The exact value of $\lim_{n \to \infty} \frac{a_n}{\sqrt[4]{n^3}}$ [duplicate]

Let $a_1=1$ and $a_{n+1}=a_n+\frac{\sqrt{n}}{a_n}$. If it exists, find $$\lim_{n \to \infty} \frac{a_n}{\sqrt[4]{n^3}}$$

Obviously, $a_n \to \infty$ and $a_n$ is strictly increasing. I tried to get rid of that $n$ in the limit and get only $a_n$ terms in the numerator so that I could substitute $a_n$ to $x \to \infty$ and maybe use L'Hospital to finish. But $a_{n+1}, a_n$ and $\sqrt{n}$ are so tightly connected that I didn't manage to do that.

Another approach was Cesaro-Stolz, writing the limit as $$\lim_{n \to \infty}\left(\sqrt[4]{\frac{\sqrt[3]{a_n^4}}{n}}\right)^3=\left(\lim_{n \to \infty}\left(\sqrt[3]{a_{n+1}^4}-\sqrt[3]{a_n^4}\right) \right)^\frac{3}{4}$$ but I encountered the same issue. I'm even wondering if there is a nice closed form for the limit.

Edit: As pointed below, if the limit exists, then its value is $\frac{2}{\sqrt{3}}$. How can we prove that it exists?

marked as duplicate by Martin R, Community♦Jan 12 '18 at 13:58

Assume the limit $$l=\lim_{n \to \infty} \frac{a_n}{n^{3/4}}$$ exists and is nonzero, then it is very easy to see that $l=2/{\sqrt{3}}$. To see why, consider $$\lim_{n\to\infty} \frac{{{a_{n + 1}} - {a_n}}}{{{{(n + 1)}^{3/4}} - {n^{3/4}}}} =\lim_{n\to\infty} \frac{{\sqrt n }}{{{{(n + 1)}^{3/4}} - {n^{3/4}}}}\frac{1}{{{a_n}}} = \frac{4}{3}\lim_{n\to\infty}\frac{{{n^{3/4}}}}{{{a_n}}} = \frac{4}{3l}$$
Hence by Stolz's theorem, $l=4/(3l)$.
• If it exists, it must be nonzero, because $a_n \geq \sqrt[4]{n^3}.$ I tried to use this inequality in order to prove that $\frac{a_n}{\sqrt[4]{n^3}}$ is eventually decreasing, but it is not strong enough – Shroud Jan 10 '18 at 14:03