Evaluation *and transcendence* of $\sum_{m=1}^{\infty}\frac1{(mp)!}$ How can we evaluate the sum $\sum_{m=1}^{\infty}\frac1{(mp)!}$, where $p$ is a fixed prime?
It is a subseries of the exponential series, but the its evaluation is out of my knowledge. Is the sum transcendental, like $e$? Any hints? Thanks beforehand.
 A: Wolfram Alpha seems to be able to do some of these sums, it doesn't look like there's a $-1$.
Following @Mindlack's comment.
Let $\omega$ be a primitive $p$'th root of unity.
$$\sum_{j=0}^{p-1}e^{\omega^j}=\sum_{j=0}^{p-1}\sum_{n=0}^\infty \frac{\omega^{jn}}{n!}=\sum_{n=0}^\infty\sum_{j=0}^{p-1} \frac{\omega^{jn}}{n!}=\sum_{p\not\mid n }^\infty\frac{(\omega^n)^p-1}{\omega^n-1}\cdot\frac{1}{n!}+\sum_{p\mid n }^\infty\cdot\frac{p}{n!}=p\sum_{m=0 }^\infty\frac{1}{(mp)!}$$
where we can use absolute convergence to rearrange the sums.
So that
$$\frac{1}{p}\sum_{j=0}^{p-1}e^{\omega^j}=\sum_{m=0 }^\infty\frac{1}{(mp)!}$$
Edit:
Then
$$\sum_{m=1 }^\infty\frac{1}{(mp)!}=-1+\frac{1}{p}\sum_{j=0}^{p-1}e^{\omega^j}$$
I don't know much of the details in showing it's transcendental. However again as @Mindlack suggests a form of the Lindemann–Weierstrass Theorem (Baker's reformulation) cited for convenience from Wikipedia  — If $a_1, ..., a_n$ are algebraic numbers, and $\alpha_1, ...,\alpha_n$ are distinct algebraic numbers, then
$$a_{1}e^{\alpha _{1}}+a_{2}e^{\alpha _{2}}+\cdots +a_{n}e^{\alpha _{n}}=0$$
has only the trivial solution $a_i=0, \forall i\in \{1,\ldots, n\}$.
A: For any positive integer $m$, prime or not, we have
$$
\begin{align}
\frac1m\sum_{k=0}^{m-1}e^{e^{2\pi ik/m}}-1
&=\frac1m\sum_{k=0}^{m-1}\sum_{j=0}^\infty\frac{e^{2\pi ijk/m}}{j!}-1\tag1\\
&=\sum_{j=0}^\infty\frac1{j!}\color{#C00}{\frac1m\sum_{k=0}^{m-1}e^{2\pi ijk/m}}-1\tag2\\
&=\sum_{j=0}^\infty\frac{\color{#C00}{[m\mid j]}}{j!}-1\tag3\\
&=\sum_{j=1}^\infty\frac1{(mj)!}\tag4
\end{align}
$$
Explanation:
$(1)$: use the Taylor series for $e^x$
$(2)$: change the order of summation
$(3)$: $\frac1m\sum_{k=0}^{m-1}e^{2\pi ijk/m}=[m\mid j]$
$(4)$: keep only the terms where $m\mid j$
Since each $e^{2\pi ik/m}$ is algebraic, Lindemann-Weierstrass says the left side of $(1)$ is transcendental.
The imaginary part of the left side of $(1)$ is
$$
\frac1m\sum_{k=0}^{m-1}e^{\cos(2\pi k/m)}\sin(\sin(2\pi k/m))=0\tag5
$$
because the $k=0$ term is $0$ and the $k$ and $m-k$ terms cancel. This is good, since we expect the sum to be real.
Thus, $(4)$ and $(5)$ say that
$$
\sum_{j=1}^\infty\frac1{(mj)!}
=\frac1m\sum_{k=0}^{m-1}e^{\cos(2\pi k/m)}\cos(\sin(2\pi k/m))-1\tag6
$$
Note that for $m\ge3$, at least graphically, the log of the sum in $(6)$ and $-\log(m!)$ are indistinguishable:

