Question 89
Let $a \gt 0$ and the sequence $(x_n)$ is defined by
$$ x_1 = a, $$
$$ x_{n+1} = x_n + \frac{\sqrt{x_n}}{n^2}, \forall n \ge 1. $$
Prove that $(x_n)$ has a finite limit when $n$ tends to infinity.
$$ x_1 = a, $$
$$ x_{n+1} = x_n + \frac{\sqrt{x_n}}{n^2}, \forall n \ge 1. $$
Prove that $(x_n)$ has a finite limit when $n$ tends to infinity.