Limit $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 > 0 $

Question

Let $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.

I did the following:

We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:

$$ a_{n+1} - a_n = \frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = \frac{5 - a_n^{3}}{3a_n^{2}}$$

We see that for each $a_n > ^3\sqrt{5}$ the sequence is decreasing. That means the proof collapses to prove $a_n > ^3\sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3\sqrt{5}$ then obviously $a_{n+1} = \frac{2a_n^{3} + 5}{3a_n^{2}} > \frac{2(\sqrt[3]{5})^{3} + 5}{3(\sqrt[3]{5})^{2}} = \sqrt[3]{5}$ so we deduced $a_{n+1} > ^\sqrt[3]{5} $.

(It's important to note $a_1$ might be less then $\sqrt[3]{5}$, hence the sequence will decrease from the second element.) So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:

$$\begin{align} L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}) = \frac{1}{3}(2L + \frac{5}{L^{2}}) = \sqrt[3]{5}\end{align}$$

I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $\sqrt[3]{5}$, or rather in this case $(\sqrt[3]{5}, \sqrt[3]{5}+\epsilon)$.

Is it correct?

EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $\sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.

Answer 1

First observe that $$ a_{n+1}=\frac{1}{3}\left(2a_n+\frac{5}{a_n^2}\right)=\frac{1}{3}\left(a_n+a_n+\frac{5}{a_n^2}\right)\ge \left(a_n\cdot a_n\cdot \frac{5}{a_n^2}\right)^{1/3}=5^{1/3}. $$ Hence, $a_n\ge 5^{1/3}$, for all $n>1$.

Next observe that $$ f(x)=\frac{1}{3}\left(2x+\frac{5}{x^2}\right)<x, $$ for all $x$ in $[5^{1/3},\infty)$, since $$ x-f(x)=\frac{1}{3}\left(x-\frac{5}{x^2}\right)=\frac{x^3-5}{3x^2}\ge 0. $$

Hence $$ 5^{1/3}\le a_{n+1}=f(a_n)\le a_n, \quad \text{for all $n>1$.} $$ Thus $\{a_n\}$ is bounded and decreasing, and therefore it is convergent.

Also, if $a_n\to a$, then, as $f$ is continuous in $(0,\infty)$, then $$ a=\lim a_n=\lim a_{n+1}=\lim f(a_n)=\lim f(a) $$ and hence the limit $a$ satisfies $$ a=f(a)=\frac{1}{3}\left(2a+\frac{5}{a^2}\right) $$ and thus $a^3=5$, which implies that $a_n\to 5^{1/3}$.

Answer 2

In order to for the sequence to converge: $$lim_{n\to\infty}(a_{n+1}-a_{n})=0$$ Then, we can plug in the recursion formula for $a_{n+1}$ and we get $$lim_{n\to\infty}(a_{n+1}-a_n)=lim_{n\to\infty}(\frac{2}{3}a_n+\frac{5}{3a_n^2}-a_n)=lim_{n\to\infty}(\frac{1}{3}+\frac{5}{3a_n^2})=3 lim_{n\to\infty}(-a_n+\frac{5}{a_n^2})=3 lim_{n\to\infty}\frac{5-a_n^3}{a_n^3}=0$$ As such, $lim_{n\to\infty} a_n=\sqrt[3]5$