TITLE: Asymptotic behaviour of a multiple integral on the unit hypercube
QUESTION [16 upvotes]: A few days ago I found an interesting limit on the "problems blackboard" of my University:
$$\lim_{n\to +\infty}\int_{(0,1)^n}\frac{\sum_{j=1}^n x_j^2}{\sum_{j=1}^n x_j}d\mu = 1.$$
The correct claim, however, is:
$$\lim_{n\to +\infty}\int_{(0,1)^n}\frac{\sum_{j=1}^n x_j^2}{\sum_{j=1}^n x_j}d\mu = \frac{2}{3}.$$
In fact, following @Tetis' approach we have:
$$I_n = \int_{(0,1)^n}\frac{\sum_{j=1}^n y_j^2}{\sum_{j=1}^{n}y_j}d\mu = \int_{(-1/2,1/2)^n}\frac{\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2+\frac{2}{n}\sum_{j=1}^n x_j}{1+\frac{2}{n}\sum_{j=1}^n x_j}d\mu,$$
now setting $x_j=-z_j$ and summing the two integrals
$$I_n = \int_{(-1/2,1/2)^n}\frac{\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2+\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu$$
follows, so:
$$ I_n-\frac{2}{3}=\int_{(-1/2,1/2)^n} \left(-\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2\right)\frac{\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu < 0,$$
$$ \left| I_n-\frac{2}{3}\right|\leq\int_{\sum x_i^2\leq\frac{n}{4}}\left(\frac{1}{2}-\frac{2}{n}\sum_{j=1}^n x_j^2\right)\frac{\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu,$$
$$ \frac{2}{3}-I_n\leq \frac{n^{n/2}}{2^{n+1}} \int_{\sum x_i^2\leq 1}\left(1-\sum_{j=1}^{n}x_j^2\right)\frac{x_1^2}{1-x_1^2}d\mu.$$
The last bound, anyway, is too crude, since the RHS is
$$ \Theta\left(\left(\sqrt{\frac{e\pi}{2}}\right)^n\frac{\log n}{n^{3/2}}\right).$$
My question now is: what is the asymptotic behaviour of $I_n$?
A second one is: can we prove $I_n\geq\frac{2}{3}-\frac{C}{n}$, for a suitable positive constant $C$, without the Central Limit Theorem?

UPDATE: After a few I came out with a proof of my own. The challenge is now to give the first three terms of the asymptotics, and possibly a continued fraction expansion for $I_n$.

REPLY [3 votes]: Another probabilistic approach, with $x_i$ being iid uniform(0,1) random variables. To evaluate $\displaystyle E\left[\frac{\sum x_i^2}{\sum x_i} \right]$, we can use a Taylor expansion in two variables. 
Suppose $s$ and $t$ are two random variables with joint density $f(s,t)$ and mean $s_0$, $t_0$. Then, for a function $g(s,t)$ well behaved around $(s_0,t_0)$ we can write:
$$ \begin{align} E[g(s,t)]&=\int g(s,t) f(s,t) ds dt \\&= g(s_0,t_0)+ \frac{1}{2}Var(s) g_{ss}(s_0,t_0)+  \frac{1}{2}Var(t) g_{tt}(s_0,t_0)+ Cov(s,t)g_{st}(s_0,t_0) + \cdots
\end{align}
$$
because the linear terms vanish. 
Then, let $$t=\frac{\sum_{i=1}^n x_i}{n} \\s=\frac{\sum_{i=1}^n x_i^2}{n}$$
And let $$g(s,t)=\frac{s}{t}$$
Then it's straightforward: $s_0 = 1/3$, $t_0 = 1/2$, $Var(t)=1/(12 n)$, $E(s t) =1/6+1/(12 n)$, $Cov(s,t) = 1/(12 n)$.
Further, $g_{ss}=0$, $g_{tt}=2 s/ t^{3}$, $g_{st}=-1/t^2$. Hence $$ E\left[\frac{\sum x_i^2}{\sum x_i} \right]=E\left[\frac{s}{t} \right] \approx \frac{s_o}{t_0} + \frac{1}{2}\frac{2 s_o}{t_0^3}Var(t) - \frac{1}{t_0^2}Cov(s,t)=\\
=\frac{2}{3} - \frac{1}{9 n}$$
What remains is to show that the next terms of the Taylor expansion are $o(n^{-1})$. One could also compute them, of course.