Define a sequence $J_n$ such that $J_1=1/2$, $J_2=k$, and $$J_n=\prod_{m=1}^{n-1} J_m$$ Determine, if it exists, the value of $k$ for which $J_n$ goes to 1 as $n$ tends to infinity.
What if $J_1$ is given an arbitrary value other than $1/2$? Is there then a way to determine the value of $J_2$ such that $J_n$ goes to 1 as $n$ tends to infinity?
We have $$\frac{J_{n+1}}{J_n}=\frac{\prod_{m=1}^{n}J_m}{\prod_{m=1}^{n-1}J_m} =J_{n-1},$$ so we have the Fibonacci-like recurrence $$J_{n+1}=J_nJ_{n-1}$$ and so $$\log_2 J_{n+1}=\log_2 J_n +\log_2 J_{n-1}.$$ We have $\log_2J_1=-1$ and want to determine $\log_2J_2$ so that $\log_2J_n\to 0$. It is well.known that $$ \log_2 J_n=c_+\cdot\lambda_+^n+c_-\lambda_-^n$$ where $\lambda_{\pm}=\frac{1\pm\sqrt 5}{2}$ and $c_{\pm}$ are determined by the first two terms via the equations $$\begin{align}\log_2J_1&=c_+\lambda_++c_-\lambda_-\\\log_2J_2&=c_+\lambda_+^2+c_-\lambda_-^2\end{align} $$ Hence we have $\log_2J_n\to 0$ iff $c_+=0$, in other words, iff $$ \log_2J_2=\lambda_-\cdot\log_2J_1.$$ With $\log_2J_1=-1$ given, we need $$ J_2=2^{-\frac{1-\sqrt 5}{2}}\approx 1.53478225382648941.$$
Let us first determine a closed form for $J_n$. With some work we arrive at $$ J_n = \frac{k^{F_{n-2}}}{2^{F_{n-3}}}$$ where $F_n$ denotes the $n$-th Fibonacci number ($F_0 = 1, F_1 = 1, F_2 = 2$ etc)
For $J_n$ to be one we want $k^{F_{n-2}} = 2^{F_{n-3}}$, or $k = 2^{\frac{F_{n-3}}{F_{n-2}}}$. As $n$ goes to infinity, this tends to $2^{1/\varphi}$ where $\varphi$ is the golden ratio.
This shows that (check) if we take $k = 2^{1/\varphi}$ then $J_n = 2^{ F_{n-2} / \varphi - F_{n-3}}$ so tends to $2^0 = 1$.
