<s>[INST] We know that $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. Prove ${x_n}$ is convergent Suppose $a>0, x_0>0$ and $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. How to prove ${x_n}$ is convergent? Can someone helps me?
I tried to prove that $x_n$ is monotonically decreasing and have a lower bound: $a^\frac{1}{3}$.
 [/INST] Easy to see that all terms of $(x_n)$ are positive.
Now by AM-GM, we have the lower bound:
$$x_{n+1} = \frac 13\left(x_n + x_n + \frac{a}{x_n^2}\right)\ge \sqrt[3] a, \quad (\forall n \in \mathbb N) \tag{1}$$
Also, from the given formula we have
$$x_{n+1}-x_n = -\frac{1}{3}x_n + \frac {a}{3x_n^2} = \frac{-x^3_n + a}{3x_n} \tag{2}$$
But from $(1)$ we have $$-x_n^3 \le -a$$
Plugging this into $(2)$, we have
$$x_{n+1}-x_n \le \frac{- a + a}{3x_n^2} = 0 \\ x_n \ge x_{n+1}$$
So, the sequence is decreasing and lower-bounded and, therefore, convergent.

Finally, let the limit be $x$. Then, we have
$$3x = 2x + \frac{a}{x^2} \\ x = \frac{a}{x^2} \implies x = \sqrt[3]a \\ \bbox[5px,border:2px solid #C0AA00]{\lim_{n\to\infty}x_n = \sqrt[3]a}$$
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