What is limit of $\frac 1 {\sqrt n} \int_0^{1} \int_0^{1} ...\int_0^{1} \sqrt {\sum_{i=1}^{n} x_i^{2}} dx_1...dx_n$ as $n \to \infty$? This question was just deleted probably because there were errors in the statement. So I am posting it and providing my answer also.
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Wy don't you start with the simple $1D$ case, then you proceed in order to find the general result? – Enrico M. Apr 09 '18 at 08:56
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I don't remember who posted this question a few minutes ago so I am posting a corrected version in my name. Apologies to the OP. – Kavi Rama Murthy Apr 09 '18 at 09:04
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Let $X_1,X_2,...$ be i.i.d. with uniform distribution on $(0,1)$. Then $ Y_n \to EX_1^{2}=\frac 1 3$ almost surely as $n \to \infty$ where $Y_n = \frac {X_1^{2}+X_2^{2}+...+{X_n^{2}}} n$ (by the strong law). Further $EY_n =\frac 1 3$ for each $n$. Hence $\{\sqrt Y_n\}$ converges almost surely and this sequence is uniformly integrable because the second moments are bounded. Hence $E\sqrt Y_n \to \frac 1 {\sqrt 3}$. This is exactly the limit we are asked to find!
Kavi Rama Murthy
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1I get 0 as an answer, can you please find a mistake in here: $$I(n)=\frac 1 {\sqrt n} \int_0^{1} \int_0^{1} ...\int_0^{1} \sqrt {\sum_{i=1}^{n} x_i^{2}} dx_1...dx_n=\frac{2\pi^{\frac{n}{2}}}{\Gamma\big(\frac{n}{2}\big)\sqrt{n}} \int_0^{1}r^{n}dr=\frac{2\pi^{\frac{n}{2}}}{(n+1)\Gamma\big(\frac{n}{2}\big)\sqrt{n}}$$ $$\lim_{n\rightarrow\infty}I(n)=0$$ – Kiryl Pesotski Apr 09 '18 at 15:02
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I think in that polar coordinate representation you are allowing x’s to have negative values which is not true. – Landon Carter Apr 26 '18 at 08:54