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If $r>1$ show that $r^n$ goes to $+ \infty$ as $n$ goes to $ \infty$

If $|r| > 1$ then

$r^2> r$.

similarly $r^4 > r^3 > r^2 > r$.

so $r^n$ as $n\to \infty$ will be equal to $\infty$

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closed as unclear what you're asking by Jonathan, vonbrand, user127.0.0.1, TMM, user63181 Mar 10 '14 at 17:15

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  • $\begingroup$ infinite will be equal to 1 ?? $\endgroup$ – imranfat Mar 10 '14 at 15:23
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So you are given $y=r^n$ with $r>1$ You could take the natural log on both sides: $lny=lnr^n=nlnr$ Now if $r>1$ then $ln1>0$ and if $n$ goes to infinity, multiplied by a quantity greater than zero, then $lny$ is infinite as well, so $y$ is infinite.

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If you suppose that $(r^n)_{n\in\Bbb N}$ has un upper bound, then exists $L\in\Bbb R$ such that $r^n\rightarrow L\Rightarrow L>0$ and $L*r=L$ (taking limit to $r^{n+1}=r*r^n$) $\Rightarrow r=1$.

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You can formalize what you think as follows:

Suppose $|r|>1$. Define the set $\{|r|^n: n\in \mathbb{N}\}$. Clearly the set is non-empty. Now suppose to the contrary that it is bounded above.

Then by the least upper bound principle $s= \sup\{|r|^n: n\in \mathbb{N}\}$ is a real number; note that is a positive real number. Then $s/|r|<s$, so $s/|r|$ is not an upper bound, there is some $n_0$ for which $s/|r|<|r|^{n_0}$. Thus $s<|r|^{n_0+1}\in\{|r|^n: n\in \mathbb{N}\}$ which contradicts that $s$ is an upper bound. So the set cannot be bounded above.

Now given a $L$, there is a $n_0$ such that $L<|r|^{n_0}$, since otherwise the set is bounded. Then for all $n\ge n_0$ we have $L<|r|^{n_0}\le |r|^{n_0}|r|^{n-n_0}=|r^n|$. Hence $r^n \to \infty$.


You can use exaclty the same argument for $|r|<1$.

Define the set $\{|r|^n: n\in \mathbb{N}\}$ which is bounded below by zero and is non empty, so it has a greatest lower bound $a\ge0$. If were the case that $a>0$, then $a<a/|r|$ which means that $a/|r|$ is greater than the greatest lower bound so there is some $n_0$ such that $|r|^{n_0}<a/|r|$. But then this means $\{|r|^n: n\in \mathbb{N}\}\ni |r|^{n_0+1}<a$, a contradiction. Thus $a=0$. Now given $\varepsilon>0$, choose $n_0$ such that $|r^{n_0}|<\varepsilon$ and for all $n\ge n_0$ we have $|r^n|=|r^{n_0}||r^{n-n_0}|\le |r^{n_0}|<\varepsilon$. Hence $r^n \to 0$

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