Asymptotic Least Squares question (with random regressors) 
Consider the DGP $y_i=x_i+\epsilon_i$, where $\epsilon_i \sim Z$. We estimate $\beta=1$ by regression without a constant term, so in $y_i=\beta x_i + \epsilon_i$. 
  
  
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*Show that this DGP does not satisfy the stability assumption ($\operatorname{plim}(\frac{1}{n}X'X) \rightarrow Q, Q$ invertible) and show that the speed of convergence of $b$ to $\beta$ is $n\sqrt{n}$ in this case. (Hint: $\sum_{i=1}^n i^2=\frac{1}{6}n(n+1)(2n+1))$
  
*Let $x_i=\frac{1}{i}$. Show that again the stability assumption is not satisfied and that the speed of convergence is $n^0$. (Hint: $\sum_{i=1}^\infty \frac{1}{i^2}=\frac{1}{6}\pi^2$).
  

What I tried for (1): $\frac{1}{n}x'x=\frac{1}{n}\sum x_i^2=\frac{1}{6}(n+1)(2n+1)$.  Which does not satisfy the assumption I guess?
Also 
\begin{align*} n^p(b-\beta) &= (x'x)^{-1} x'\epsilon\\
 &= \left(\frac{1}{n^p} x'x\right)^{-1} x'\epsilon\\ &= \frac{6n^p}{n(n+1)(2n+1)} x'\epsilon\end{align*}
And I think that I need to show that if I would take the limit $n\rightarrow \infty$, the first term would evaluate to a constant if $p=\frac{3}{2}$. I don't however see why this is so ($p=3$ would follow from my derivation I believe).
What I tried for (2). $\frac{1}{n}x'x=\frac{1}{n}\sum_{i=1}^n \frac{1}{i^2}$. Again this doesn't satisfy the assumption? And I'm not sure about the second part of this subquestion. 
Thank you in advance for your help :)!
 A: The original question appears ill-posed, regarding part 1).
For arbitrary regressor $X$ the stability condition may or may not be satisfied. The "hint" provided could imply a regressor $X=1,2,...$ which is not random, and for which obviously the stability condition is not satisfied, since the average moment "matrix" is a 2nd-degree polynomial in $n$ which does not converge.  
If indeed this is the regressor implied, then regarding speed of convergence we have
\begin{align*} n^p(b-\beta) &= n^p\left(\frac 1n X'X\right)^{-1} \frac 1n X'\epsilon\\
 &= \frac{6n^p}{(n+1)(2n+1)} \cdot \left(\frac 1n\sum_{i=1}^ni\epsilon_i\right)\\
&=\frac{6n^{p+1/2}}{(n+1)(2n+1)} \cdot \left(\frac 1{\sqrt n}\sum_{i=1}^n(i/n)\epsilon_i\right)\end{align*}
Now the last term converges to a zero-mean normal (see for example Hamilton's "Time-Series Analysis", the chapter about Deterministic Time Trends), while in order for the first to converge to a finite non-zero constant we need $p+1/2 = 2$ (to match orders of magnitude of numerator and denominator). So $p=3/2$ and so the speed of convergence is $n^{3/2} = n\sqrt n$.
Part 2) can be worked analogously.  
ADDENDUM
The sequence $\{(i/n)\epsilon_i\}$ is a martingale difference (m.d.s) because
a) $E[(i/n)\epsilon_i] = 0 $ and
b) $E[(i/n) \epsilon_i\mid \epsilon_{i-1}, \epsilon_{i-2},... ] = 0$
In order then for $\frac 1{\sqrt n}\sum_{i=1}^n(i/n)\epsilon_i$ to obey the Central Limit Theorem we need more over the following conditions (denote $\sigma^2$ the variance of $\epsilon$ and $v_i$ the variance of $(i/n)\epsilon_i$):
c) $v_i=E[(i/n)\epsilon_i]^2>0$ for all $i$
d) $\frac 1n \sum_{i=1}^nv_i \rightarrow v>0$
e) $E|(i/n)\epsilon_i|^r<\infty$ for some $r>2$ and all $i$
f) $\frac 1n \sum_{i=1}^n[(i/n)\epsilon_i]^2 \rightarrow v$
Condition c) is obviously satisfied.
For condition e) assume that the 4th moment of $\epsilon$ exists and it is finite (usually an innocuous assumption related to the real world phenomenon under study), set $r=4$ and the condition is satisfied.
For condition d) we have
$$\frac 1n \sum_{i=1}^nv_i = \frac 1n \sum_{i=1}^nE[(i/n)\epsilon_i]^2 = \frac {\sigma^2}{n^3} \sum_{i=1}^ni^2 = \frac {\sigma^2}{n^3}\frac{1}{6}n(n+1)(2n+1) \rightarrow \frac {\sigma^2}{3}$$
so condition d) is satisfied. I 'll leave condition f) unproven, it needs a roundabout way -but it holds, as demonstrated in the reference I have provided. Behind these technical requirements the intuition is the usual one associated with the CLT: the variance does not explode, neither does it tend to zero (although this intuition is just a step in understanding - there exists the generalized CLT for processes with infinite variance).
