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Fix $a > 1$. I want to prove that $$ \lim_{n \to \infty} a^n = \infty. $$ When $a = 2$, then this follows from that $2^n \ge n$ for all $n \ge 1$, which can be proved easily by induction. However, $a^n \ge n$ is not true in general. I tried to prove that $a^n \ge n$ for all $n \ge 1/(a-1)$. The inductive step is okay but the base step is hard. What is the simplest way to prove the above? Thank you.

D.L
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  • $a^n=\exp(n\ln(a))\geq 1+ n\ln(a)$ and $\ln(a)>0$ is a constant – C Squared Oct 20 '21 at 13:50
  • Thank you. But for that idea, I have to develop the whole theory for the exponential and logarithm. – D.L Oct 20 '21 at 13:58
  • I suggest to prove Bernoulli's inequality $(1+x)^n\ge 1+xn$ which holds for $x>-1$ and non-negative integer $n$ by induction. The rest is easy. – Peter Oct 20 '21 at 14:30
  • If $0<a<1$ then $a_n$ exponentially decreases, while for $a>1$ it exponentially increases (and is of course monotone). – Meitar Oct 20 '21 at 14:42
  • There are tons of proof techniques in the link above. Try adapting them to your problem: if not, you can ask a new question. Also, Approach0 is very useful at finding duplicates and similar questions. – Toby Mak Oct 23 '21 at 04:08
  • I could prove it by Bernoulli's inequality suggested by Botnakov N. and Peter. Thanks. – D.L Oct 24 '21 at 08:04

2 Answers2

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$$x = a-1, a^n = (1 + x)^n = 1 + nx + ... \ge n x = (a-1)n.$$

Botnakov N.
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  • Thank you. So I only have to prove $(1+x)^n \ge nx$ by induction. I will try. – D.L Oct 20 '21 at 13:55
  • I meant this way: $(1+x)^n = 1+nx+\frac{n(n-1)}2 x^2 + ... $ by binomial theorem. All terms are nonnegative and hence $(1+x)^n \ge nx$. – Botnakov N. Oct 20 '21 at 13:58
  • Well, then I have to prove the binomial theorem first. I will try that too. Thanks. – D.L Oct 20 '21 at 14:00
  • @D.L: You don't need the explicit statement of the binomial theorem in the sense that you don't need to know what the coefficients of $x^2, x^3, \ldots$ are. Just that they are nonnegative. – Aryaman Maithani Oct 20 '21 at 14:02
  • Oh, I see. I only need to find the coefficient of $x$. It will be much easier. – D.L Oct 20 '21 at 14:05
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Let $a_n := a^n$.

  1. Note that $(a_n)_n$ is increasing.
  2. For the sake of contradiction, assume that $(a_n)_n$ is bounded. In this case, it must be convergent. Let the limit be $A$.
  3. Note that $$a_{n + 1} = a a_n.$$ Since $(a_n)_n$ converges, we may take limit on both sides above to get $A = aA$.
  4. Since $a \neq 1$, we get $A = 0$. But this is absurd since $a_n > 1$ for all $n$.
  5. Thus, $(a_n)_n$ is increasing and unbounded. This finishes the proof.
Aryaman Maithani
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  • Thank you. So I have to prove that $\lim a_{n+1} = \lim a_n$, and that increasing unbounded sequences diverge. Well, I think it requires quite long proof. – D.L Oct 20 '21 at 13:56
  • @D.L: Oh, if those are not things that you are allowed to take for granted, then I'll point out that I'm also using: "Monotonic + Bounded $\Rightarrow$ Convergence". – Aryaman Maithani Oct 20 '21 at 14:00
  • Oh, I see. It requires only the monotone convergence theorem. Thanks. – D.L Oct 20 '21 at 14:08