Let $(x_n)$ be a sequence defined for $n≥1$ so that $x_{n+1}\leq x_n^2 \leq x_{n-1}$ holds for every $n \ge 2$. Is it convergent? In case of convergence, what is $\lim_{n\to \infty} x_n$?


Note that all terms $x_n$ are nonnegative.

Let $y_n := x_{2n+1}$. Then $(y_n)$ is decreasing (or "nonincreasing" as some say) and nonnegative, hence convergent, say to $\ell$. Then $x_{2n}^2\to \ell^2$ by the squeeze theorem.

Now let $z_n:=x_{2n}$. By the same argument, $z_n\to\ell^2$ and $y_n\to \ell^4$.

Altogether, $\ell=\ell^4$. Hence $\ell=0$ or $\ell=1$. In either case $\ell=\ell^2=\ell^4$ and the entire sequence converges to $\ell$.

Both choices are possible: $x_n = 0$ for all $n$ gives $0$; $x_n=1$ for all $n$ gives $1$.


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