Show that the sequence defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}$ converges

Question

Show that the sequence $(x_n)$ defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}, x_0=3,x_1=2$ converges. Here i wil try to prove that $(x_n)$ converges considering all details. So. i would like to have a critical feedback on my proof and would like to know if i missed something or did much more than it should be.. Here $\mathbf{R}^*$ will denote $\mathbf{R}\ \ \not \ \{0\}$.

Consider a function $f(x)=\sqrt[3]{x}$. First, we show that it's defined on $\mathbf{R}$ and then we will show that $f(x)$ is an increasing function on $\mathbf{R}$.

We remark that $f(x)$ is differntiable on $\mathbf{R}^*$. Let $a \in \mathbf{R}^*$. By definition of differentiable function the $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ msut exist:

$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\frac{1}{3\sqrt[3]{a^2}}$ and so it exists $\forall a \in \mathbf{R}^*$ as we chose an arbitrary $a$. So $f(x)$ is differentiable on $\mathbf{R}^*$ and so is continious on $\mathbf{R}^*$.

We show now that $f(x)$ is continious on $0$. To show that we pass by the definition :

$\forall \varepsilon>0 \ \exists \delta>0 \ \forall x \in \mathbf{R}$: $|x|<\delta \implies |f(x)|=|\sqrt[3]{x}|<\varepsilon$

To satisfy the definition, one can take $\delta=\varepsilon^3$, so it holds and we can conclude that $f(x)$ is continious on $\mathbf{R}$.

Let $a,b \in \mathbf{R}$ such that $a<b$. As $f(x)$ is continious on $\mathbf{R}$ we can write:

$a<b \iff \sqrt[3]{a}<\sqrt[3]{b} \iff f(a)<f(b)$

As $a,b$ are arbitrary numbers (satisfying $a<b$), we can conclude that $f(x)$ is strictly incrasing function.

Now, as all preliminary results are proven, we can show that $(x_n)$ is a convergent sequence. For that, we show that $(x_n)$ is monotonic and bounded.

$\bullet$ $(x_n)$ is decreasing: Suppose $x_{n+1}<x_n$. The previous inequality holds for $n=1$ so we can suppose that it works up to a certain $n\in \mathbf{N}^*$. We show nowthat it holds for $n+1$:

$x_{n+2}<x_{n+1} \iff \sqrt[3]{x_{n+1}+x_n}<\sqrt[3]{x_n+x_{n-1}}$. But, as $f(x)=\sqrt[3]{x}$ is strictly increasing function and by induction hypothesis $x_{n+1}<x_n$, we can conclude that $x_{n+2}<x_{n+1}$ and so $x_{n+1}<x_n$ holds $\forall n\in\mathbf{N}^*$. So $(x_n)$ is decreasing.

$\bullet$ $(x_n)$ is bounded below: Suppose $x_n>1$. It holds for $n=1$ so we can suppose that it works up to a certain $n\in \mathbf{N}^*$. We want to show now that t holds for $n+1$:

$x_{n+1}>1 \iff \underbrace{x_{n}}_{>1}+\underbrace{x_{n-1}}_{>1}>1$ which is true, because $(x_n)$ is decreasing. So we conclude that $x_n>1 \ \forall n\in \mathbf{N}^*$.

So, we have shown that $(x_n)$ converges.

Answer 1

$$x_{n+1}=\sqrt[3]{x_{n}+x_{n-1}};\;x_0=3;\;x_1=2$$ $(x_n)$ is a decreasing sequence

Proof. (by strong induction)

for $n=1$

$x_2=\sqrt[3]{x_0+x_1}=\sqrt[3]{5}<x_1$

Now suppose that it is true for all integers $1,2,\ldots, n$, and let's prove it for $(n+1)$

$$x_{n+2}=\sqrt[3]{x_{n+1}+x_{n}}$$

$$x_{n+2}^3=x_{n+1}+x_n<x_n+x_{n-1}=x_{n+1}^3$$ thus $x_{n+2}<x_{n+1}$. Proved.

The sequence is decreasing and is bounded above by $x_n\le 3$

$x_n>1$ for any $n$.

$x_1=2>1$. Suppose $x_n>1$ for $0,1,2,\ldots,n$.

$x_{n+1}=\sqrt{x_n+x_{n+1}}>\sqrt[3]{1+1}>1$. Proved.

Thus the sequence is bounded above and below $1<x_n<3$, therefore it converges

$x_n\to x$ as $n\to\infty$

To find the limit let's use the definition

$$x=\sqrt[3]{x+x}\to x^3=2x\to x=\sqrt 2$$

The sequence converges to $\sqrt 2$.

Answer 2

You asked for feedback on your proof, here are some thoughts:

• To prove $f$ is continuous, you say that it's differentiable. But you do not prove that $f$ is differentiable directly - you only state the derivative. You could show that the derivative exists with this value, but that proof would be very similar to proving $f$ is continuous directly (e.g. using the binomial theorem). So it would be simpler to just do that.

• To prove $f$ is increasing, you just say "$f$ is continuous, therefore [$f$ is increasing]". There is no proof in your question that $f(x) = \sqrt[3]{x}$ is an increasing function.

• Your proofs that $(x_{n})$ is decreasing and bounded below by 1 appear to be correct. Note that they only rely on the fact that $f$ is increasing and that $f(1) = 1$. You haven't used the fact that $f$ is continuous anywhere.

• However, the property that $f$ is continuous does allow you to find the limit. In general, if $x_n \to x$ and $f$ is continuous, then $f(x_n) \to f(x)$. Applying this to the equation $x_{n+1} = f(x_n+x_{n-1})$ shows that the limit satisfies $x = f(2x)$. In other words, $x^3 = 2x$, so either $x=0$, $x=-\sqrt{2}$ or $x=\sqrt{2}$. But since you argued already that $x > 1$, the limit must be $\sqrt{2}$.

Answer 3

After showing that the limit exists and call it $x$. We show next that: $x \ge \sqrt{2}$. Using the decreasing property just established we have: $x_n = \sqrt[3]{x_{n-1} + x_{n-2}} > \sqrt[3]{2x_n} \implies x_n^3 > 2x_n \implies x_n(x_n^ 2 - 2) > 0 \implies x_n > \sqrt{2}, \forall n \ge 1 \implies x \ge \sqrt{2} \implies x = \sqrt{2}$, because it’s shown that $x = 0$ or $x = \sqrt{2}$.