# If $a>1$, prove that $\lim_{n \rightarrow \infty } a^n = \infty$

I want to know a rigorous method to prove that

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

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

• How can you prove that inequality? It does not seem obvious to me. thanks! Jul 30 '14 at 15:57
• @flawr : use binom thm. Jul 30 '14 at 15:57
• Remember the binomial theorem? $$(1+x)^n =\sum _{i=0} ^n \frac{n!}{(n-i)!i!}x^i$$ Jul 30 '14 at 15:58
• Oh thanks, you have to see that first=) Jul 30 '14 at 16:00
• ...or, another option, use Bernoulli inequality here: $(1+a-1)^n > 1+ n(a-1)$
– Alex
Jul 30 '14 at 16:50

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 $B<a^{k+1}$.

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.

$$a^{n+1} = a^n a > a^n$$ 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 : $$a^n but $$a^ C> Ba > B$$

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

so the limit is infinite.