Vijetnam 2011

For all $n\ge 1$, we have $$x_{n+1}=\frac{2(n+1)}{n^2}\cdot \sum _{i=1}^n{x}_i=\frac{2(n+1)}{n^2}\left(\frac{(n-1{)}^2}{2n}+1\right)x_n=\frac{(n+1)(n^2+1)}{n^3}{x}_n.$$ Consequently $\frac{x_{n+1}}{n+1}=\left(1+\frac{1}{n^2}\right)\cdot \frac{x_n}{n}\forall n\ge 1$. Hence, for all $n\ge 2$ $$y_n=x_{n+1}-x_n=\left(\frac{(n+1)(n^2+1)}{n^3}-1\right)x_n=\frac{n^2+n+1}{n^2}\cdot \frac{x_n}{n}=\left(1+\frac{n+1}{n^2}\right)\prod _{k=1}^{n-1}\left(1+\frac{1}{k^2}\right).(1)$$ Whence, with the notice that $y_1=x_2-x_1=3$, we have $y_n>0\forall n\ge 1$, $y_1<y_2$ and for all $n\ge 3$ $$\frac{y_n}{y_{n-1}}=\frac{n^2+n+1}{n^2}\cdot \frac{(n-1{)}^2}{(n-1{)}^2+n}\cdot \left(1+\frac{1}{(n-1{)}^2}\right)=1+\frac{2}{n^4-n^3+n^2}>1.$$ Consequently $(y_n)$ is an increasing sequence. (2)

Since for all $n\ge 2$ we have $n+1<n^2$ and ${\prod }_{k=1}^{n-1}\left(1+\frac{1}{k^2}\right)\le {\left(1+\frac{{\sum }_{k=1}^{n-1}\frac{1}{k^2}}{n-1}\right)}^{n-1}$, it follows from (1) that $$y_n<2{\left(1+\frac{{\sum }_{k=1}^{n-1}\frac{1}{k^2}}{n-1}\right)}^{n-1}\forall n\ge 2.(3)$$ But ${\sum }_{k=1}^{n-1}\frac{1}{k^2}<1+{\sum }_{k=2}^{n-1}\frac{1}{k(k-1)}=1+{\sum }_{k=2}^{n-1}\left(\frac{1}{k-1}-\frac{1}{k}\right)=2-\frac{1}{n-1}<2\forall n\ge 3$ Hence, (3) implies $y_n<2{\left(1+\frac{2}{n-1}\right)}^{n-1}<2e^2\forall n\ge 2$. Hence $(y_n)$ is bounded from above. Together with (2) this implies that $(y_n)$ is convergent. ■