Let $X_1, X_2, ... , X_n$ be i.i.d random variables with common density function: $f(x)=\frac{\alpha}{x^{\alpha+1}}I(x\geq 1)$ where $I(x\geq 1)$ is the indicator function and $\alpha >0$. Now define: $$S_n=\left[ \prod_{i=1}^{n}X_i \right ]^{n^{-1}}$$ Prove that $\lim_{n\rightarrow \infty}S_n=e^{\alpha^{-1}}$, and use the law of large numbers and the continuous mapping theorem to conclude about the convergence in probability of $\ln(S_n)$.
I honestly don't have the slightest inkling on how to go about evaluating that limit. However, since the value is given in the problem statment, my intuition tells me that $\ln(S_n)$ converges in probability to $\alpha^{-1}$, but I'm not certain.
The second one is actually more like a hint, because: $$\ln(S_n) =\frac{1}{n}\sum_{k=1}^n\ln(X_k)$$ Now we find the density of $Y_k:=\ln(X_k) $. Let $y>0$, $$P(Y_k<y) =P(\ln(X_k) <y)=P(X_k<e^y)$$ The Fundamental Theorem of Calculus yields: $$f_Y(y) =e^y \frac{\alpha} {e^{(\alpha+1)y} }=\alpha e^{-\alpha y} $$ Hence $Y$ is exponentially distributed. Now by the SLLN we have $$\ln(S_n) \to E[Y_1]=\frac{1}{\alpha}\ \ \ \ \text{ a. s.} $$ And a. s. convergence implies convergence in probability. Moreover by the continuous mapping theorem we have...?
Can you finish?