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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 $


Proof:

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 $a>1$

Then

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

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

$\lim_{n\rightarrow\infty}a^{n}=\infty$


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

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    $\begingroup$ 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
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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, $$a^n=(1+\varepsilon)^n\geq n\varepsilon+1.$$

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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.

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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

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

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  • $\begingroup$ 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
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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 $\frac{x}{a-1}$.

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

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  • $\begingroup$ 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
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    $\begingroup$ " (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
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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.

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