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.

  • $\begingroup$ I just have a question: i also could have shown that $(x_n)$ is bounded below by 0. But, if i calculate the limit (solving an equation), i have 3 different values: $L=0,L=\sqrt{2}$. We can drop the solution $"-"\sqrt{2}$ but how can i be sure that it is $"+"\sqrt{2}$ and not 0 in my case? It is kinda immediat if i calculate first 5-6 values of the sequence, but still i would like to know how to be sure for 100% and have a rigorous proof. $\endgroup$
    – Daniil
    Feb 14, 2021 at 17:34
  • $\begingroup$ The query's title indicates that it is sufficient to determine that the sequence is convergent - calculating the actual limit of the sequence is therefore irrelevant. Consequently, all that is required is that [1] the sequence is strictly decreasing and [2] there exists some lower bound for the sequence (regardless of what that lower bound is). From this perspective, a lower bound of $0$ works fine. $\endgroup$ Feb 14, 2021 at 17:41
  • $\begingroup$ @user2661923 Yes, it wasn't asked to calculate a limit but still by curioisty why couldn't I ask this question... $\endgroup$
    – Daniil
    Feb 14, 2021 at 17:42
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    $\begingroup$ Assuming that it is established that the sequence is strictly decreasing and (therefore) converges to some non-negative value $L$, one approach is to recognize that the $\lim_{n\to\infty} x_n = L = \lim_{n\to\infty} \left(x_{(n-1)} + x_n\right)^{(1/3)}.$ Another approach is to guess the value of $L$, show that $\forall n, x_n > L$, and further show that the assumption that the limit equals some other value $L_1 > L$ leads to a contradiction. $\endgroup$ Feb 14, 2021 at 17:51
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    $\begingroup$ My bad, I didn't actually scrutinize your analysis. If you have analysis that demonstrates that there is a finite set $S$ such that $L$ must be an element of $S$, then yes, one way of completing the problem is to prove that there is only 1 value (i.e. element) in the set $S$ that will work. $\endgroup$ Feb 14, 2021 at 18:19

3 Answers 3


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

Proof. (by strong induction)

for $n=1$


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


$$x_{n+2}^3=x_{n+1}+x_n<x_n+x_{n-1}=x_{n+1}^3$$ 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 $\sqrt 2$.

  • 1
    $\begingroup$ @archuser The fact that $f(x)=\sqrt[3]{x}$ is strictly increasing is common knowledge. The feedback on your proof is implicit in my answer's length: 8 lines vs 33. $\endgroup$
    – Raffaele
    Feb 14, 2021 at 21:20
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    $\begingroup$ @archuser Your proof is verbose. You prove that $\sqrt[3]{x}$ is continuous, that $x^3$ is increasing. You even compute the derivative from the definition! The inferior bound is trivially zero... as user2661923 commented three hours ago. My feeling is that you confused rigorous with pedantic $\endgroup$
    – Raffaele
    Feb 14, 2021 at 21:37
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    $\begingroup$ @archuser were you looking for feedback or for unconditionally praise? $\endgroup$
    – Raffaele
    Feb 14, 2021 at 21:44
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    $\begingroup$ @Raffaele: Your $x$ could be $0$ why not ?. You still have to “show” that $x > 0$. ! $\endgroup$
    – DeepSea
    Feb 14, 2021 at 21:55
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    $\begingroup$ @DeepSea You are right! $\endgroup$
    – Raffaele
    Feb 14, 2021 at 22:02

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 $\sqrt{2}$.

  • $\begingroup$ Thank you very much! That is the answer that I was waiting for! $\endgroup$
    – Daniil
    Feb 15, 2021 at 10:26

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 $x = \sqrt{2}$.


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