Good morning i was thinking about this problem and I make this. I need someone review my exercise and say me if that good or bad. Thank!

Problem: Prove if $a>1$ then $\lim_{n\rightarrow\infty}a^{n}=\infty $


Suppose $\left\{ a^{n}\right\} $ is monotonically increasing. In other words $a^{n}<a^{n+1}< a^{n+2}...$ and Suppose $\left\{ a^{n}\right\} $ is Bounded set then $\left\{ a^{n}\right\} $ converge. By definition $\lim_{n\rightarrow\infty}a^{n}=L$.

We know this

$\left(a^{n+1}-a^{n}\right)=a^{n}(a-1)$ , $(a-1)>0$. Because $a>1$


$a^{n}(a-1)>(a-1)\Rightarrow a^{n}>1$

Exist $N ∈ \mathbb{N} $ such that $a^{N}$ > $L$ and $\left\{ a^{n}\right\} $ is non bounded set Then $\left\{ a^{n}\right\} $ diverge and


But, i don't sure it is fine, please help.

  • 4
    $\begingroup$ en.wikipedia.org/wiki/Bernoulli%27s_inequality $\endgroup$ – Gabriel Romon Jul 9 '16 at 15:20
  • $\begingroup$ Your proof is not good. How can you conclude that $a^n(a-1)>(a-1)$?Maybe you already assume that $a^n>1$?And the following statement is also not completely correct. I think you may take the following answers to complete your proof. $\endgroup$ – Deepleeqe Jul 9 '16 at 15:47
  • $\begingroup$ a>1 so $a^{n+1} = a(a^n) > a^n$ solves Deepleeqe's objection. My concern is there exist N so that $a^N > L$. Why? that seems like you are assuming what you wish to prove. $\endgroup$ – fleablood Jul 9 '16 at 22:12

An other way

Let $a>1$. $$a^n=e^{n\ln(a)}\underset{\ln(a)>0}{>}n\ln(a)\underset{n\to \infty }{\longrightarrow }\infty .$$

An other way (using Bernoulli)

Since $a>1$, there is $\varepsilon>0$ s.t. $$a=1+\varepsilon.$$ Using Bernoulli, $$a^n=(1+\varepsilon)^n\geq n\varepsilon+1.$$


Simply use this version of Bernoulli's inequality:

For any $a>0$, one has $\quad a^n-1\ge n(a-1)$

to show than $a^n$ can be made larger than any prescribed number.


You can achieve your proof using contradiction: If the limit $L$ of $(a^n)$ is finite then

$$\lim_{n\to\infty} a^{n+1}-a^n=0=\lim_{n\to\infty} a^n(a-1)=L(a-1)\implies L=0$$ which is a contradiction.

An alternative proof is: let $h>0$ such that $a=1+h$ so

$$a^n=(1+h)^n\ge 1+nh\xrightarrow{n\to\infty}+\infty$$

  • $\begingroup$ My proof are bad? Sure? $\endgroup$ – Bvss12 Jul 9 '16 at 15:34
  • $\begingroup$ @Battani Probably by the fact that $L(a-1)=0=\lim(a^{n+1}-a^n)$ and that $a-1\ne 0$. $\endgroup$ – BigbearZzz Jul 9 '16 at 16:32

What you need to show is that for any $x > 0$, there is an $N > 0$ such that $a^N > x$.

Given that you have shown the difference between elements is greater than $a - 1$, this means that you can choose $N$ to be any integer larger than $\frac{x}{a-1}$.

  • $\begingroup$ But my proof is bad, sure? Thanks. $\endgroup$ – Bvss12 Jul 9 '16 at 15:24

A problem I see with your proof is that $$ (a-1)>0 $$ does NOT implies that $$ a^n(a-1)>(a-1). $$

This means that you cannot just cancel $(a-1)$ on both sides to get $a^n>1$. Anyway, the fact can be easily proved by induction so this is not the real problem here.

The main problem with your proof is that you seemed to claim that if $a^n>1$ for all $n\in\Bbb N$, then there exists an $N\in\Bbb N$ such that $$ a^N>L\ . $$ This does not follow logically from your previous points.

If you somehow think that I misunderstood you, you'll have to be more explicit in each of your steps.

  • $\begingroup$ Yes it does imply that. A positive times something greater than 1 (which a^n clearly is) is more than the same positive times 1. $\endgroup$ – Jacob Wakem Jul 9 '16 at 21:13
  • 1
    $\begingroup$ " (which a^n clearly is)" if it was that "clear" we could have simply said "a^n is clearly monotonically increasing". To claim (a - 1) > 1 implies a^n(a - 1) > a-1 we have to show a^n > 1. Which is easy. But easy in such a way that the entire discussion should have been avoided. $\endgroup$ – fleablood Jul 9 '16 at 22:17
  • $\begingroup$ @JacobWakem If you read my post carefully, you'll see that I never denied the truthfulness of the statement $a^n>1$. The OP tried to deduce $a^n(a-1)>(a-1)$ from $(a-1)>0$, without assuming that $a^n>1$ in the beginning, that's what I denied. $\endgroup$ – BigbearZzz Jul 10 '16 at 1:54
  • $\begingroup$ @BigbearZzz Ah I see. Maybe its just an intuitive leap, I don't know. $\endgroup$ – Jacob Wakem Jul 10 '16 at 2:08

Your phrasing is very bad — for instance, you say 'suppose $\{a^n\}$ is monotonically increasing', but in fact that's not something you should suppose; it's something that's true, and you should prove it.

You've got a good idea in the next step: once you know that $\{a_n\}$ is monotonic increasing, you can use the Monotone Convergence theorem to derive a contradiction. Unfortunately, as others have noted your attempted proof from there is mathematical gibberish.

Instead, you can use an argument like this: Suppose that $\{a^n\}$ were bounded. Then by the Monotone Convergence theorem it has a limit $L$. By the definition of limit, for every $\epsilon$ we can choose an $n_0$ such that $|L-a^n|\lt\epsilon$ for all $n\gt n_0$. Now, the plan is to find a particular $\epsilon$ for which this breaks down. If we pick some specific $m\gt n_0$ and the 'right' epsilon, then the idea is that $a^m$ is 'close enough' to $L$ that $a^{m+1}=a\cdot a^m$ is guaranteed to be larger than $L$ (what size does $\epsilon$ have to be for you to guarantee this?); then $a^{m+2}=a\cdot a^{m+1}\gt aL$. But this implies that $|a^{m+2}-L|\gt (aL-L)=(a-1)L$, and if $\epsilon$ is chosen correctly, then this supplies the contradiction and proves that the assumption of boundedness must be wrong.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.