The convergence of a sequence This problem is form the Romanian GM.
Let $(x_{n})_{n\geq1}$ be a sequence of real numbers defined by $x_{1}=1$ and $$x_{n+1}=x_{n}+ \frac{\sqrt{n}}{x_{n}}$$ for all $n \geq 1$. Study the convergence of the sequence $$(\frac{x_{n}}{n^{\frac{3}{4}}})_{n\geq1}$$
I only managed to show from the given relation that $(x_{n})_{n\geq1}$ is strictly increasing and tends to $+\infty$.
 A: Well, $$x_{n+1}=x_{n}+ \frac{\sqrt{n}}{x_{n}}$$ implies
$$x^2_{n+1}=x^2_{n}+ 2\sqrt{n}+\frac{n}{x^2_{n}},$$ so $$x^2_n=2\sum^{n-1}_{k=1}\sqrt{k}+O(\sqrt{n})=\frac43\,n^{3/2}+O(\sqrt{n}),$$
meaning the limit is $\displaystyle\frac2{\sqrt{3}}$, as @clark announced.
If that was a bit rushed: we get $$x^2_{n+1}\ge x^2_{n}+ 2\sqrt{n},$$ first, so
$$x^2_n\ge2\sum^{n-1}_{k=1}\sqrt{k}=\frac43\,n^{3/2}+O(\sqrt{n}),$$ and consequently $$\sum^{n-1}_{k=1}\frac{k}{x^2_{k}}=O(\sqrt{n}).$$ And $$x^2_n=\frac43\,n^{3/2}+O(\sqrt{n})=\frac43\,n^{3/2}\left(1+O\left(\frac1n\right)\right)$$ happens to be equivalent with $$x_n=\sqrt{\frac43}\,n^{3/4}\left(1+O\left(\frac1n\right)\right),$$ so
$$\lim_{n\to\infty}\frac{x_n}{n^{3/4}}=\sqrt{\frac43}=\frac2{\sqrt{3}}.$$
I'm sorry answers at MSE were never meant to explain entire concepts, but if you are unsure concerning the notation, check Big O notation, please. 
A: $$x_{n+1}=x_{n}+ \frac{\sqrt{n}}{x_{n}}\implies
x^2_{n+1}-x^2_{n}=2\sqrt{n}+\frac{n}{x^2_{n}}.$$
Using Stolz's theorem twice: first, consider limit
$$\lim_{n\to\infty}\frac{x^2_n}{n}
=\lim_{n\to\infty}(x^2_{n+1}-x^2_n)
=\lim_{n\to\infty}\left(2\sqrt{n}+\frac{n}{x_n^2}\right)=\infty,$$
which implies
$$\lim_{n\to\infty}\frac{n}{x^2_n}=0.$$
Second,
$$\lim_{n\to\infty}\frac{x^2_n}{\sqrt{n^3}}
=\lim_{n\to\infty}\frac{x^2_{n+1}-x^2_n}{\sqrt{(n+1)^3}-\sqrt{n^3}}
=\lim_{n\to\infty}\frac{2\sqrt{n}+\frac{n}{x_n^2}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\\
=\lim_{n\to\infty}\frac{\sqrt{(n+1)^3}+\sqrt{n^3}}{(n+1)^3-n^3}\cdot\left(2\sqrt{n}+\frac{n}{x_n^2}\right)\\
=\lim_{n\to\infty}\frac{2\sqrt{n^3}}{3n^2}\cdot\left(2\sqrt{n}+\frac{n}{x_n^2}\right)=\frac{4}{3}.$$
So
$$\lim_{n\to\infty}\frac{x_n}{n^{3/4}}=\frac2{\sqrt{3}}.$$
