\pi i\frac nk}-1}\\ &=0 \end{align} $$ if $\frac nk\in\mathbb{Z}$, then $$ \begin{align} \sum_{j=0}^{k-1}e^{2\pi ij\frac nk} &=\sum_{j=0}^{k-1}1\\ &=k \end{align} $$ Thus, $$ \begin{align} \sum_{n=0}^\infty\frac1{(kn)!} &=\sum_{n=0}^\infty\overbrace{\left(\frac1k\sum_{j=0}^{k-1}e^{2\pi ij\frac nk}\right)}^{1\iff\frac nk\in\mathbb{Z}}\frac1{n!}\\ &=\frac1k\sum_{j=0}^{k-1}\sum_{n=0}^\infty\frac{\left(e^{2\pi ij/k}\right)^n}{n!}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\large e^{2\pi ij/k}}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)+i\sin(2\pi j/k)}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)}\left(\cos(\sin(2\pi j/k))+i\sin(\sin(2\pi j/k))\right)\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)}\cos(\sin(2\pi j/k)) \end{align} $$ The last step follows since the imaginary parts of the $j$ and $k-j$ terms cancel.

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  • $\begingroup$ Well, they cancel because the value is obviously real :) $\endgroup$ – Thomas Andrews Mar 22 '16 at 17:22
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    $\begingroup$ @ThomasAndrews: indeed, but since we have introduced complex numbers, we should show that the computed answer is actually real. $\endgroup$ – robjohn Mar 22 '16 at 17:26
  • $\begingroup$ I can't understand how did you write the first equality after "Thus,". Could you explain this a bit for me, please? $\endgroup$ – user486600 Aug 18 '18 at 6:49
  • $\begingroup$ @user486600: for a proof that the quantity in parentheses is $[\,k\mid n\,]$ see the Preliminaries of this answer. $\endgroup$ – robjohn Aug 18 '18 at 11:56
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Sum expressed by a special function:

$$\color{red}{\sum _{n=0}^{\infty } \frac{1}{(k n)!}}=\sum _{n=0}^{\infty } \frac{1}{\Gamma (k n+1)}=\color{red}{E_{k,1}(1)}$$

where: $\color{red}{E_{k,1}(1)}$ is generalized Mittag-Lefflere function.

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