Cesaro-Stolz limit We have $x_{n+1}=x_n+\dfrac{\sqrt n}{x_n}$ with $x_1=1$ and we have to compute $\displaystyle\lim_{n\to\infty} \dfrac{x_n}{n^{3/4}}$. I already proved that $x_n$ is strictly increasing and tends to infinity and I know that we can use the Cesaro-Stolz Lemma to solve it but it hasn't been successful for me. I tried to apply it for $\displaystyle\lim_{n\to\infty} \dfrac{(x_n)^4}{n^3}$, but got stuck.
 A: Let us start by assuming that $\lim_{n\to +\infty}\frac{x_n}{n^{3/4}}=L>0$. By Cesàro-Stolz,
$$ L = \lim_{n\to +\infty}\frac{x_{n+1}-x_n}{(n+1)^{3/4}-n^{3/4}}=\lim_{n\to +\infty}\frac{\sqrt{n+1}}{x_{n+1}\left(\frac{3}{4}(n+1)^{-1/4}+O((n+1)^{-5/4})\right)}=\frac{4}{3L}$$
hence $L=\frac{2}{\sqrt{3}}$ and we just need to prove that the wanted limit exists. Let $y_n=x_n^2$. We have
$$ y_{n+1} = y_n + 2\sqrt{n} + \frac{n}{y_n} \tag{1}$$
and it is not difficult to prove by induction that
$$ \forall n\geq1,\qquad y_{n+1}-y_n \geq \frac{4}{3}\left[\left(n+\tfrac{1}{2}\right)^{3/2}-\left(n-\tfrac{1}{2}\right)^{3/2}\right] \tag{2}$$
leading to $y_{n}\geq \frac{4}{3}\left[(n-1/2)^{3/2}-(1/2)^{3/2}\right]$. On the other hand, by plugging this inequality back into $(1)$ we also get
$$ y_{n+1}-y_n \leq \frac{4}{3}\left[\left(n+\tfrac{1}{2}\right)^{3/2}-\left(n-\tfrac{1}{2}\right)^{3/2}\right]+2\left[\left(n+\tfrac{1}{2}\right)^{1/2}-\left(n-\tfrac{1}{2}\right)^{1/2}\right]\tag{3} $$
and $y_n = \frac{4}{3} n^{3/2}+ O(n^{1/2})$, implying $x_n = \frac{2}{\sqrt{3}}n^{3/4}+O(n^{-1/4})$, is proved.
It follows that $L=\frac{2}{\sqrt{3}}$, also without invoking Cesàro-Stolz.
