Given the following function:
$W_n = \frac{1}{\sqrt{n}}\Pi_{k=1}^{\infty}\log(U_k)$ where $U_k$ is uniformly distributed from $1$ to $e$.
Does $\{W_n\}_{n\geq 1}$ converge in distribution?
I found that: $\log(\Pi_{k=1}^{\infty}U_k)$ is equivalent to $\sum_{k=1}^\infty \log(U_k)$ which can be called $S_n$.
$W_n$ can then be written as $\frac{S_n}{\sqrt{n}}$. The answers says that this function has the distribution $\mathcal{N}(\sqrt{n}E[\log(U_k)],V[\log(U_k])$ where $V$ is variance. The final result is that it does not converge because its mean is dependent on $n$.
I am having trouble understanding where the multiplication by $\sqrt{n}$ comes from.
I know that $\frac{S_n-n\mu}{\sqrt{n\sigma^2}}$ has the distribution $\mathcal{N}(0,1)$ by the central limit theorem. I also saw in my book that $\frac{S_n-n\mu}{\sqrt{n}}$ has the distribution $\mathcal{N}(0,\sigma^2)$.
I am just having trouble understanding how modifying that equation has the corresponding effect on the distribution.
I believe by $S_n$, you intend to mean $S_n = \sum_{k=1}^n \log(U_k).$
\begin{align} E\left[ \frac{S_n}{\sqrt{n}}\right] &= \frac1{\sqrt{n}}E\left[\sum_{k=1}^n \log(U_k) \right] \\ &= \frac1{\sqrt{n}}\left(n E(\log(U_1 \right)))\\ &= \sqrt{n} E(\log(U_1)) \end{align}
\begin{align} V\left[ \frac{S_n}{\sqrt{n}}\right] &= \frac1{n}V\left[\sum_{k=1}^n \log(U_k) \right] \\ &= \frac1{n}\left(n V(\log(U_1 \right)))\\ &= V(\log(U_1)) \end{align}