I have a sequence of real numbers for which I wish to prove convergence.

$\displaystyle s_{n+1} = \sqrt[3]{s_{n}^{2} + s_{n-1} + 2}$ and $\displaystyle 0 < s_{0} < s_{1} < 2$

So, if the sequence $\displaystyle \{s_{n}\}$ converges to a limit $\displaystyle s$ then

$\displaystyle s^3 - s^2 - s - 2 = 0 \; \Rightarrow \; \left( s - 2 \right)\left( s^2 + s + 1 \right) = 0 \; \Rightarrow \; s = 2$

Also, we have that $\displaystyle t^3 - t^2 - t - 2 < 0 \; \Rightarrow \; t < s \forall t > 0$

But I have yet to prove convergence.

$\displaystyle \begin{align*}s_{n+1}^3 - s^3 &= -s^3 + s_{n}^{2} + s_{n-1} + 2 \\ &= -\left( s^3 - s^2 - s - 2 \right) + \left( s_{n}^{2} - s^2 \right) + \left( s_{n-1} - s \right) \\ &= \left( s_{n}^{2} - s^2 \right) + \left( s_{n-1} - s \right) \\ \end{align*}$

So if $\displaystyle s_{n-1}$ and $\displaystyle s_{n}$ are both less than $\displaystyle s$, then $\displaystyle s_{n+1} < s$ and our initial conditions give us $\displaystyle s_{n} < s \; \forall n \in \mathbb{N}$.

And thus the sequence $\displaystyle \{s_{n}\}$ is bounded above.

So now I need to prove that sequence $\displaystyle \{s_{n}\}$ is increasing.

$\displaystyle \begin{align*} s_{n+1}^3 - s_{n}^3 &= -s_{n}^3 + s_{n}^{2} + s_{n-1} + 2 & \text{but} \\ s_{n} &< s \; \Rightarrow \; s_{n}^3 - s_{n}^{2} - s_{n} + 2 < 0 \\ \Rightarrow \; s_{n}^3 - s_{n}^{2} + 2 &< s_{n} & \text{so} \\ s_{n+1}^3 - s_{n}^3 &> s_{n-1} - s_{n} \\ \end{align*}$

But that doesn't show that $\displaystyle s_{n+1}^3 - s_{n}^3 > 0$ as I require.

Can you help?