Let {$a_{n}$} be a sequence with $0 \lt a_1 \lt 1$, and $a_{n+1}=a_n(1-a_{n})$. Prove $\lim_{n\to \infty}n a_{n}=1$.
Solution 1:
Since $(a_n)_n$ is eventually decreasing to $0^+$ then $(1/a_n)_n$ is a strictly increasing and divergent sequence, and we can use Stolz-Cesaro theorem: $$\lim_{n\to \infty}n a_{n}=\lim_{n\to \infty}\frac{n}{1/a_n} \stackrel{\text{SC}}{=}\lim_{n\to \infty}\frac{(n+1)-n}{1/a_{n+1}-1/a_n}=\lim_{n\to \infty}\frac{1}{\frac{1}{a_n(1-a_n)}-\frac{1}{a_n}}= \lim_{n\to \infty}(1-a_n)=1.$$