Let $(x_n)_n $ such that $\displaystyle\frac {2018}{x_n} =\left(\frac {x_{n+1}}{x_n}\right)^{(n+1)/2}$, $\forall n\geq 1$.
I have to study the convergence of the sequence.
I have no idea how to start.
2026-03-25 06:00:33.1774418433
Could be $(x_n)_n $ convergent?
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Let $y_n=x_n^n$. Then the recursion is equivalent to $$y_{n+1}=2018^2y_n. $$ Find a closed form for $y_n$ in terms of $y_0$ from this and draw conclusions about $x_n$.