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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.

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up vote 6 down vote

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 $\frac{1}{\sqrt{3}}$.

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Clever solution, but I would like a more elementary one, with some explicit bound on $\frac{1}{\sqrt{3}}-I_n$. Nonetheless, I will accept this answer if nothing different come out. –  Jack D'Aurizio yesterday
    
@JackD'Aurizio +1 Ha ha, I was going to say the opposite. This is not a tricky solution, it is the obvious approach, at least to a probabilist. –  Byron Schmuland yesterday
    
@math110 : I bet you were interested, too. –  Jack D'Aurizio yesterday
    
@ByronSchmuland: I think that even Mathematics is a matter of taste, sometimes :) –  Jack D'Aurizio yesterday
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