= Solution
The <strong law of large numbers> gives $S_n/n=\overline X\to1/\theta$ almost surely, and hence in probability. Therefore
$$
\widehat\theta_\pi
=\frac{1+\alpha/n}{\overline X+\beta/n}
\xrightarrow{p}\theta
$$
by the <continuous mapping theorem>. Thus the Bayes estimator has <consistency (statistics)>.
It differs from the maximum-likelihood estimator by
$$
\widehat\theta_\pi-\widehat\theta_{\rm MLE}
=\frac{\alpha S_n-n\beta}{S_n(S_n+\beta)}.
$$
Since $S_n/n\to1/\theta$, the numerator is $O_p(n)$ and the denominator is $O_p(n^2)$, so this difference is $O_p(n^{-1})=o_p(n^{-1/2})$. The limit in part a and <Slutsky theorem> now give
$$
\boxed{
\sqrt n(\widehat\theta_\pi-\theta)
\xrightarrow dN(0,\theta^2).}
$$
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