<s>[INST] 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.
 [/INST] 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 $0$.
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