Compute $S_n=\sum\limits_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}$ It is tagged as an open problem in the book Fractional parts,series and integrals. If this proof is valid , I don't have any idea how to get it published so I posted it here .
$\displaystyle \sum_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!} = \; ?$
I am posting a proof for the closed form of the above series, please let me know if there are flaws,I came by some special cases of the above sum, that is for the case of $2$ & $3$ variables. They are .
$$ \displaystyle \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{ab}{(a+b)!} \;=\;\frac{2}{3}e $$
$$ \displaystyle \sum_{a=1}^\infty \sum_{b=1}^\infty\sum_{c=1}^\infty \frac{abc}{(a+b+c)!} \;=\;\frac{31}{120}e $$
This led me to solve the general version of the sum for any number of variables,So if $S$ is our sum then,
$$\displaystyle \begin{align} S &= \sum_{k=n}^\infty\frac{1}{k!}\;\left( \sum_{a_1+a_2+\cdots+a_n=k}a_1 a_2\cdots a_n\right) \end{align}$$
This was achieved by setting $\sum_{i=1}^n a_i =k$, and what remains to calculate is the inner sum enclosed by brackets.
We start by investigating the lower cases , suppose we have only two variables $a_1,a_2$ with $a_1+a_2=k$ then
$$\displaystyle \sum_{a_1+a_2=k}(a_1 a_2) =\sum_{N=1}^{k-1} N(k-N)=\frac{k(k-1)(k+1)}{3!}=\binom{k+1}{3}$$
Now if we take the case of $3$ variables where $a_1+a_2+a_3=k$ , we can achieve the sum as :
$$ \displaystyle \sum_{a_1+a_2+a_3=k} a_1 a_2 a_3 = \sum_{N=1}^{k-2} N\binom{k+1-N}{3}= \frac{k(k-1)(k+1)(k-2)(k+2)}{5!}$$ 
Similarly for $4$ variables it turns out to be ,
$$ \displaystyle \sum_{a_1+a_2+a_3+a_4=k}a_1 a_2 a_3 a_4 = \frac{k(k-1)(k+1)(k-2)(k+2)(k-3)(k+3)}{7!} $$
I believe for the same reason that that ,
$$ \displaystyle \sum_{a_1+a_2+\cdots+a_n=k}a_1 a_2\cdots a_n = \frac{k}{(2n-1)!}\prod_{m=1}^{n-1}(k-m)(k+m)$$
This is indeed tough to prove by induction , but I guess it can be proved due to the great symmetry and pattern this sequence follows. I haven't tried but will try to update a proof on this asap, but till then it's reasonable to conjecture this.
Lastly we have that ,
$$\displaystyle \begin{align} S &= \sum_{k=n}^\infty \frac{1}{k!} \left(\frac{k}{(2n-1)!}\prod_{m=1}^{n-1}(k-m)(k+m)\right) \\ &= \frac{1}{(2n-1)!}\sum_{k=n}^\infty \frac{1}{k.k!} (k)_n (k)^n \\ &= \frac{1}{(2n-1)!}\sum_{k=n}^\infty \frac{1}{k.k!} \left(\sum_{r=1}^{n}s(n,r)k^r\right) \left(\sum_{t=1}^n {n\brack t}k^t\right) \\ &= \frac{1}{(2n-1)!}\sum_{r,t=1}^n (-1)^{n+r} {n\brack r}{n\brack t}\left(\sum_{k=n}^\infty \frac{k^{r+t-1}}{k!}\right) \end{align}$$ 
Now using Dobinski's Formula we have finally,
$$\displaystyle \sum_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}\\ = \frac{1}{(2n-1)!}\sum_{r=1}^n\sum_{t=1}^n (-1)^{n+r} {n\brack r}{n\brack t} \left[eB_{r+t-1}-\sum_{m=1}^{n-1}\frac{m^{r+t-1}}{m!}\right] $$
where $B_n$ is the n-th Bell Number. 
Edit:
After some investigation it was clear that the constant term in the final closed form which always disappeared whenever you calculate the sum for a specific $n$ and you are left with a rational multiple of $e$ was no magic but some logic. I proved it by induction. 
Firstly, if we separate the answer into two parts and take the constant term which doesn't have $e$ , we get
$$\displaystyle \sum_{r=1}^n \sum_{t=1}^n \sum_{m=1}^{n-1} (-1)^{n-r}{n\brack r}{n\brack t}\frac{m^{r+t-1}}{m!} $$ 
A little modification and interchange of sums will give the result in terms of the Pochammer symbol.
$$ \displaystyle \sum_{m=1}^{n-1} \frac{(m)_n m^{t-1}}{m!} =0$$
This sum is eventually equal to zero and is easy to prove by induction.
Thus the answer is  :
$$\displaystyle \sum_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}\\ = e\left[\frac{1}{(2n-1)!}\sum_{r=1}^n\sum_{t=1}^n (-1)^{n+r} {n\brack r}{n\brack t} B_{r+t-1}\right]$$
 A: I did not check your computation thoroughly, but your idea of resummation followed by the use of Stirling numbers definitely seems to work.
Alternatively, here is another answer in finite sum:
$$ \sum_{a_1, \cdots, a_n \geq 1} \frac{a_1 \cdots a_n}{(a_1 + \cdots + a_n)!}
= \left( \sum_{k=0}^{n}\binom{n}{k} (-1)^{n+k-1} \sum_{j=0}^{n+k-1} \frac{(-1)^j}{j!} \right) e. \tag{1} $$
This formula also tells that the sum is always a rational multiple of $e$.

Proof of (1). We begin by recalling the multivariate beta identity. Let
$$\Delta^{n-1} = \{ (x_1, \cdots, x_n) \in \mathbb{R}^n : x_i \geq 0 \text{ and } x_1 + \cdots + x_n = 1 \}$$
denote the $(n-1)$-simplex. Then
$$ \mathrm{B}(a_1,\cdots,a_n) := \int_{\Delta^{n-1}} \left( \prod_{k=1}^{n} x_k^{a_k-1} \right) \, dx_1 \cdots dx_{n-1} = \frac{\Gamma(a_1)\cdots\Gamma(a_n)}{\Gamma(a_1+\cdots+a_n)}. $$
This is essentially equivalent to the usual beta identity. Now denoting the sum by $S_n$, we have
\begin{align*}
S_n
&= \sum_{a_1,\cdots,a_n \geq 1} \frac{1}{a_1+\cdots+a_n} \left( \prod_{k=1}^{n} \frac{a_k}{\Gamma(a_k)} \right) \mathrm{B}(a_1, \cdots, a_n) \\
&= \sum_{a_1,\cdots,a_n \geq 1} \left( \int_{0}^{1} u^{a_1+\cdots+a_n-1} \, du \right) \left( \prod_{k=1}^{n} \frac{a_k}{\Gamma(a_k)} \right) \left( \int_{\Delta^{n-1}} \left( \prod_{k=1}^{n} x_k^{a_k-1} \right) \, d\mathrm{x}\right) \\
&= \int_{\Delta^{n-1}} \int_{0}^{1} \left( \prod_{k=1}^{n} \sum_{a_k=1}^{\infty} \frac{(u x_k)^{a_k-1} a_k}{(a_k - 1)!} \right) \, u^{n-1} du d\mathrm{x}.
\end{align*}
The inner sum can be easily computed by the formula
$$ \sum_{a=0}^{\infty} \frac{a+1}{a!}z^a = e^z (1+z), $$
and hence we obtain
\begin{align*}
S_n
&= \int_{\Delta^{n-1}} \int_{0}^{1} u^{n-1} e^u (1 + ux_1)\cdots(1 + ux_n) \, du d\mathrm{x} \\
&= \sum_{I \subset \{1,\cdots,n\}} \int_{0}^{1} u^{n+|I|-1} e^u \left( \int_{\Delta^{n-1}} \prod_{i\in I}x_i \, d\mathrm{x} \right) \, du \\
&= \sum_{k=0}^{n} \binom{n}{k} B(\underbrace{2,\cdots,2}_{k\text{ terms}}, \underbrace{1,\cdots,1}_{n-k\text{ terms}}) \int_{0}^{1} u^{n+k-1} e^u \, du.
\end{align*}
Evaluating the last sum gives the expression $\text{(1)}$ as desired.

Addendum. I checked the preface of the book and found the quote:

Each chapter contains a section of difﬁcult problems, motivated by other problems in the book, which are collected in a special section entitled “Open problems”....

That being said, they are truly intended as exercises for readers!
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\sum_{a_{1},a_{2},\cdots,a_{n} = 1}^{\infty}
{a_{1}a_{2}\cdots a_{n} \over \pars{a_{1} + a_{2} + \cdots + a_{n}}!} =
\sum_{a_{1},a_{2},\cdots,a_{n} = 1}^{\infty}a_{1}a_{2}\cdots a_{n}
\sum_{k = 0}^{\infty}{\delta_{k,a_{1} + a_{2} + \cdots + a_{n}} \over k!}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{1 \over k!}
\sum_{a_{1},a_{2},\cdots,a_{n} = 1}^{\infty}a_{1}a_{2}\cdots a_{n}
\oint_{\large\verts{z} = 1^{-}}
{1 \over z^{k + 1 - a_{1} - a_{2} + \cdots - a_{n}}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{1 \over k!}\oint_{\large\verts{z} = 1^{-}}
{1 \over z^{k + 1}}\pars{\sum_{a = 1}^{\infty}a\,z^{a}}^{n}
\,{\dd z \over 2\pi\ic} =
\sum_{k = 0}^{\infty}{1 \over k!}\oint_{\large\verts{z} = 1^{-}}
{1 \over z^{k + 1}}\bracks{z \over \pars{1 - z}^{2}}^{n}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{1 \over k!}\oint_{\large\verts{z} = 1^{-}}
{\pars{1 - z}^{-2n} \over z^{k + 1 - n}}\,{\dd z \over 2\pi\ic} =
\sum_{k = n}^{\infty}{1 \over k!}{-2n \choose k - n}\pars{-1}^{k - n} =
\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over \pars{k + n}!}{-2n \choose k}
\\[5mm] = &\
\sum_{k = 0}^{\infty}{1 \over \pars{k + n}!}
{2n + k - 1 \choose k} =
\sum_{k = 0}^{\infty}{1 \over \pars{k + n}!}
{2n + k - 1 \choose 2n - 1}
\\[5mm] = &\
{1 \over \pars{2n - 1}!}
\sum_{k = 0}^{\infty}{\pars{2n + k - 1}! \over \pars{k + n}!\,k!}
\quad\bbox[#ffe,10px,border:1px dotted navy]{so\,\,\, far\,!!!}
\end{align}
