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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 $g$...?)

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up vote 7 down vote accepted

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 $$\lim\limits_{n\to\infty} a_n=\lim\limits_{n\to\infty}\mathrm E(R_n)=\mathrm E(\lim\limits_{n\to\infty} R_n)=\dfrac{\mathrm E(f(X_1))}{\mathrm E(g(X_1))}=\dfrac{\displaystyle\int_0^cf(x)\mathrm dx}{\displaystyle\int_0^cg(x)\mathrm dx}.$$

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Thanks, Didier. This was the 1st part of my question. The 2nd part involves replacing the x_i with f(x_i) in the numerator and g(x_i) in the denominator (and removing the constant), where f and g are not polynomials. They are rather complicated functions so I am not including their definition here. However, numerical simulations demonstrate the convergence. I add this now to the main question above. – Omri Jan 16 '12 at 22:43
Out of personal curiosity and perhaps for the enhancement of this question's usefulness, could you provide those functions and their context? You do not have to, this is simply a suggestion. – 000 Jan 16 '12 at 23:06
Omri: Do not do that (adding new questions to a post). – Did Jan 17 '12 at 6:33
OK Didier, and thanks a lot for the continued answer... – Omri Jan 17 '12 at 7:39
Didier, one last point: if the n i.i.d random variables are not uniformly distributed (actually, a Beta), so that the integrand is multiplied by the corresponding pdf (and a proper normalization replaces c^n), does your final limit still apply? – Omri Jan 17 '12 at 10:23

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