Find a sequence $r_n \to \infty$ such that $r_n(\hat{\theta} - \theta)$ has a non-degenerate limiting distrubtion.
Where $X_1,...,X_n$ is iid with pdf $f(x) = \frac{2x}{\theta^2}1(0 \le x \le \theta)$
As it turns out the the MLE $\hat{\theta} = \max X_i = X_{(n)}.$
I am not sure what to use to find this rate $r_n$.
I showed $X_{(n)}$ is consistent so it converges in probability to $\theta$ thus is converges in distribution. So this means $(X_{(n)} - \theta)$ will converge to zero in distribution.
It seems I need to find $\alpha$ such that $n^\alpha(X_{(n)}-\theta)$ converges in distribution to something other than zero.
I was trying to solve using expectation so $n^\alpha(E[X_{(n)} - \theta]) = n^\alpha \bigg(\frac{2n}{2n+1}-1\bigg)\theta$ and solve for $\alpha$ such that in the limit this isn't 0.
Continuing from where you left off, we see that $$n^\alpha\left(\frac{2n}{2n+1}-1\right)\theta = \left(\frac{-n^\alpha}{2n+1}\right)\theta$$ which tends to a nonzero finite value if and only if $\alpha = 1.$ Therefore, our candidate for $r_n$ is $r_n = n$
Now we consider $Y_n = n\left(\theta - X_{(n)}\right).$ Since $X_{(n)} \leq \theta$, we see that $Y_n \geq 0$, and $$t > 0, n >\frac{t}{\theta} \\ \implies \\ \begin{align*}P(Y_n \geq t) &= P\left(X_{(n)} \leq \theta - \frac{t}{n}\right) \\ &= \left[P\left(X_1\leq\theta-\frac{t}{n}\right)\right]^n \\ &= \left(\frac{(\theta-t/n)^2}{\theta^2}\right)^n \\ &= \left(1-\frac{t}{n\theta}\right)^{2n} \\ &\to e^{-2t/\theta} \quad \text{as } n\to \infty\end{align*}$$ so $Y_n$ converges in distribution to an exponential with mean $\theta/2.$ Your expression is $-Y_n$, so it converges in distribution to the negative of such an exponential.