What is my limit?

If the limit $a$ exists, it must be a fixed point of the iteration $\begin{aligned} & & a \stackrel{!}{=} \frac{2}{3} \left( a + \frac{1}{a^2}\right) \\ \Rightarrow & & \frac{1}{3} a = \frac{2}{3 a^2} \\ \Rightarrow & & a^3 = 2 \\ \Rightarrow & & a = \sqrt[3]{2} \end{aligned}$ Now all we have to do is check if the sequence is convergent. Suppose, $\sqrt[3]{2} < a_n \leq 2$ . Then it follows $\begin{aligned} a_{n+1} - a_n &= \frac{2}{3} \left( a_n + \frac{1}{a_n^2} \right) - a_n \\ &= \frac{1}{3} \frac{2 - a_n^3}{a_n^2} \\ &< 0 \\ a_{n+1} - \sqrt[3]{2} &= \frac{2}{3} \left( a_n + \frac{1}{a_n^2} \right) - \sqrt[3]{2} \\ &= \frac{2}{3 a_n^2} \left( a_n^3 - \frac{3}{2^{2/3}} a_n^2 + 1 \right) \\ &=\frac{2}{3 a_n^2} \left( a_n + \frac{1}{2^{2/3}} \right) \left( a_n - \sqrt[3]{2}\right)^2 \\ &> 0 \end{aligned}$ so that $\sqrt[3]{2} < a_{n+1} < a_n \leq 2$ . By proof by induction it follows, that the sequence is bounded and monotonically decreasing, so that it is also convergent. QED.