Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$. Show that $a_{n+1} = a_{n} + a_{n}^{\frac{1}{3}}$ satisfies $\lim_{n\to\infty}\frac{a_n}{pn^q}=1$ for some real valued $p, q$.
I had hoped that I could use a strategy similar to Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$, wherein the author found that $\lim_{n\to\infty}\sqrt{a_{n+1}}-\sqrt{a_n}=\frac{1}{2}$ and used the telescoping sum $\sum_{k=1}^{n-1}\sqrt{a_{k+1}}-\sqrt{a_k}\approx\frac{1}{2}n$ to approximate $\sqrt{a_n}$. However, $a_{n+1}^{\frac{1}{3}}-a_{n}^{\frac{1}{3}}=\frac{a_{n+1}-a_{n}}{a_{n+1}^{\frac{2}{3}}+a_{n+1}^{\frac{1}{3}}a_{n}^{\frac{1}{3}}+a_{n}^{\frac{2}{3}}}=
\frac{a_{n}^\frac{1}{3}}{a_{n+1}^{\frac{2}{3}}+a_{n+1}^{\frac{1}{3}}a_{n}^{\frac{1}{3}}+a_{n}^{\frac{2}{3}}}=
\frac{1}{\frac{a_{n+1}^{\frac{2}{3}}}{a_n^{\frac{1}{3}}}+a_{n+1}^{\frac{1}{3}}+a_{n}^{\frac{1}{3}}}
$ which converges to $0$. I also tried $a_{n+1}^q-a_n^q$ for different positive and negative values of $q$ to no avail.
If anyone has any hints, I'd appreciate it. Thanks.

If $a_0=0$ then the result is not true.
I'm assuming that $a_0>0$.
It's easy to check that $\{a_n\}$ is non-decreasing. So it either diverges to $+\infty$ or converges to a finite number $l>0$.
If the latter, then $l$ must verify $l=l+l^{\frac 1 3}$, which is a contradiction. Thus the sequence diverges to $+\infty$.
Let $\alpha$ be a positive number. Then
$$\begin{split}
a_{n+1}^\alpha &= \left(a_n + a_n^{\frac 1 3} \right)^\alpha\\
&= a_n^\alpha \left( 1+a_n^{-\frac 2 3}\right)^\alpha\\
&=a_n^\alpha\left( 1+\alpha a_n^{-\frac 2 3}+o\left ( a_n^{-\frac 2 3}\right)\right)\\
&=a_n^\alpha + \alpha a_n^{\alpha -\frac 2 3}+o\left ( a_n^{\alpha-\frac 2 3}\right)
\end{split}$$
Thus, selecting $\alpha=\frac 2 3$ yields
$$a_{n+1}^{\frac 2 3} = a_n^{\frac 2 3} + \frac 2 3 +o(1)$$
Summing this gives
$$a_n^{\frac 2 3}=\frac{2n}3+o(n)$$
Equivalently,
$$a_n=\left(\frac {2n}3\right)^{\frac 3 2}+o\left(n^{\frac 3 2}\right)$$
$$\boxed{\lim_{n\rightarrow+\infty} \frac{a_n}{\left(\frac {2n}3\right)^{\frac 3 2}}=1}$$
