Let $\ x_{n+1}= \ x_n + \dfrac {n^{1/2}}{\ x_n}$, where $n\geq 1$ with $x_1=1$. Study the convergence of $\dfrac{x_n}{n^{3/4}}$.
Firstly, this is increasing as $x_1=1$ and $x_n$ is bigger than $1$, but if I try to use Stolz-Cesaro's lemma it fails. Why? Is there a more sofisticated way to attack this?
Stolz-Cesaro's lemma is a useful tool here! If $L\geq 0$ is the limit of $\dfrac{x_n}{n^{3/4}}$ then $$L=\lim_{n\to \infty}\dfrac{x_n}{n^{3/4}}\stackrel{\text{SC}}{=}\lim_{n\to \infty}\dfrac{x_{n+1}-x_n}{(n+1)^{3/4}-n^{3/4}}=\lim_{n\to \infty}\dfrac{\frac{n^{1/2}}{x_{n}}}{\frac{3n^{3/4}}{4n}}=\frac{4}{3}\lim_{n\to \infty}\dfrac{n^{3/4}}{x_n}=\frac{4}{3L}.$$ which implies that $3L^2=4$, that is $L=\frac{2}{\sqrt{3}}$. It remains to show that the limit exists (prove that the positive sequence $\dfrac{x_n}{n^{3/4}}$ is eventually decreasing)!