If $a_{n+1}=a_n(1-a_n)$ is given, prove that $\lim_{n\to\infty} a_n=0$ and that $\lim_{n\to\infty}(na_n)=1$ if $a_1 \in(0,1)$.

Question

Now, I have proven that sequence $a_n$ is monotonically decreasing and that its range is $a_n \in (0,1)$, so that $\lim_{n\to\infty} a_n=0$, but I'm really struggling to prove the other part of the question.

Answer 1

We have that

$$na_n =\frac{n}{\frac1{a_n}}\to 1$$

indeed by Stolz-Cesaro

$$\frac{n+1-n}{\frac1{a_{n+1}}-\frac1{a_n}}=\frac{1}{\frac1{a_{n+1}}-\frac1{a_n}} \to 1$$

with

$$\frac1{a_{n+1}}-\frac1{a_n}=\frac{a_{n}-a_{n+1}}{a_{n+1}a_{n}}=\frac{a_n^2}{a_n^2(1-a_n)}=\frac1{1-a_n} \to 1$$