# Limit of sequence with relation $x_{n+1}=x_n-x_n^2$ [duplicate]

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.

## marked as duplicate by rtybase, max_zorn, Lord Shark the Unknown, Eric Wofsey real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 27 at 6:58

• 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? – Sasha Jan 26 at 22:28
• I'd start by observing that $$\frac1{x_{n+1}}=\frac{1}{x_n}+1+O(x_n).$$ – Lord Shark the Unknown Jan 26 at 22:30
• @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 – ZFR Jan 26 at 22:56

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.
• Yes, see the other solution. Or prove that $\sum_{k=1}^n1/k=o(n)$. @ZFR – Lord Shark the Unknown Jan 27 at 10:44
$$\lim_{n\to\infty}nx_n=\lim_{n\to\infty}\frac{n}{\frac{1}{x_n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}\frac{x_nx_{n+1}}{x_n-x_{n+1}}=\lim_{n\to\infty}\frac{x_n^2-x_n^3}{x_n^2}=\lim_{n\to\infty}(1-x_n)=1$$