VMO

a. First, we prove by induction that $0<x_n<\frac{3}{2n}$, $\forall n\ge 1$. The base case $n=1$ is trivial. For $n=2$, we have $$0<x_1^2(3-2x_1)<3x_1^2<\frac{3}{4}\Rightarrow 0<x_2<\frac{3}{4}.$$ Assume that $0<x_k<\frac{3}{2k}$ for some $k\ge 2$, by the AM-GM inequality, $$0<x_k^2(3-2k{x}_k)=\frac{1}{k^2}(k{x}_k)(k{x}_k)(3-2k{x}_k)\le \frac{1}{k^2}<\frac{3}{2(k+1)}.$$ Therefore, $0<x_{k+1}<\frac{3}{2(k+1)}$. The inductive step is completed. By the Squeeze theorem, we have ${\lim }_{n\to \infty }{x}_n=0$.

b. It is clear that $x_{n+1}=3x_n^2-2n{x}_n^3\le 3x_n^2$ implies $$x_{n+2}\le 3x_{n+1}^2\le 27x_n^4\le \frac{3^7}{16n^4}=\frac{C}{n^4},\forall n\ge 1.$$ For all $n>2$, we obtain that $$\begin{gathered} y_n=\sum _{i=1}^{n}i{x}_i=x_1+2x_2+\sum _{i=1}^{n-2}(i+2)x_{i+2} \\ \le x_1+2x_2+C\sum _{i=1}^{n-2}\frac{i+2}{i^4}\le x_1+2x_2+3C\sum _{i=1}^{n-2}\frac{1}{i^2}. \end{gathered}$$ Hence, $(y_n)$ is bounded above. Note that $(y_n)$ is increasing, we conclude that $(y_n)$ is convergent. $\square$