Limiting Distribution of $\left(\prod\limits_{i=1}^{n} U_i\right)^{1/n}$ with $(U_i)$ i.i.d. uniform $(0,\theta)$ 
Let $(U_i)$ i.i.d. uniform $(0,\theta)$ and
  $$T_n=\left(\prod_{i=1}^{n} U_i\right)^{1/n}$$
Compute the limiting distribution of the sequence $(T_n)$.

My try:
$$
F_{T_n}(t)
=\mathsf P(T_n \leq t)=\mathsf P\left(\left(\prod_{i=1}^{n} U_i\right)^{1/n}\leq t\right)
=\mathsf P\left(\prod_{i=1}^{n} U_i\leq t^n\right)$$
hence
$$
F_{T_n}(t)
=\mathsf P\left(\log\prod_{i=1}^n U_i\leq \log t^n\right)
=\mathsf P\left(\sum_{i=1}^n \log U_i \leq n\log t\right)
$$
that is,
$$
F_{T_n}(t)=\mathsf P\left(V_n \leq \log t\right)=F_{V_n}(\log t)
$$
where $$V_n=\frac1n\sum_{i=1}^n \log U_i$$
Since $E(\log U_1)=\log\theta-1$ and $\log U_1$ is square integrable, by the CLT, for some positive $\sigma^2$,
$$\sqrt{n}\left(V_n-(\log \theta-1\right))\stackrel{d}{\rightarrow}\mathsf N(0,\sigma^2)$$
Then
$$ \lim_{n\rightarrow\infty}F_{V_n}(v)=  
\begin{cases} 
1 & v\gt \text{log }\theta -1 \\
0 & v\lt \text{log }\theta -1 \\
\end{cases} $$ 
hence
$$ \lim_{n\rightarrow\infty}F_{T_n}(t)=  
\begin{cases} 
1 & t\gt \ell \\
0 & t\lt \ell \\
\end{cases} $$ 
where $$\ell=\theta e^{-1}$$
Thus, $F_{T_n}(t)\to F_T(t)$ where $P(T=\ell)=1$, at every point $t$ where $F_T$ is continuous, that is, at every point $t\ne\ell$. By a well-known theorem, this suffices to show that $T_n\to T$ in distribution, where $P(T=\ell)=1$, that is, $T_n\to\theta e^{-1}$ in distribution (hence also in probability).
Thus $T_1$, $T_2$, . . . converges to a degenerate random variable with pmf
$$f_T(t)=I_{\{\theta e^{-1}\}}(t)$$
 A: This method right here follows approximations and need to be made more rigorous as stressed out by @Chris Janjigian. Indeed the usage of the CLT below is an approximation for large $n$.
Consider the random variable
\begin{equation}
 T = \frac{\sum \ln U_i}{n}
\end{equation}
The mean and variance of $\ln U_i$ are 
\begin{equation}
 \mu =E(\ln U)= \int_{-\infty}^{\infty} f_U(u) \ln u  \ du = \frac{1}{\theta} \int_{0}^{\theta} \ln u \ du = \ln \theta - 1
\end{equation}
\begin{equation}
 E(\ln^2 U)= \int_{-\infty}^{\infty} f_U(u) \ln^2 u \ du = \frac{1}{\theta} \int_{0}^{\theta} \ln^2 u \ du = \ln^2 \theta - 2 \ln \theta + 2
\end{equation}
So 
\begin{equation}
 \sigma^2 = \operatorname{var}(\ln U)
 =
 E(\ln^2 U) - \mu^2
 =
 1
\end{equation}
Now
\begin{equation}
 \mathsf P\left(\frac{\sum_{i=1}^n \ln U_i}{n} \leq \ln t\right) \simeq \mathsf P \left(\sqrt{n}( T - \mu )\leq \sqrt{n}( \ln t -\mu)\right)
\end{equation}
But by the Central limit theorem, $\sqrt{n}( T - \mu ) \rightarrow N(0,\sigma^2) = N(0,1)$ So
\begin{equation}
 \mathsf P\left(\frac{\sum_{i=1}^n \ln U_i}{n} \leq \ln t\right)\simeq
 \frac{1}{\sqrt{2\pi}}
 \int_{-\infty}^{\sqrt{n}(\ln t - \mu)}
  \exp(-\frac{1}{2}x^2)
  dx
\end{equation}
Let $y = e^{x\frac{1}{\sqrt{n}}+\mu}$, then $dy = \frac{1}{\sqrt{n}}y dx$, we get
\begin{equation}
 \mathsf P\left(\frac{\sum_{i=1}^n \ln U_i}{n} \leq \ln t\right)\simeq
 \frac{1}{\sqrt{2\pi}}
 \int_{0}^{t}
  \exp(-\frac{1}{2}x^2)
  \frac{\sqrt{n}}{y}
  dy
\end{equation}
But $x = \sqrt{n}(\ln y - \mu)$ so we can re-arrange as 
\begin{equation}
 \mathsf P\left(\frac{\sum_{i=1}^n \ln U_i}{n} \leq \ln t\right) 
 \simeq
 \frac{1}{\sqrt{2\pi\frac{1}{n}}}
 \int_{0}^{t}
 \frac{1}{y}
  \exp(-\frac{1}{2\frac{1}{n}}(\ln y -\mu)^2)
  dy
\end{equation}
which is a Log-Normal distribution with mean $\ln \theta - 1$ and variance $\frac{1}{n}$.
