$0 \le x_0 \le \frac{1}{2}$ , and $x_{n+1}=x_n-\dfrac{4x_n^3}{n+1}$

When I take $x_0=\sqrt{\frac{1}{12}}$, it converges very very slow. I can see it is monotonic decreasing but don't know how to find its limit.


Let's study $$x_{n+1}^{-2} - x_n^{-2} = \left( x_n - \frac{4x_n^3}{n+1} \right)^{-2} - x_n^{-2} = x_n^{-2} \left( \left( 1- \frac{4x_n^2}{n+1}\right)^{-2} -1\right)$$

So $$x_{n+1}^{-2} - x_n^{-2} = x_n^{-2} \left( \frac{8x_n^2}{n+1} + o\left(\frac{x_n^2}{n+1} \right) \right) = \frac{8}{n+1} + o\left( \frac{1}{n+1} \right)$$

So you get that $$x_{n+1}^{-2} - x_n^{-2} \sim \frac{8}{n+1}$$

The series $\sum \frac{8}{n+1}$ diverges, so we can sum to obtain $$x_{n+1}^{-2} \sim \sum_{k=1}^n \frac{8}{k+1}$$

You obtain finally $$x_{n+1} \sim \frac{1}{\sqrt{\sum_{k=1}^n \frac{8}{k+1}}}$$

In particular, the limit is indeed $0$.

  • $\begingroup$ That's just amazing! $\endgroup$ – Zhenduo Cao Sep 30 '18 at 19:16
  • $\begingroup$ Yes, the technic to study $u_{n+1}^{\alpha} - u_n^{\alpha}$ is very powerful ! $\endgroup$ – TheSilverDoe Sep 30 '18 at 19:20

If $x_0=0$ or $x_0=\frac12$, then $x_1=0$ and so $x_n=0$ for all $n\ge1$.

If $0\lt x_n\lt\frac12$, then $$ \begin{align} x_{n+1} &=x_n-\frac{4x_n^3}{n+1}\\ &=x_n\left(1-\frac{4x_n^2}{n+1}\right)\\[3pt] &\in(0,x_n) \end{align} $$ This means that $x_n$ is decreasing and bounded below; therefore, it converges.

Suppose the limit is $a$. Then, because $x_n$ is decreasing and bounded below by $0$, $$ \begin{align} x_0-x_m &=\sum_{n=0}^{m-1}(x_n-x_{n+1})\\ &=\sum_{n=0}^{m-1}\frac{4x_n^3}{n+1}\\ &\ge\sum_{n=0}^{m-1}\frac{4a^3}{n+1}\\ &=4a^3H_m \end{align} $$ Thus, $$ a^3\le\frac{x_0}{4H_m}\to0. $$

  • $\begingroup$ That's quite clear $\endgroup$ – Zhenduo Cao Oct 1 '18 at 20:39

Notice that $x_{n+1}=\left(1-\frac{x_n^2}{n+1}\right)x_n$ and, therefore, $$x_{n+1}=x_0\prod_{k=0}^{n-1}\left(1-\frac{4x_k^2}{n+1}\right)$$

Now, recall this standard result on infinite products:

Let $0\le u_n<1$; then $\lim\limits_{n\to\infty} \prod\limits_{k=0}^n (1-u_k)\ne 0$ if and only if $\sum\limits_{k=0}^\infty u_k<\infty$.

If $\inf_n x_n=\beta>0$ then we'd have $\sum_{k=0}^\infty\frac{4x_k^2}{n+1}\ge 4\beta^2\sum_{k=1}^\infty \frac1n=\infty$. But by the previous theorem, this conclusion implies $\lim_{n\to\infty} x_n=0$.

  • $\begingroup$ Thanks. Really powerful result. $\endgroup$ – Zhenduo Cao Oct 1 '18 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.