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Prove: $$\lim_{n\to\infty}r^n = +\infty\,, r > 1;$$ $$\lim_{n\to\infty}r^n = 0\,, 0 \le r < 1.$$

I am not quite sure how to prove this, but once someone proves it I will make sure to ask questions if I'm in doubt. Thank you very much! :)

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Proofs of the second fact are on this ProofWiki page. Proofs that the first follows from the second are on this page. –  Lord_Farin Oct 12 '12 at 21:01
    
You assume $r>0$? –  lhf Oct 12 '12 at 21:03

3 Answers 3

up vote 1 down vote accepted

Perhaps one can add something to the answer by lhf for the case $0\lt r\lt 1$.

For such an $r$, let $r=\dfrac{1}{1+x}$. Then $x \gt 0$. Now go to the answer by lhf.

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I first saw the proof that $r^n \to 0$ if $0 \le r < 1$ and $r^n \to \infty$ if $r > 1$ in Courant and Robbin's "What is Mathematics?" many years ago. As stated by others, the proof uses Bernoulli's inequality and, implicitly, Archimede's axiom about the real numbers.

They also have a simple proof that $n^{1/n} \to 1$ as integer $n \to \infty$.

One nice thing about these proofs is that all the bounds are explicit and easily computable.

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Hint: Use Bernoulli's inequality: $$ (1+x)^n \ge 1 + nx \quad \mbox{for $x\ge 0$} $$ This can be proven by induction.

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And of course this would work for both of them :) –  N. S. Oct 12 '12 at 21:06

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