Expected Euclidean norm of a random vector Let $\mathbf{v} = (v_1,\ldots,v_n) \in \mathbb{R}^n$ be a vector, whose coordinates are i.i.d random variables with zero mean and standard deviation $\sigma$.
Using Jensen's inequality, we obtain
$$
{\mathbb{E}[\|\mathbf{v}\|]} \leq \sqrt{ \mathbb{E}[\|\mathbf{v}\|^2 } = \sqrt{ \sum_{i=1}^n \mathbb{E}[v_i^2] } = \sigma \sqrt{n}.
$$
Question: Is this bound asymptotically sharp, i.e., does it hold that $\lim_{n \to \infty} \frac{\mathbb{E}[\|\mathbf{v}\|]}{{\sqrt{n}}} = \sigma  $?
According to this answer, the strong law of large numbers "suggests" that this is the case. I am not sure, whether "suggests" is meant here in a heursitic sense, or whether it is supposed to mean "implies".
I am by no means an expert in probability theory, and therefore unfortunately fail to see, how the statement $\lim_{n \to \infty} \frac{\mathbb{E}[\|\mathbf{v}\|]}{{\sqrt{n}}} = \sigma$ would follow from the law of large numbers. As far as I understand it, the strong law of large numbers (together with the continuous mapping theorem) only implies that
$$
\sqrt{\frac{v_1^2+\ldots+v_n^2}{n}} \xrightarrow[]{a.s.} \sigma.
$$
But as shown in this answer, almost sure convergence does not necessarily imply convergence of the mean.
 A: You would need to use uniform integrability and the theorem that $X_{n}\xrightarrow{L^{1}} X$ if and only if $\{X_{n}\}$ is uniformly integrable and $X_{n}\xrightarrow{P} X$.
Let $X_{n}=v_{n}^{2}$  and $S_{n}=\sum_{i=1}^{n}X_{i}$ . Then, you already have $\sqrt{\frac{S_{n}}{n}}\xrightarrow{a.s} \sigma$ by the Strong Law (even weak law works here) and hence  $E[\sqrt{\frac{S_{n}}{n}}]\to \sigma$ if and only if $\sqrt{\frac{S_{n}}{n}}$ is uniformly integrable.
If you look at my answer here , I have shown that $\frac{S_{n}}{n}$ is uniformly integrable. Hence you directly have that $E[\frac{S_{n}}{n}]\to\sigma^{2}$.
Then you also have that the sequence $\sqrt\frac{S_{n}}{n}$ is uniformly integrable.
Proof:-We use the following result .
Let $\{X_{n}\}$ be a sequence such that $\{E[|X_{n}|^{p}]\}$ is bounded for some $p>1$ , then $\{X_{n}\}$ is uniformly integrable. The proof of this is by the Markov inequality.
So here you have as  $E[|\sqrt{\frac{S_{n}}{n}}|^{2}]\to\sigma^{2}$  and hence $E[|\sqrt{\frac{S_{n}}{n}}|^{2}]$ is bounded. And thus $\sqrt{\frac{S_{n}}{n}}$ is uniformly integrable.
Else you can use that if $\{X_{n}\}$ is uniformly integrable then , $\{\sqrt{|X_{n}|}\}$ is uniformly integrable. You'll need to use the Cauchy-Shwartz inequality . Or directly you can use the fact that the $L^{p}$ norm is weaker than the $L^{q}$ norm for finite measure spaces when $p<q$ (the proof of which uses the Holder's Inequality) .
Thus you have the required result that $E[\sqrt{\frac{S_{n}}{n}}]\to\sigma$ .
