Whoa! That's Insane!

First we simplify $\displaystyle A_n$ :

The limits of the integrals tell us that $\displaystyle 0<\prod_{i=1}^{n} a_i <1$ since $\displaystyle 0<a_i<1$ for $\displaystyle i=1,2,3 \cdots n$

This reminds us of the convergent geometric series : $\sum_{m=0}^{\infty} x^m = \frac{1}{1-x}$ which is equivalent to $\displaystyle \sum_{m=1}^{\infty} x^m = \frac{x}{1-x}$

putting $\displaystyle x=\prod_{i=1}^{n} a_i$ we get:

$\displaystyle \frac{\prod_{i=1}^{n} a_i}{1-\prod_{i=1}^{n} a_i}= \sum_{m=1}^{\infty} (\prod_{i=1}^{n} a_i) ^m.$

$\displaystyle => A_n = \int_{0}^{1} \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \cdots \int_{0}^{1} \sum_{m=1}^{\infty} (\prod_{i=1}^{n} a_i) ^m da_1da_2da_3 \cdots da_n$

Since our limits of integration are independent of m, we can take out the summation notation:

$\displaystyle A_n =\sum_{m=1}^{\infty} \int_{0}^{1} \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \cdots \int_{0}^{1} (\prod_{i=1}^{n} a_i) ^m da_1da_2da_3 \cdots da_n$

Now it is a well known fact that if the variables of a multiple integral are seperated, then the whole multiple integral is the product of the integrals with respect to each variable. In other words :

$\displaystyle \int \int \int f(x)f(y)f(z) dxdydz = \int f(x) dx \times \int f(y) dy \times \int f(z) dz$ (Example on a triple integral).

Therefore:

$\displaystyle A_n =\sum_{m=1}^{\infty} \int_{0}^{1} \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \cdots \int_{0}^{1} (\prod_{i=1}^{n} a_i) ^m da_1da_2da_3 \cdots da_n$

$\displaystyle =\sum_{m=1}^{\infty} \int_{0}^{1} \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \cdots \int_{0}^{1} (\prod_{i=1}^{n} {a_i}^m) da_1da_2da_3 \cdots da_n$

$\displaystyle =\sum_{m=1}^{\infty} \prod_{i=1}^{n} \int_{0}^{1} {a_i}^m da_i$

$\displaystyle = \sum_{m=1}^{\infty} \prod_{i=1}^{n} \left.\frac{{a_i}^{m+1}}{m+1}\right|_0^1$

$\displaystyle A_n = \sum_{m=1}^{\infty} \prod_{i=1}^{n} \frac{1}{m+1} = \sum_{m=1}^{\infty} (\frac{1}{m+1})^n=\sum_{m=2}^{\infty} \frac{1}{m^n}$

We are seeking $\displaystyle \sum_{n=2}^{\infty} A_n = \sum_{n=2}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^n} = \sum_{m=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{m^n}$

We deal now with a convergent geometric series : $\displaystyle \sum_{n=2}^{\infty} (\frac{1}{m})^n = \frac{\frac{1}{m^2}}{1-\frac{1}{m}} = \frac{1}{m(m-1)}$

therefore the desired answer is : $\displaystyle \sum_{m=2}^{\infty} \frac{1}{m(m-1)} = \sum_{m=2}^{\infty} ( \frac{1}{m-1} - \frac{1}{m} )= \boxed {1}$ (a telescoping series).

We have that $A_n = -1+\zeta(n),$ since: $\int_{[0,1]^n}\frac{x_1\cdot\ldots\cdot x_n}{1-x_1\cdot\ldots\cdot x_n}\,d\mu = \sum_{k=1}^{+\infty}\int_{[0,1]^n}x_1^k\cdot\ldots\cdot x_n^k\,d\mu = \sum_{k=1}^{+\infty}\frac{1}{(k+1)^n}$ hence $\sum_{n=2}^{+\infty}A_n = \sum_{n=2}^{+\infty}\sum_{m=2}^{+\infty}\frac{1}{m^n}=\sum_{m=2}^{+\infty}\frac{1}{m(m-1)}=1.$

Integrate Ln x from 0 to 1 = -1; apply all common techniques of integration;

Sum of all converging harmonics from 2 to infinity in G.P.;

Finally,

1/ (2 - 1) - 1/ 2 + 1/ (3 - 1) - 1/ 3 + 1/ (4 - 1) - 1/ 4 + 1/ (5 - 1) - 1/ 5 + .... = 1/ (2 - 1) = 1