For $Z_n:=\exp (X_1 \dotsm X_n)$, does $\frac{\left(Z_n-E\left(Z_n\right)\right)}{\log (n) \sqrt{n}}$ converge to 0 as $n \to\infty$ with prob 1? $X_1, \dotsc, X_n$ be independent valued random variables (not necessarily identically distributed) taking values on $[0,1]$.
For $Z_n:=\exp (X_1 \dotsm X_n)$, does $\frac{\left(Z_n-E\left(Z_n\right)\right)}{\log (n) \sqrt{n}}$ converge to 0 as $n \to\infty$ with prob 1?
And does it work for $X_{i}$ in whole line?
 A: Yes of course to the first question, and no of course to the second question.
Indeed, if the $X_i$'s take values only in $[0,1]$, then $X_1\cdots X_n$ takes values only in $[0,1]$. So, then $Z_n$ takes values only in $[1,e]$ and for
$$Y_n:=\frac{Z_n-EZ_n}{\sqrt n\,\ln n}$$
we have $|Y_n|\le\frac{2e}{\sqrt n\,\ln n}$, so that $Y_n\to0$ with probability $1$.

As for the second question, suppose that the $X_i$'s are independent and $P(X_i=1)=\frac12=P(X_i=e)$ for each $i$. Then
$$B_n:=\ln(X_1\cdots X_n)=\sum_{i=1}^n \ln X_i$$
has the binomial distribution with parameters $n,1/2$. So,
$$Z_n=e^{e^{B_n}}$$
and
$$EZ_n=\sum_{i=1}^n e^{e^i}\binom ni\frac1{2^n}\sim \frac{e^{e^n}}{2^n}$$
as $n\to\infty$. On the other hand, $P(B_n\le n/2)\to1/2$ by the central limit theorem. So,
$$P(Z_n\le e^{e^{n/2}})\to1/2$$
and hence
$$P\Big(|Y_n|\ge\frac{e^{e^n}(1-o(1))}{2^n \sqrt n\,\ln n}\Big)
\ge P\Big(Y_n\le-\frac{e^{e^n}(1-o(1))}{2^n \sqrt n\,\ln n}\Big)
\ge\frac12-o(1).$$
So, $Y_n\not\to0$ with probability $1$. In fact, by the 0--1 law, $\lim\sup_{n\to\infty}|Y_n|=\infty$ with probability $1$.
For an illustration, below are the connected graphs $\{(n,\frac1n\,\ln\ln|y_n|)\colon n=2,\dots,20\}$ for two simulated realizations $(y_2,\dots,y_n)$ of the random sequence $(Y_2,\dots,Y_n)$ in the latter setting:

These graphs confirm that $\lim\sup_{n\to\infty}|Y_n|=\infty$ with probability $1$.
