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

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

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 $x = \sqrt{2}$.

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 $\sqrt{2}$.