Does the following limit exist?
$$\lim\limits_{i \to \infty} \frac 1 {\sqrt i} \int_0^1...\int_0^1 \sqrt{\sum_{1}^i z_i^2}dz_1...dz_i$$
I do not understand this question as in is xi considered as a random vector a a sequence of random variables? Thanks in advance.
I believe the correct question is what is $$lim_{n \to \infty} \frac 1 {\sqrt n} \int_0^{1} \int_0^{1} ...\int_0^{1} \sqrt {\sum_{i=1}^{n} z_i^{2}} dz_1...dz_n$$ The answer is $\frac 1 {\sqrt 3}$. To prove this let $\{X_n\}$ be i.i.d. with uniform distribution on $(0,1)$. Then $$\frac 1 {\sqrt n} \int_0^{1} \int_0^{1} ...\int_0^{1} \sqrt {\sum_{i=1}^{n} z_i^{2}} dz_1...dz_n=E\sqrt {\frac {X_1^{2}+X_2^{2}+..+X_n^{2}} n} $$ By Strong Law $\frac {X_1^{2}+X_2^{2}+..+X_n^{2}} {n} \to EX_1^{2}=\frac 1 3$ almost surely. Hence $\sqrt {\frac {X_1^{2}+X_2^{2}+..+X_n^{2}} n} \to \frac 1 {\sqrt 3}$ almost surely. To prove that the expectations converge it is enough to prove that $E {\frac {X_1^{2}+X_2^{2}+..+X_n^{2}} n} $ is bounded. [ If $Y_n \to Y$ almost surely and $\sup EY_n^{2} <\infty$ then $EY_n \to EY$]. But $E {\frac {X_1^{2}+X_2^{2}+..+X_n^{2}} n} =\frac 1 3$ for every $n$.