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I want to know a rigorous method to prove that

If $a>1$, $\displaystyle\lim_{n \rightarrow \infty } a^n = \infty$

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$a^n=(1+a-1)^n\geq n (a-1) \to \infty$

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  • $\begingroup$ How can you prove that inequality? It does not seem obvious to me. thanks! $\endgroup$
    – flawr
    Jul 30 '14 at 15:57
  • $\begingroup$ @flawr : use binom thm. $\endgroup$
    – Soham
    Jul 30 '14 at 15:57
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    $\begingroup$ Remember the binomial theorem? $$(1+x)^n =\sum _{i=0} ^n \frac{n!}{(n-i)!i!}x^i$$ $\endgroup$
    – pppqqq
    Jul 30 '14 at 15:58
  • $\begingroup$ Oh thanks, you have to see that first=) $\endgroup$
    – flawr
    Jul 30 '14 at 16:00
  • $\begingroup$ ...or, another option, use Bernoulli inequality here: $(1+a-1)^n > 1+ n(a-1)$ $\endgroup$
    – Alex
    Jul 30 '14 at 16:50
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Alternative proof. The sequence $b_n = a^{n}$ is increasing. If it has an upper bound, then it has a least upper bound, $B$. Clearly, $B>a>0$. Then, since $B/a<B$, $B/a<a^k$ for some $k$, and then $B<a^{k+1}$.

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Alternately, if $a > 1$, then $a^2 - a = a(a-1) = \delta > 0$. Then $a^3 - a^2 = a\cdot a(a-1) > a(a-1) = \delta$. Inductively, we have that $a^n - a^{n-1} > \delta$, so that

$$\begin{align} a^n - a &= a^n + (- a^{n-1} + a^{n-1}) + (-a^{n-2} + a^{n-2}) + \ldots \\ &= (a^n - a^{n-1}) + (a^{n-1} - a^{n-2}) + \ldots \\ &\geq \delta + \delta + \ldots \\ & = n\delta. \end{align}$$

And as $\delta > 0$, we know that $n\delta \to \infty$. So $a^n$ gets unboundedly larger than $a$, and thus $a^n \to \infty$ as well.

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$ a^{n+1} = a^n a > a^n $ so this is an increasing sequence of positive numbers. It has either an infinite limite or a finite positive limite. We have to demonstrate it's the first case by finding a paradox ( reductio ad absurdum):

If the sequence has an upper finite limit ,let's call him B, then for each n : $a^n <B$ but $ a^ C> Ba > B$

Where c is : $ C > lg[a](B) + 1 $

so the limit is infinite.

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