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Let $s_1 = \frac{1}{4}$ and $s_{n+1} = s_n(1-s_n)$ for all $n\geq2$. Show that $\lim_{n\to\infty} ns_n =1.$

Can someone please help me to solve this problem? I have couldn't figure out.

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1) $s_1>s_2>s_3>\dotsb$. let $\lim s_n =l$, then $$l=l(1-l)$$ $l=0$. that is $$\lim_{n\to\infty} s_n =0$$ so $$\lim_{n\to\infty} \frac{s_{n+1}}{s_n}=\lim_{n\to\infty}(1-s_n)=1$$ 2)Stolz's theorem

\begin{equation} \begin{split}\lim_{n\to\infty} \frac1{ns_n} &=\lim_{n\to\infty} \dfrac{\frac1{s_n}}n\\&=\lim_{n\to\infty} \left(\frac1{s_{n+1}}-\frac1{s_n}\right)\\ &=\lim_{n\to\infty} \frac{s_n-s_{n+1}}{s_{n+1}s_n}\\ &=\lim_{n\to\infty} \frac{s_n^2}{s_{n+1}s_n}\\ &=\lim_{n\to\infty} \frac{s_n}{s_{n+1}}\\&=1& \end{split} \end{equation}

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