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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.

person
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    As a heuristic, I'd expect the behavior to be roughly like the solution to the diff eq $y' = y^{1/3}$. That has solution $y' y^{-1/3} = 1 \implies 3/2 y^{2/3} = x+C \implies y = (\frac{2}{3} (x+C))^{3/2}$ which is asymptotically like $(2/3)^{3/2} x^{3/2}$. So, I would expect the finite difference equation to have solutions asymptotic to $(2/3)^{3/2} n^{3/2}$, i.e. $p = (2/3)^{3/2}$ and $q = 3/2$. – Daniel Schepler Aug 26 '22 at 00:31

2 Answers2

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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, $$a_n=\left(\frac {2n}3\right)^{\frac 3 2}+o\left(n^{\frac 3 2}\right)$$ $$\boxed{\lim_{n\rightarrow+\infty} \frac{a_n}{\left(\frac {2n}3\right)^{\frac 3 2}}=1}$$

Stefan Lafon
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  • Thanks! That sort of generalizes what was done for $a_{n+1}=a_n + \sqrt{a_n}$. The same technique you used generalizes to any r where $a_{n+1}=a_n + a_n^r$. Must r < 1 for it to be true that $(1 + a_n^{r - 1} )^\alpha = (1 + \alpha a_n^{r-1}+O(a_n^{r-1})$ ? If so, can we get around that somehow? – person Aug 26 '22 at 05:30
  • Well, in the case of $r=1$, the equation would amount to $a_{n+1} = 2 a_n$ which has solution $a_n = c \cdot 2^n$. For $r>1$, I'd expect the $a_n^r$ term to dominate and give an asymptotic behavior of $\log a_n \sim c \cdot 2^n$ (which doesn't necessarily imply that $a_n \sim c^{2^n}$ for any constant $c$). – Daniel Schepler Aug 26 '22 at 15:54
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Hint: It suffices to show that there exists a real number $q$ such that $\lim_{n\rightarrow \infty}{{a_n}\over{n^q}}$ converges. Once this is done, call the value $p$, and then divide both sides by $p$.

Andrew Sotomayor
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