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Let the sequence of real numbers is defined as follows: $x_1=\frac{1}{2}$ and $x_{n+1}=x_n-x_n^2$. Show that $\lim_{n\to \infty}nx_n=1$.

I've shown that the limit of $x_n$ is zero since this sequence is bounded and monotone. How to show that $nx_n\to 1$ as $n\to \infty$?

I have no ideas how to handle this problem.

It would be interesting to see approach.

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marked as duplicate by rtybase, max_zorn, Lord Shark the Unknown, Eric Wofsey real-analysis Jan 27 at 6:58

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  • $\begingroup$ Let $y_n = n x_n$. Then find the recurrence for $y_n$. The initial condition is known. Can you apply the same method to find the limit? $\endgroup$ – Sasha Jan 26 at 22:28
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    $\begingroup$ I'd start by observing that $$\frac1{x_{n+1}}=\frac{1}{x_n}+1+O(x_n).$$ $\endgroup$ – Lord Shark the Unknown Jan 26 at 22:30
  • $\begingroup$ @Sasha, i cannot answer you remark from the top of my head but i have done this approach before posting a question and if i am not mistaken this approach was not useful $\endgroup$ – ZFR Jan 26 at 22:56
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We have $$\frac1{x_{n+1}}=\frac1{x_n(1-x_n)}=\frac1{x_n}+1+x_n+x_n^2+\cdots$$ (a geometric series). Thus $$\frac1{x_{n+1}}>\frac1{x_n}+1$$ and so $$\frac1{x_n}\ge n-1+\frac1{x_1}=n+1.$$ Therefore $x_n=O(1/n)$. Then $$\frac1{x_{n+1}}=\frac1{x_n}+1+O(1/n)$$ and so $$\frac1{x_n}=n+O(\ln n).$$ That's enough.

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  • $\begingroup$ Quite nice solution? But is it possible to donit without logarithm? $\endgroup$ – ZFR Jan 27 at 1:19
  • $\begingroup$ Yes, see the other solution. Or prove that $\sum_{k=1}^n1/k=o(n)$. @ZFR $\endgroup$ – Lord Shark the Unknown Jan 27 at 10:44
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By Stolz-Cesaro's Lemma we get that
$\lim_{n\to\infty}nx_n=\lim_{n\to\infty}\frac{n}{\frac{1}{x_n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}\frac{x_nx_{n+1}}{x_n-x_{n+1}}=\lim_{n\to\infty}\frac{x_n^2-x_n^3}{x_n^2}=\lim_{n\to\infty}(1-x_n)=1$

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