# Convergence in probability: $\lim_{n\to\infty}\int_0^1\cdots\int_0^1\frac{x_1^2+x_2^2+\cdots+x_n^2}{x_1+x_2+\cdots +x_n}dx_1\cdots dx_n=\frac23$

How to prove the following: $$\lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{x_1^2+x_2^2+ \cdots +x_n^2}{x_1+x_2+ \cdots +x_n} dx_1 dx_2 \cdots dx_n = \frac23$$

I would really appreciate if you could help me!

Let $X_1, X_2,\dots$ be independent, uniform$(0,1)$ random variables. By the law of large numbers we have $$\begin{eqnarray*} {X_1+\cdots + X_n\over n}&\to& \mathbb{E}(X)={1\over 2}\\ {X_1^2+\cdots + X^2_n\over n}&\to& \mathbb{E}(X^2)={1\over 3}\\ \end{eqnarray*}$$ in probability as $n\to\infty$. Therefore $${X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}={X_1^2+\cdots +X^2_n\over n}\cdot{n\over X_1+\cdots +X_n}\to {2\over 3}$$ in probability as $n\to \infty$. The ratio random variables ${X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}$ are bounded below by zero and above by one. This guarantees convergence of the expectations, as well. So $$\mathbb{E}\left({X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}\right)\to{2\over 3}$$ which is the required result.