Problem: Let {$a_{n}$} be a sequence with $0 \lt a_1 \lt 1$, and $a_{n+1}=a_n(1-a_{n})$. Prove $\lim_{n\to \infty}n a_{n}=1$.

My try: I have figured out {$a_{n}$} is decreasing and converges to $0$. But how can I relate it to {${na_{n}}$} ?

Any hint to start, thanks!

  • $\begingroup$ I think this problem was given at Putnam, 1966. $\endgroup$ – mickep Dec 10 '16 at 12:24

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: $$\lim_{n\to \infty}n a_{n}=\lim_{n\to \infty}\frac{n}{1/a_n} \stackrel{\text{SC}}{=}\lim_{n\to \infty}\frac{(n+1)-n}{1/a_{n+1}-1/a_n}=\lim_{n\to \infty}\frac{1}{\frac{1}{a_n(1-a_n)}-\frac{1}{a_n}}= \lim_{n\to \infty}(1-a_n)=1.$$


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