# 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 :

$$\displaystyle \sum_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}\\ = e\left[\frac{1}{(2n-1)!}\sum_{r=1}^n\sum_{t=1}^n (-1)^{n+r} {n\brack r}{n\brack t} B_{r+t-1}\right]$$

• Is it an accident that $n=2,3$ give zero constant term, or is there further simplification that can be done? What is the result when $n=4$? – Thomas Andrews Jul 3 '17 at 6:15
• @ThomasAndrews , It doesn't seem to be an accident. the case of 4 variables give $\frac{179}{2520}e$ . It turns out that the constant terms get cancelled magically – Aditya Narayan Sharma Jul 3 '17 at 6:19

## 2 Answers

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!

• That's great ! I think our expressions for the coefficients must be equal then. – Aditya Narayan Sharma Jul 3 '17 at 7:56
• The section 3.7 which is termed as 'Open Problems' , under which in 3.137 the author conjectures this open problem to be a rational multiple of $e$ . Since it's under the section open problems and is one of those few open problems stated there, then it should be an open problem as much as I think. Isn't it ? – Aditya Narayan Sharma Jul 3 '17 at 8:25
• @AdityaNarayanSharma Except that, as Sangchul tried to delicately explain, this is not an open problem in the usual sense of the term. – Did Jul 3 '17 at 8:49
• @AdityaNarayanSharma, Although it seems quite unlikely, it might be the case that the proposed series is studied by no none so far. But even in that case, a problem for which two guys carried out a simple solution within a couple of hours does not seem to qualify as an 'open problem' in my personal opinion. – Sangchul Lee Jul 3 '17 at 8:50
• @AdityaNarayanSharma Even more disqualifying than the time needed to solve it is, with no offense to Sangchul, the fact that one of the first ideas that comes to mind to approach it, works. – Did Jul 3 '17 at 9:02

$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ \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}

• Nice to see a contour integral approach, the last sum is indeed equal to $e$ times the rest of the answer. – Aditya Narayan Sharma Jul 4 '17 at 4:39