Iterated integral question Show $$\lim_{n \to\infty} \int_0^1 \cdots  \int_0^1  \int_0^1 \frac{ x_1^2 + \cdots +  x_n^2}{x_1 + \cdots + x_n} \, dx_1 \cdots dx_n = \frac 2 3.$$
Not sure how to start off this iterated integral question, any help would be appreciated. 
 A: We follow the idea of @peter, and also prepare for an integration by parts. We try, however, to keep the argument as elementary as possible.
$$\begin{aligned}
I_n & =\int_{[0,1]^n}\frac{x^2+x_2^2+\cdots+x_n^2}{x+x_2+\cdots+x_n}\,dx\,dx_2\,\ldots\,dx_n\\
& = n\int_0^{+\infty}\int_0^{1}x^2e^{-\lambda x}\,dx \biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^{n-1}\,d\lambda\\
& = n\int_0^{+\infty}\biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr)\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^{n-1}\,d\lambda\\
& = \int_0^{+\infty}\biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr)
\biggl(\frac{e^{-\lambda}}{\lambda}-\frac{1-e^{-\lambda}}{\lambda^2}\biggr)^{-1}
\frac{d}{d\lambda}\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\,d\lambda\\
\end{aligned}
$$
It happens that
$$
\biggl(\frac{2}{\lambda^3}-\frac{e^{-\lambda}(2+2\lambda+\lambda^2)}{\lambda^3}\biggr)
\biggl(\frac{e^{-\lambda}}{\lambda}-\frac{1-e^{-\lambda}}{\lambda^2}\biggr)^{-1}
=\frac{\lambda}{e^\lambda-1-\lambda}-\frac{2}{\lambda}.
$$
Thus, integrating by parts,
$$
\begin{aligned}
I_n & =\biggl[\biggl(\frac{\lambda}{e^\lambda-1-\lambda}-\frac{2}{\lambda}\biggr)
\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\biggr]_0^{+\infty}\\
& \quad-
\int_0^{+\infty}\biggl(\frac{2}{\lambda^2}-\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr)
\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n
\,d\lambda\\
& =\frac{2}{3}-
\int_0^{+\infty}\biggl(\frac{2}{\lambda^2}-\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr)
\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n
\,d\lambda.
\end{aligned}
$$
We need to show that the last integral tends to $0$ as $n\to+\infty$. 
The absolute value of the first factor is bounded by some constant $C$.
Indeed, close to $0$ a Maclaurin expansion shows that it is of size 
$1/18+O(\lambda)$. Moreover, it is continuous for $0<\lambda<+\infty$ and tends to $0$ 
as $\lambda\to +\infty$. For the other factor, one can use that
$$
0<\frac{1-e^{-\lambda}}{\lambda}\leq \frac{2}{2+\lambda}\quad\text{for $\lambda>0$.}
$$
Thus,
$$
\begin{aligned}
0 & < \biggl|\int_0^{+\infty}\biggl(\frac{2}{\lambda^2} -\frac{\lambda^2}{(e^\lambda-1-\lambda)^2}-\frac{\lambda-1}{e^{\lambda}-1-\lambda}\biggr)
\biggl(\frac{1-e^{-\lambda}}{\lambda}\biggr)^n\,d\lambda\biggr|
\\
& \leq \int_0^{+\infty} M\biggl(\frac{2}{2+\lambda}\biggr)^n\,d\lambda =\frac{2M}{n-1}\to 0\quad\text{as $n\to+\infty$.}
\end{aligned}
$$
We conclude that $\lim_{n\to+\infty}I_n=2/3$. Of course, Peter will receive the bounty.
A: Let $X_1,\ldots,X_n$ be independent random variables, each distributed uniformly on the interval $[0,1]$. Your question is then equivalent to
$$
\lim_{n\to\infty}\mathbb E\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n}=\frac{2}{3}.
$$
We will deduce this from the Strong Law of Large Numbers and the Bounded Convergence Theorem. Consider an infinite iid sequence $(X_i)_{i=1}^{\infty}$ of uniform $[0,1]$ random variables on a probability space $\Omega$. By the Strong Law of Large Numbers, each of the events
$$
\left\{\omega\in\Omega\colon\lim_{n\to\infty}\frac{X_1(\omega)+\cdots+X_n(\omega)}{n}=\frac{1}{2}\right\}
$$
and
$$\left\{\omega\in\Omega\colon\lim_{n\to\infty}\frac{X_1(\omega)^2+\cdots+X_n(\omega)^2}{n}=\frac{1}{3}\right\}
$$
occurs with probability 1. Therefore, it holds with probability 1 that
$$
\lim_{n\to\infty}\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n}=\frac{2}{3}.
$$
Taking expectations gives that
$$
\mathbb E\lim_{n\to\infty}\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n}=\frac{2}{3}.
$$
Since $X_i^2\leq X_i$ for all $i$, the quantity inside the limit is bounded above by $1$. Thus by the Bounded Convergence Theorem we may interchange the expectation with the limit, and therefore
$$
\lim_{n\to\infty}\mathbb E\frac{X_1^2+\cdots+X_n^2}{X_1+\cdots+X_n}=\frac{2}{3}.
$$
A: Suppose $X_1,X_2,X_3,\ldots$ are independent random variables, each uniformly distributed on the interval $[0,1].$ Then for each value of $n$ we have $\operatorname{E}(X_n^2) = 1/3.$ The weak law of large numbers says
$$
\operatorname*{l.i.p.}_{n\to\infty} \frac{X_1^2 + \cdots + X_n^2} n = \frac 1 3
$$
where $\operatorname{l.i.p.}$ means "limit in probability", and that is defined by saying
$$
\text{for every } \varepsilon>0\  \lim_{n\to\infty} \Pr\left( \left| \frac{X_1^2+\cdots + X_n^2} n - \frac 1 3 \right| < \varepsilon \right) = 1.  
$$
Similarly
$$
\operatorname*{l.i.p.}_{n\to\infty} \frac{X_1+\cdots + X_n} n = \frac 1 2.
$$
In general, $\Pr(A\cap B) \ge \Pr(A) + \Pr(B) - 1.$ Thus
\begin{align}
& \Pr\left( \left| \frac{X_1^2+\cdots + X_n^2} n - \frac 1 3 \right| < \varepsilon \text{ and } \left| \frac{X_1+\cdots + X_n} n - \frac 1 2 \right| < \varepsilon \right) \\[10pt]
\ge {} & \Pr\left( \left| \frac{X_1^2+\cdots + X_n^2} n - \frac 2 3 \right| < \varepsilon\right) + \Pr\left( \left| \frac{X_1+\cdots + X_n} n - \frac 1 2 \right| \right) - 1.
\end{align}
Next you need to say that if one number is near $1/3$ and another near $1/2$, then the quotient is near $2/3.$
This sketch of an argument leaves a lot of details to be filled in.
A: My proof based on calculation of asymptotic on integral.
Let us consider the integral at lagre fixed $n$
$$
\int_0^1 \cdots  \int_0^1\, dx_1 \cdots dx_n \frac{ x_1^2 + \cdots +  x_n^2}{x_1 + \cdots + x_n} =
n\int_0^1 \cdots  \int_0^1 \, dx_1 \cdots dx_n \frac{ x_1^2}{x_1 + \cdots + x_n}  =
$$
$$
n\int_0^1 \cdots  \int_0^1 \, dx_1 \cdots dx_n \int_0^\infty d\lambda\,\,  x_1^2 e^{-\lambda(x_1 + \cdots + x_n)} $$
After  integration over $x_2\cdots x_n$ we obtain
$$
n\int_0^1 dx_1 x_1^2 \int_0^\infty d\lambda\,\Bigr(\frac{1-e^{-\lambda}}{\lambda}\Bigr)^{n-1} e^{-\lambda x_1}
$$
After that we can calculate asymptotic of integral over $\lambda$. 
We can present integral over $\lambda$ in the following form
$$
\int_0^\infty d\lambda e^{-n f(\lambda)} G(\lambda)
$$
where $f(\lambda)=\log \lambda-\log(1-e^{-\lambda})$ and $G(\lambda)=\dfrac{\lambda e^{-\lambda x_1}}{1-e^{-\lambda}} $
Since the function $f$ is a monotonically decreasing function we can obtain asymptotic of its integral using integral by parts. 
$$
\int_0^\infty d\lambda e^{-n f(\lambda)} G(\lambda)=\frac{G(0)}{n f'(0)}+o\Bigr(\frac{1}{n}\Bigr)
$$
And we obtain 
$$
2\int_0^1 dx_1 x_1^2=\frac{2}{3}
$$
