Limit of $\frac1{c^n}\iint_{[0,c]^n}\frac{f(x_1) +f(x_2) +\cdots+f(x_n)}{g(x_1) +g(x_2) +\cdots+g(x_n)}\,dx_1dx_2\cdots dx_n$ when $n\to\infty$ A. How do I prove the following sequence converges as $n$ goes to $\infty$ for any $c$, and how do I find the limit?
$$
\begin{align} a_1 &=\frac{1}{c} \int _0^c\frac{x_1 }{1+x_1}\;dx_1 \\  \\
a_2 & =\frac{1}{c^2 } \int _0^c\int _0^c\frac{x_1 +x_2 }{2+x_1 +x_2 }\;dx_2\;dx_1  \\  \\
a_3 & =\frac{1}{c^3 } \int _0^c\int _0^c\int _0^c\frac{x_1 +x_2 +x_3 }{3+x_1 +x_2 +x_3 } \;dx_3\;dx_2\;dx_1 \end{align} 
$$
and so on for $a_n$ $\dots$ 
B. Similarly, with this, where $f$ and $g$ are not polynomials [I verified the convergence numerically]:
$$
\begin{align} a_1 &=\frac{1}{c} \int _0^c\frac{f(x_1) }{g(x_1)}\;dx_1 \\  \\
a_2 & =\frac{1}{c^2 } \int _0^c\int _0^c\frac{f(x_1) +f(x_2) }{g(x_1) +g(x_2) }\;dx_2\;dx_1  \\  \\
a_3 & =\frac{1}{c^3 } \int _0^c\int _0^c\int _0^c\frac{f(x_1) +f(x_2) +f(x_3) }{g(x_1) +g(x_2) +g(x_3) } \;dx_3\;dx_2\;dx_1 \end{align} 
$$
and so on for $a_n$ $\dots$ (perhaps I need to include the very long definition of $f$ and $g$...?)
 A: Let $S_n$ denote the sum of $n$ i.i.d. random variables uniformly distributed on $(0,c)$. Then $a_n=\mathrm E\left(\dfrac{S_n}{n+S_n}\right)$. By the strong law of large numbers, $\dfrac{S_n}n\to\dfrac{c}2$ almost surely, hence 
$$\lim\limits_{n\to\infty} a_n=\dfrac{c/2}{1+c/2}=\dfrac{c}{2+c}.$$
Edit In case B (added afterwards to the question) one should consider the sums $S_n$ and $T_n$ of $n$ random variables $f(X_i)$ and $g(X_i)$, for some i.i.d. random variables $(X_n)_{n\geqslant1}$ uniformly distributed on $(0,c)$. Then $a_n=\mathrm E(R_n)$ with $R_n=S_n/T_n$ and, by the strong law of large numbers, $S_n/n\to\mathrm E(f(X_1))$ and $T_n/n\to\mathrm E(g(X_1))$ hence $R_n\to\mathrm E(f(X_1))/\mathrm E(g(X_1))$ almost surely. 
If, for example, $g\gt0$ almost everywhere on $(0,c)$, $R_n$ is well defined. If, for example, $|f|\leqslant Ag$ almost everywhere for some finite $A$, $(R_n)_{n\geqslant1}$ is uniformly integrable hence
$$\lim\limits_{n\to\infty} a_n=\lim\limits_{n\to\infty}\mathrm E(R_n)=\mathrm E(\lim\limits_{n\to\infty} R_n)=\dfrac{\mathrm E(f(X_1))}{\mathrm E(g(X_1))}=\dfrac{\displaystyle\int_0^cf(x)\mathrm dx}{\displaystyle\int_0^cg(x)\mathrm dx}.$$
