# Proof of Convergence

• Dec 23rd 2013, 06:50 PM
Archie
Proof of Convergence
I have a sequence of real numbers for which I wish to prove convergence.
$s_{n+1} = \sqrt[3]{s_{n}^{2} + s_{n-1} + 2}$ and $0 < s_{0} < s_{1} < 2$

So, if the sequence $\{s_{n}\}$ converges to a limit $s$ then
$s^3 - s^2 - s - 2 = 0 \; \Rightarrow \; \left( s - 2 \right)\left( s^2 + s + 1 \right) = 0 \; \Rightarrow \; s = 2$
Also, we have that $t^3 - t^2 - t - 2 < 0 \; \Rightarrow \; t < s \forall t > 0$

But I have yet to prove convergence.
\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 $s_{n-1}$ and $s_{n}$ are both less than $s$, then $s_{n+1} < s$ and our initial conditions give us $s_{n} < s \; \forall n \in \mathbb{N}$.
And thus the sequence $\{s_{n}\}$ is bounded above.

So now I need to prove that sequence $\{s_{n}\}$ is increasing.
\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 $s_{n+1}^3 - s_{n}^3 > 0$ as I require.

Can you help?
• Dec 23rd 2013, 07:18 PM
SlipEternal
Re: Proof of Convergence
Have you learned that $f:\mathbb{R} \to \mathbb{R}$ is continuous if and only if $\{f(a_n)\}$ converges for all sequences $\{a_n\}$ which converge in $\mathbb{R}$? Well, you know that $f(x) = \sqrt[3]{x}$ is continuous. So, let $r_n = s_n^3$. If you can show that $r_n$ converges, then since $f$ is continuous, $f(r_n) = s_n$ converges, as well.

Anyway, to show convergence, you will want to end with something like:

Given $\varepsilon>0$, let $N$ be a positive integer such that $N > \text{expression}$ (where the expression is some function of $\varepsilon$).

Then show that for all $n > N$, $|s_n - 2| < \varepsilon$.
• Dec 23rd 2013, 07:52 PM
Archie
Re: Proof of Convergence
The text I'm looking at ("A First Course in Mathematical Analysis", Burkill) hasn't introduced the concept of a function at this point. So although your idea would work, I'm looking for something using much more basic ideas.

I don't intend to go into $\epsilon$ notation for this, since I have already shown that if the sequence converges, it must converge to 2. Given that, all I need is to show that it does converge. And the standard approach in the text is to show that the sequence is bounded (which I've done) and that it is monotonic. The combination of these results being sufficient for convergence. Another approach would be to show that $\left| s_n - 2 \right|$ converges to zero, but that doesn't look very practical here, the analysis of $s_{n+1}^3 - s^3$ proving only that the sequence is bounded.
• Dec 23rd 2013, 09:44 PM
SlipEternal
Re: Proof of Convergence
Suppose $s_0 = 0.1, s_1 = 1.9$. Then $s_2 = \sqrt[3]{1.9^2 + 0.1 + 2} \approx 1.78736 < s_1$, so the sequence is not necessarily monotonically increasing. You might need to prove that $|s_n - 2|$ converges to zero.
• Dec 23rd 2013, 10:07 PM
SlipEternal
Re: Proof of Convergence
You can, however, use the squeeze theorem. Let $r_{n+1} = \sqrt[3]{r_n^2 + r_{n-1} + 2}, r_0 = r_1 = 0$. Let $t_{n+1} = \sqrt[3]{t_n^2 + t_{n-1} + 2}, t_0 = t_1 = 2$. Obviously, $r_n < s_n < t_n$ for all $n$. It is pretty easy to show that $t_n = 2$ for all $n$. So, if you can show that $r_n$ converges to $2$, then by the squeeze theorem, $s_n$ also converges to $2$. Now, you can show that $r_n$ is bounded and monotonically increasing.
• Dec 23rd 2013, 10:38 PM
SlipEternal
Re: Proof of Convergence
To show that $r_n$ converges, you know that $r_1 = 0, r_2 = \sqrt[3]{2} \approx 1.26, r_3 = \sqrt[3]{\sqrt[3]{4}+2} \approx 1.53$, so $r_1 < r_2 < r_3$. Assume that for all positive integers greater than or equal to 3 and up to $n$, $2 > r_n > r_{n-1} > 1$ and $r_{n-1}>r_{n-2} > 0$ (which is true for $r_1,r_2,r_3$). Then $r_{n+1}^3 - r_n^3 = (r_n^2 + r_n + 2) - (r_{n-1}^2 + r_{n-2} + 2) = (r_n^2 - r_{n-1}^2) + (r_{n-1} - r_{n-2}) > 0$, so $r_{n+1} > r_n$. Hence, by mathematical induction, $r_n$ is monotone increasing. Then, since it is bounded, it converges. Finally, as you showed above, if it converges, it converges to 2. So, by the Squeeze Theorem, so too does $s_n$.
• Dec 24th 2013, 11:34 PM
Archie
Re: Proof of Convergence
Haha, I hadn't realised that the initial values were sensitive like that. The original question has something like $s_0 = 1, s_1 = 1.3$

One thing I have started to think about is that
\begin{align*}
s_{n+1}^3 - 2 &= s_{n}^2 + s_{n-1} + 2 - 2 \\
&= s_{n}^2 + s_{n-1} \\
\end{align*}

So $s_{n} > 0, s_{n-1} > 0$ (which is easily seen from the definition of $s_{n+1}$) gives $s_{n} \ge \sqrt[3]{2} \forall n > 0$

But then we also have
\begin{align*}
s_{n+1}^3 - 4 &= s_{n}^2 + s_{n-1} + 2 - 4 \\
&= s_{n}^2 + s_{n-1} - 2 \\
\end{align*}

So $s_{n} > \sqrt[3]{2}, s_{n-1} > \sqrt[3]{2}$ gives $s_{n} \ge 4 \forall n > 2$.

And this is obviously something that can be repeated ad nauseam. I have yet to decide how useful it is though (if at all).
• Dec 24th 2013, 11:40 PM
Archie
Re: Proof of Convergence
Ah! This is the trick I missed. Thanks a lot for your efforts.