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Let $(x_n)$ be a sequence with $x_1=1 $ and $x_{n+1}=x_n+\frac{1}{x_n^2}$ for all integers n. I want to formally prove that this sequence is not limited.

I started with showing that $x_n$ is strictly monotonically increasing because:

$x_{n+1}-x_n=\frac{1}{x_n^2} > 0$

Now I want to show that for all real numbers $a$ I will find an integer $n_0$ thus $x_n>a$ for all integers $n_0$. But I am stuck at this point. Can someone give me a hint how to go on? :)

Bernard
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Blue2001
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4 Answers4

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Since $(x_n)$ is a non decreasing sequence; either $(x_n)$ converges, either its limit is $+\infty$. If $(x_n)$ converges towards $\ell\geqslant x_1=1$ we would have $\ell=\ell+\frac{1}{\ell^2}$ which is not, thus $\lim\limits_{n\rightarrow +\infty}x_n=+\infty$.

Tuvasbien
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  • Why do we know that $l = l+1/l^2$ is true? This would imply that $x_{n+1}=l$ and $x_n= l$ – Blue2001 Dec 10 '20 at 18:20
  • Let $f(x)=x+\frac{1}{x^2}$, then $x_{n+1}=f(x_n)$ and since $f$ is continuous, we have $f(\ell)=\ell$. – Tuvasbien Dec 10 '20 at 18:31
  • We don't have continuous functions in our course :/ – Blue2001 Dec 10 '20 at 18:34
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    Well, you can see it as $\lim_{n\rightarrow +\infty}x_{n+1}=\lim\limits_{n\rightarrow +\infty}x_n$, that is to say $\lim\limits_{n\rightarrow +\infty}\left(x_n+\frac{1}{x_n^2}\right)=\lim\limits_{n\rightarrow +\infty}x_n$ which leads to the desired equality. – Tuvasbien Dec 10 '20 at 18:35
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    Can I prove it like this?: $x=lim(x_{n+1}) = lim (x_n) + lim(\frac{1}{x_n^2}) = x + 1/x^2 $ – Blue2001 Dec 10 '20 at 18:36
  • Oh thank you very much! :) – Blue2001 Dec 10 '20 at 18:36
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Here's a direct proof of divergence showing the rate of divergence.

$x_{n+1}=x_n+\frac{1}{x_n^2}, x_1>0 $

$\begin{array}\\ x_{n+1}^3 &=(x_n+\frac{1}{x_n^2})^3\\ &=x_n^3+3+\frac{3}{x_n^3}+\frac1{x_n^6}\\ &\gt x_n^3+3\\ \end{array} $

so

$x_n^3 \gt 3(n-1) $ or $x_n \gt \sqrt[3]{3(n-1)} $.

marty cohen
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  • I don't understand the last thing. Why can you conclude that $x_n^3 > 3(n-1)$ and where does the $n$ come from? – Blue2001 Dec 10 '20 at 18:13
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Well, first we can prove that the sequence is positive, then, let $ n\in\mathbb{N}^{*} $, for any $ k\in\mathbb{N} $, we have : \begin{aligned} x_{k+1}&=x_{k}+\frac{1}{x_{k}^{2}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{k+1}^{3}&=x_{k}^{3}+3+\frac{3}{x_{k}^{3}}+\frac{1}{x_{k}^{6}}\\&\geq x_{k}^{3}+3 \\ \Longrightarrow\ \sum_{k=1}^{n-1}{\left(x_{k+1}^{3}-x_{k}^{3}\right)}&\geq 3\left(n-1\right)\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{n}^{3}&\geq 3n-2\end{aligned}

Thus : $$ \lim_{n\to +\infty}{x_{n}}=+\infty $$

CHAMSI
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Note that for $m > 0$, we have $$x_{n+m} - x_n = \sum\limits_{k=1}^m\frac{1}{x_{n+k-1}^2} \geq \frac{m}{x_n^2}$$

But if $x_n$ converges, then $|x_{n+m} - x_n| \to 0$. In particular, if $(x_n)$ is bounded above, say by $C$ (WLOG, $C > 0$), then we have $$x_{n+m} - x_n \geq \frac{m}{C^2} > 0,$$ so in particular $$x_{m} \geq 1 + \frac{m-1}{C^2} \to \infty$$ for all $m$, and hence $(x_m) \to \infty$, contradicting our assumption that $(x_n)$ is bounded above. Thus, it is not so bounded.

user3482749
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