, 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 $\frac{x_{n+1}}{n+1} = \left(1 + \frac{1}{n^2}\right) \cdot \frac{x_n}{n} \quad \forall n \ge 1$.
Hence, for all $n \ge 2$
$$y_n = x_{n+1} - x_n = \left( \frac{(n+1)(n^2+1)}{n^3} - 1 \right) x_n = \frac{n^2+n+1}{n^2} \cdot \frac{x_n}{n} = \left(1 + \frac{n+1}{n^2}\right) \prod_{k=1}^{n-1} \left(1 + \frac{1}{k^2}\right). \quad (1)$$
Whence, with the notice that $y_1 = x_2 - x_1 = 3$, we have $y_n > 0 \quad \forall n \ge 1$, $y_1 < y_2$ 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
$$y_n < 2 \left( 1 + \frac{\sum_{k=1}^{n-1} \frac{1}{k^2}}{n-1} \right)^{n-1} \quad \forall n \ge 2. \quad (3)$$
But $\sum_{k=1}^{n-1} \frac{1}{k^2} < 1 + \sum_{k=2}^{n-1} \frac{1}{k(k-1)} = 1 + \sum_{k=2}^{n-1} \left(\frac{1}{k-1} - \frac{1}{k}\right) = 2 - \frac{1}{n-1} < 2 \quad \forall n \ge 3$
Hence, (3) implies $y_n < 2 \left(1 + \frac{2}{n-1}\right)^{n-1} < 2e^2 \quad \forall n \ge 2$.
Hence $(y_n)$ is bounded from above. Together with (2) this implies that $(y_n)$ is convergent. ■