I've got the following recursive sequence $\{a_n\}_0^\infty$, defined by: $$a_0 = 0$$ $$a_{n+1} = \sqrt[3]{a_n - 2}$$
I have to prove that it's strictly decreasing, i.e. $\forall n\in\mathbb N_0: a_n > a_{n+1}$.
The usual approach I wanted to follow:
- Find the fixed points
- Prove the sequence is bounded
- Prove it's strictly monotonic using the above two.
I found myself at a deadend when trying to solve the fixed point equation, namely: $$x^3 - x + 2 = 0$$ since its solution is enormously huge and ugly (kudos to Cardano for figuring out the neat formula for solving depressed cubic equations btw). So this brought me to the idea that maybe that was not the intention of the author, so I thought about induction:
- Check for $n = 0$: $$a_0 = 0 > \sqrt[3]{-2} = a_1$$
- Assume for a fixed $n$: $$a_n > a_{n+1}$$
- Prove for $n+1$: $$a_{n+1} > a_{n+2} $$ $$\Leftrightarrow \sqrt[3]{a_n - 2} > \sqrt[3]{a_{n+1} - 2}$$ $$\Leftrightarrow a_n > a_{n+1}$$
since $f(x) = \sqrt[3]{x - 2}$ is a strictly increasing continuous bijection from $\mathbb R$ to $\mathbb R$.
But I somehow feel uncomfortable with my solution since it doesn't feel on par with the current topic we're studying (namely analysis of recursive sequences). What are your thoughts? Is everything correct or is there a smarter thing to do? Thank you in advance!