Asymptotic behaviour of the integral of the quadratic mean of the coordinates on the hypercube I have to compute the limit $\lim_{n\to +\infty}I_n$, where:
$$\qquad I_n=\int_{[0,1]^n}\sqrt{\frac{1}{n}\sum_{i=1}^n x_i^2}\,d\mu.$$
I believe that its value is just $\frac{1}{\sqrt{3}}$, since the mean value of $x_i^2$ over the unit hypercube is $\frac{1}{3}$. 
Numerical experiments agree with this conjecture. 
Moreover, since the square root is a concave function, Jensen's inequality gives
$$ I_n \leq \frac{1}{\sqrt{3}}$$
and $\{I_n\}_{n\in\mathbb{N}}$ looks to be a monotonic sequence, but an extra insight is needed to prove the conjecture, maybe a sort of converse of Jensen's inequality or a clever application of Fubini's theorem.
 A: There's one tricky way to calculate such a limit: use Strong Law of Large Numbers
Let $X_1, X_2, . . . , X_n$ be independent random variables, each having a uniform distribution over $[0,1]$. You have:
$$\qquad I_n=\int_{[0,1]^n}\sqrt{\frac{1}{n}\sum_{i=1}^n x_i^2}\,d\mu=E\left[\sqrt{\frac{1}{n}\sum_{i=1}^{n}}X_{i}^2\right]$$
From Strong Law of Large Numbers:
$$\sqrt{\frac{1}{n}\sum_{i=1}^{n}X_{i}^2} \to \sqrt{E[X_1^{2}]}=\frac{1}{\sqrt{3}}$$
Now because $0\leq \sqrt{\frac{1}{n}\sum_{i=1}^n x_i^2}\leq 1$ from Dominated convergence theorem you have that the limit is $\frac{1}{\sqrt{3}}$.
A: The evaluation of limit is essentially same as here.
We have, $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{3}$
We can show $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{n}\right)^k \,dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{3^k}$, in the same way as the linked answer.
Thus, $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{n}\right)^{1/2}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{\sqrt{3}}$ follows from Weierstrass-Approximation of the function $f(x) = \sqrt{x}$, $x \in [0,1]$.
I'm not aware of the Asymptotics of these multiple-integral limits.
Edit: Set $E_n(\delta) := \left\{ (x_1,x_2,\cdots,x_n) \in [0,1]^n : \left| \dfrac{\sum_{i=1}^n x_i^2}{n} - \dfrac{1}{3} \right| > \delta \right\}$
Then, $\displaystyle \frac{1}{\sqrt{3}}-I_n \le \int_{[0,1]^n} \left|\sqrt{\frac{\sum\limits_{i=1}^n x_{i}^2}{n}} - \frac{1}{\sqrt{3}} \right|\,dx < \sqrt{3}\int_{[0,1]^n} \left|\frac{\sum\limits_{i=1}^n x_{i}^2}{n} - \frac{1}{3} \right|\,dx $
$$ < \sqrt{3}\left(\int_{E_n(\delta)}\,dx + \int_{[0,1]^n\setminus E_n(\delta)}\delta\,dx \right)$$
$$ < \sqrt{3}\delta + \sqrt{3}\int_{E_n(\delta)}\,dx$$
Since, by Chebychev's Inequality: $\displaystyle \int_{E_n(\delta)}\,dx \le \frac{1}{n^2\delta^2} \int_{[0,1]^n}\left(\sum\limits_{i=1}^n \left(x_i^2-\frac{1}{3}\right)\right)^2\,dx = \frac{1}{n^2\delta^2} \int_{[0,1]^n}\sum\limits_{i=1}^n \left(x_i^2-\frac{1}{3}\right)^2\,dx = \frac{c}{n\delta^2}$
Therefore, choosing $\delta = O(n^{-1/3})$ we have $\displaystyle 0\le  \frac{1}{\sqrt{3}} - I_n < \frac{C}{n^{1/3}}$ for some constant $C$.
A: http://en.wikipedia.org/wiki/Hoeffding%27s_inequality gives a way to bound the difference between $\frac{1}{\sqrt{3}}$ and the given integral. If we take $X_1,\ldots,X_n$ as indipendent random variables with a uniform distribution over $[0,1]$, then $X_1^2,\ldots,X_n^2$ are bounded indipendent random variables with density function
$$ f(x)=\frac{1}{2\sqrt{x}}\cdot\mathbb{1}_{(0,1]}$$
and mean value $\frac{1}{3}$. If we set $\bar{X}=\frac{1}{n}\sum_{j=1}^{n}X_j^2$, the Hoeffding's inequality gives:
$$ \mathbb{P}\left[\left|\bar{X}-\mathbb{E}[\bar{X}]\right|\geq t\right]\leq 2\exp\left(-2t^2 n\right),$$
hence by choosing $t=\sqrt{\frac{\log n}{4n}}$ we have:
$$ \mu\left(\left\{\bar{x}\in[0,1]^n : \left|-\frac{1}{3}+\frac{1}{n}\sum_{i=1}^{n}x_i^2\right|\geq\sqrt{\frac{\log n}{4n}}\right\}\right)\leq\frac{2}{\sqrt{n}},$$
so:
$$\int_{[0,1]^n}\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_i^2}d\mu\geq\left(1-\frac{2}{\sqrt{n}}\right)\sqrt{\frac{1}{3}-\sqrt{\frac{\log n}{4n}}}$$
and
$$\int_{[0,1]^n}\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_i^2}d\mu\geq\frac{1}{\sqrt{3}}\left(1-\sqrt{\frac{\log n}{n}}\right)$$
for any $n\geq e^{64}.$
