LQD_QN2020 wrote: Let $(x_n)$ define by $x_1\in \left(0;\dfrac{1}{2}\right)$ and $x_{n+1}=3x_n^2-2nx_n^3$ for all $n\ge 1$. a) Prove that $(x_n)$ convergence to $0$. From $x_1\in(0,\frac 12)$, we easily get $x_2\in(0,\frac 12)$ and $x_3\in(0,\frac 14)$ So we have $x_n\in(0,\frac 3{4n})$ for $n=3$ Then, considering $x_n\in(0,\frac 3{4n})$, we get $x_{n+1}\ge 0$ and $x_{n+1}\le 3x_n^2\le \frac {27}{16n^2}\le \frac 3{4(n+1)}$ And so $x_n\in(0,\frac 3{4n})$ $\implies$ $x_{n+1}\in(0,\frac 3{4(n+1)})$ Hence the convergence with limit $0$

LQD_QN2020 wrote: Let $(x_n)$ define by $x_1\in \left(0;\dfrac{1}{2}\right)$ and $x_{n+1}=3x_n^2-2nx_n^3$ for all $n\ge 1$. a) Prove that $(x_n)$ convergence to $0$. b) For each $n\ge 1$, let $y_n=x_1+2x_2+\cdots+n x_n$. Prove that $(y_n)$ has a limit. See here for a complete solution also including part b).

$Question \quad 1$. Write for an integer $n\geq 1$, $$x_{n+1}=\dfrac{x_n}{2n}\times (2nx_n)\times (3-2nx_n) $$Hence, by AM-GM inequality, we get $$|x_{n+1}|\leq \dfrac{|x_n|}{2n}\times \bigg(\dfrac{2nx_n+3-2nx_n}{2}\bigg)^2 $$If there's an index $n_0$ such that $x_{n_0}=0$, then the sequence will vanishing and its limit is clearly $0$. Otherwise, we get for all $n\geq 2$, $$|x_{n+1}|\leq \dfrac{9}{16}\times |x_n|,\quad\quad\quad\quad\quad (*) $$Finally, the sequence $(x_n)$ converges and its limit is $0$. $Question \quad 2$. From $(*)$, we get for all $n\geq 1$ $$\dfrac{(n+1)|x_{n+1}|}{n|x_n|}\leq \dfrac{9(n+1)}{n^2} $$$$\dfrac{(n+1)|x_{n+1}|}{n|x_n|}\to 0\quad\quad \text{when}\quad\quad n\to \infty $$It follows by D'alembert's criterion that the serie $\displaystyle \sum_n nx_n$ converges.