Suppose that a random sample of size n is taken from the Bernoulli distribution with parameter θ, which is unknown, and that the prior distribution of θ is a beta distri bution for which the mean is $\mu_o$. Show that the mean of the posterior distribution of θ will be a weighted average having the form γ$_n \bar{X_n}$ + (1 − γ$_n$)μ$_o$, and show that γ$_n$ → 1 as n → ∞.
So, I calculate $ \gamma_n $ and I got $ \gamma _n = \frac{n}{n+\alpha + \beta }$ where $ \alpha \; and \; \beta $ are the parameters of the prior distribution of theta. But I couldn't prove the limit part. Can someone give me a hint? thanks!
Hint: Rewrite it as
$$ \frac{n}{n+\alpha+\beta}=\frac{1}{1+\frac{\alpha+\beta}{n}} $$
What happens to $\frac{\alpha+\beta}{n}$ as $n$ tends to infinity? What happens to the expression then?