Let $V_n$ be a value after $n$ time, and at $n = 0, V_0 = 1$, and then it changed by a factor $U_n$ in $n$-th interval. which means that $V_n = U_1 *...* U_n$ and each $U$ is a independent uniform distributed in $(0,k)$ with parameter $k$.
Now by the law of large number that $(V_n)^{1/n}$ approaches to a number $L$ when $n\to\infty$.
How can we evaluate $L$ with $k$?
I tried:
to separate the $V_n$ with the log into $U$'s but not sure how to proceed the next step.
Thank you!
You asked a follow-up question, which I was planning to answer when you deleted it. What I was intending to say was:
$\log(K)-\log(U_n) \sim Exp(1)$
so $\log(U_n)$ has mean $\log(K)-1$ and variance and standard deviation $1$
so, since $\log(V_0)=0$, we have $\frac1n \log(V_n) \xrightarrow{LLN} \log(K)-1$ and $V_n^{1/n} \xrightarrow{LLN} \frac{K}{e}$ and this is $L$
and we have $\frac1{\sqrt{n}}\left(\log(V_n) -n(\log(K)-1)\right) \xrightarrow{d} \mathcal N(0,1)$ in distribution using the CLT
so $\left(\frac{V_n}{L^n}\right)^{1/n} \xrightarrow{d} Lognormal(0,1)$ in distribution
meaning $W \sim Lognormal(0,1)$ with mean $\sqrt{e}$ and variance $e^2-e$ and $E[W^n]=e^{n^2/2}$