how to prove that this sequence converges to $0$? $0 \le x_0 \le \frac{1}{2}$ ,  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.
 A: 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 $0$.
A: 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,
$$
a^3\le\frac{x_0}{4H_m}\to0.
$$
A: 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 $\lim_{n\to\infty} x_n=0$.
