Is it possible to solve this recurrence relation? For any real $0<x\leq1$, let $E(x)=1$.
For any reals $0<a_1,a_2,\ldots,a_n\leq1$ with $a_1+a_2+\cdots+a_n\leq1$,
let $E(a_1,a_2,\ldots,a_n)=1+\displaystyle\sum_{k=1}^n\dfrac{a_k}{1-a_k} E(a_1,a_2,\ldots,a_{k-1},a_{k+1},a_{k+2},\ldots,a_n)$ for $n>1$.
For example, $E(a_1,a_2)=1+\dfrac{a_1}{1-a_1}E(a_2)+\dfrac{a_2}{1-a_2}E(a_1)=1+\dfrac{a_1}{1-a_1}+\dfrac{a_2}{1-a_2}$, 
and $E(a_1,a_2,a_3)=1+\dfrac{a_1}{1-a_1}E(a_2,a_3)+\dfrac{a_2}{1-a_2} E(a_1,a_3)+\dfrac{a_3}{1-a_3}E(a_1,a_2)$ $=1+\dfrac{a_1}{1-a_1} \left(1+\dfrac{a_2}{1-a_2}+\dfrac{a_3}{1-a_3}\right)+\dfrac{a_2}{1-a_2} \left(1+\dfrac{a_1}{1-a_1}+\dfrac{a_3}{1-a_3}\right)+\dfrac{a_3}{1-a_3} \left(1+\dfrac{a_1}{1-a_1}+\dfrac{a_2}{1-a_2}\right)$
Is it possible to find the closed form of $E(a_1,a_2,\ldots,a_n)$ ?
Thanks in advance.
 A: For any integer $n \ge 1$, let $F_n$ be the $n$-ary function defined below.
$$F_n(u_1,u_2,\ldots,u_n)\stackrel{def} = 
\int_0^\infty \left[ \prod_{k=1}^n(1 + tu_k) - t^n \prod_{k=1}^n u_k \right]  e^{-t} dt\tag{*1}$$
In particular $F_2(u,v) = 1 + u + v$.
Integrate the first piece of $(*1)$ by part, we find
$$\begin{align}
\verb/RHS/(*1) 
&= \left[- e^{-t} \prod_{k=1}^n(1+tu_k)\right]_{t=0}^\infty
+ \sum_{k=1}^n u_k \int_0^\infty \left[\prod_{\ell=1,\ne k}^n (1+tu_\ell)\right] e^{-t} dt
- n! \prod_{k=1}^n u_k\\
&= 1 + \sum_{k=1}^n u_k \int_0^\infty 
\left[\prod_{\ell=1,\ne k}^n (1+tu_\ell) - t^{n-1}\prod_{\ell=1,\ne k}^n u_\ell\right] e^{-t}dt\\
&= 1 + \sum_{k=1}^n u_k F_{n-1}(u_1,\ldots,u_{k-1},u_{k+1},\ldots,u_n)
\end{align}
$$
Substitute $u_k$ by $\frac{a_k}{1-a_k}$, we find above relations reduces to 
the one satisfied by $E$. Notice 
$$F_2\left(\frac{a}{1-a},\frac{b}{1-b}\right) = 1 + \frac{a}{1-a} + \frac{b}{1-b} = E(a,b)$$
By induction, we find for any $n \ge 2$.
$$\begin{align}
E(a_1,\ldots,a_n) 
&= F_n\left(\frac{a_1}{1-a_1},\ldots,\frac{a_n}{1-a_n}\right)\\
&= \frac{1}{\prod_{k=1}^n (1-a_k)}\int_0^\infty 
\left[\prod_{k=1}^n(1 + (t-1)a_k) - t^n \prod_{k=1}^n a_k\right] e^{-t} dt
\end{align}
$$
For positive integer $m \le n$, let $e_m$ be the $m^{th}$ elementary polynomial associated with the $n$ numbers $a_1, \ldots a_n$.
$$e_m \stackrel{def}{=} \prod\limits_{1\le j_1 < j_2 < \cdots < j_m \le n} \prod_{i=1}^m a_{j_i}$$
In terms of the elementary polynomials, we can simplify the integrand above
$$\prod_{k=1}^n(1 + (t-1)a_k) - t^n \prod_{k=1}^n a_k
= \left( 1 + \sum_{k=1}^{n} (t-1)^n e_k \right) - t^n e_n$$
As a result, we get
$$E(a_1,\ldots,a_n) = \frac{1}{\prod_{k=1}^n(1-a_k)}\left[ 1 + \sum_{k=1}^{n-1} D_k e_k + (D_n - n!) e_n \right]\tag{*2}$$
where
$$D_n = \int_0^\infty (t-1)^n e^{-t} dt = n!\left(\sum_{s=0}^n\frac{(-1)^s}{s!}\right) = \left\lfloor \frac{n!}{e} + \frac12 \right\rfloor$$
are the $n^{th}$ derangement number.
Please note that given $n$ numbers $a_1, \ldots, a_n$, formula $(*2)$ allow one
to compute $E(a_1,\ldots,a_n)$ in polynomial time. This is because one can compute all the elementary polynomials $e_k$ one need by multiplication of $n$ monomials 
$$1 + \sum_{k=1}^n t^k e_k = \prod_{k=1}^n ( 1 + ta_k)$$
and that take $O(n^2)$ scalar multiplications/additions. For comparison, 
a naive evaluation of $E(a_1,\ldots,a_n)$ using its recursive definition 
need $O(n!)$ operations.
A: Let's denote $\frac{a_k}{1-a_k}=f(k)$ for brevity .
From the look of the recurrence I made the following nice combinatorial interpretation which instantly shows the answer:
We start forming a permutation of $1,2,\ldots ,k$ . At every step we may either choose a new element not already chosen or to say STOP at which the process terminates . We must say a STOP before forming a complete permutation .
I'll denote a STOP simply with an S .
For example for $n=3$ the possible sequences are :
$S$ ; $1,S$ ; $2,S$ ; $3,S$ ; $1,2,S$ ; $2,1,S$ ; $1,3,S$ ; $3,1,S$ ; $2,3,S$ and $3,2,S$ 
Now if one of the sequences is $x_1,\ldots , x_l,S$ then in the sum we will add :
$$f(x_1)f(x_2)\ldots f(x_l)f(S)$$ with the convention that $f(S)=1$ .
It's easy to see that this interpretation gives the same results as the sum (in the recurrence that $1$ represents the STOP and $f(k)$ the number $k$ that we choose if we don't stop .)
Now with this interpretation it's pretty simple to solve the problem :
Let's count how many times does a term $f(x_1)f(x_2)\ldots f(x_l)$ appears in the sum . It should be obvious that this term appears only for the sequences of the form :
$$\pi(1),\pi(2),\ldots,\pi(n),S$$ where $\pi$ is only a permutation of those $l$ numbers . There are $l!$ such permutations so such a term is to be found $l!$ times in the sum .
I don't think this can be simplified more than this . 
To give a complete answer : 
If $A=\left \{a_1,a_2,\ldots,a_n \right \}$ then your expression is :
$$\sum_{I \subset A} \mid I \mid ! \prod_{x \in A} \frac{x}{1-x}$$ 
