Determining a consistent estimator/asymptotic relative efficiency Question:
Let $X_1,\ldots,X_n$ be i.i.d. as $N(0,\sigma^2)$.
a) Show that $\delta_1 = k \sum_{i=1}^n |X_i|/n$ is a consistent estimator of $\sigma$ if and only if $ k = \sqrt{\pi/2}$. 
b) Determine the asymptotic relative efficiency of $\delta_1$ with respect to $\delta_2 = \sqrt{\sum X_i^2/n}$
Attempt:
a)
i) Given $ k = \sqrt{\pi/2}$. We must find a variance that converges to 0.
Calculating the variance of $\delta_1 $:
$$
\begin{align}
& = E\left[ \left(\sqrt{\frac\pi2} \sum_{i=1}^n |X_i|/n\right)-\mu\right]^2 \\[8pt]
& = \frac \pi 2 E\left[ \left(\sum_{i=1}^n |X_i|/n\right)-\mu\right]^2
\end{align}
$$
I'm stuck here, however, this should come to:
$$\frac{\pi}{2}\frac{1}{n}\left[\sigma^2 + \frac{2}{\pi}\sigma^2\right]$$
which converges to zero.
ii) Since $\delta_1$ converges in probability to $\sigma$, and since k must be unique,  k must equal $\sqrt{\pi/2}$
b) I know this involves the Central Limit Theorem; however, I'm not able to piece the whole thing together.
I know that
$$E(|X_i|) = \int_{-\infty}^\infty |x|\frac{1}{\sqrt{2\pi\sigma}}\exp\left(\frac{-x}{2\sigma^2}\right)\,dx = \sqrt{\frac{2}{\pi}}\sigma$$
and $\operatorname{Var}(|X_i|) = \sigma^2 - \frac{2}{\pi}\sigma^2$
Also,
$$\sqrt{n}(\delta_1 - \sigma)\xrightarrow[]{L} N\left(0,\left(\frac{\pi}{2}-1\right)\sigma^2\right)$$
Not sure how to proceed.
 A: Please consider: If $X_1,\ldots,X_n$ are identically and independently distributed, then by the strong law of large numbers: 
$\delta_1 \rightarrow kE|X|$ with $E|X|={\sigma}\sqrt{2/\pi}.$ So $\delta_1$ will converge to $\sigma$ iff 
$ k = \sqrt{\pi/2}$. 
Now we'll need $Var(|X|)=E(X^2)-E^2(|X|)=\sigma^2(1-2/\pi).$
The variance of $\delta_1 $ can be computed directly since $\delta_1$ is a linear combination of independent random variables. We obtain:
$$Var(\delta_1)=k^2\frac{Var(|X|)}n=\frac{\sigma^2}n\left(\frac{\pi}2-1\right).$$
Now, from the central limit theorem:
$\sqrt{n}(\delta_1-\sigma)\to N(0,\frac{\sigma^2(\pi-2)}{2})$ 
Consider $\sum X_i^2/n.$ Its expectation is clearly $\sigma^2$
and $Var(X^2)=E(X^4)-E^2(X^2)=3\sigma^4-\sigma^4=2\sigma^4$
Apply the CLT to  $\sum X_i^2/n.$
$\sqrt{n}(\sum X_i^2/n-\sigma^2)\to N(0,2\sigma^4)$
Now we will use the delta method to find the appropriate CLT for $\delta_2.$ Let $g(x)=\sqrt{x}.$ The delta method allows us to conclude:
$$\sqrt{n}(\delta_2-g(\sigma^2))\to N(0,2\sigma^4[g'(\sigma^2)]^2) $$
$$\sqrt{n}(\delta_2-\sigma)\to N(0,2\sigma^4/(4\sigma^2)) $$
So now we can find the asymptotic relative efficiency, $ARE$, of $\delta_1$ and $\delta_2.$
$$ARE=\frac{Var(\delta_2)}{Var(\delta_1)}=\frac{\sigma^2/2}{\sigma^2(\pi-2)/2}=\frac{1}{\pi-2}\approx0.87$$
This means I can use estimator $\delta_1$ with sample size $n_1$ or I could use estimator $\delta_2$ with sample size $n_2=ARE*n_1$ and obtain the same estimation accuracy (for large sample sizes).
