Show that a^n is unbounded if a>1

[Mr Davis 97]

Homework Statement

Prove that if $a > 1$, then the sequence $a^n$ is unbounded.

Homework Equations

The Attempt at a Solution

We need to show that $\forall M \in \mathbb{R}~\exists N \in \mathbb{N},~~ a^n > M$. To do this I thought maybe we could let $N = \operatorname{ceil}(\log_a(M))+1$, so that we are guaranteed that $a^N > M$, but I don't think this will work for several reasons. One, $M$ could be negative. And two, I formally haven't defined the notion of logarithm yet. What approach should I take then?

 

[Eclair_de_XII]

One, $M$ could be negative.

Why don't you just put absolute values around $M$, then, when defining $N$?

 

[Mr Davis 97]

Why don't you just put absolute values around $M$, then, when defining $N$?

Becuase, the sequence has to be greater than an arbitrary real number, not just positive real numbers, I think. Also, that still doesn't solve the issue that I haven't formally defined logarithms yet and so can't use them.

 

[Eclair_de_XII]

I mean, put absolute value signs around $M$ in your expression for $N$:

$N=\text{ceil}(\log_a(M))+1$

Surely, there is nothing stopping you doing that? This way, $M$ can still be negative and for the value of $N$, $a^N > M$.

 

[Ray Vickson]

Homework Statement

Prove that if $a > 1$, then the sequence $a^n$ is unbounded.

Homework Equations

The Attempt at a Solution

We need to show that $\forall M \in \mathbb{R}~\exists N \in \mathbb{N},~~ a^n > M$. To do this I thought maybe we could let $N = \operatorname{ceil}(\log_a(M))+1$, so that we are guaranteed that $a^N > M$, but I don't think this will work for several reasons. One, $M$ could be negative. And two, I formally haven't defined the notion of logarithm yet. What approach should I take then?

If all you want to prove is unboundedness (and are not concerned with how "fast" the sequence increases) you can use the elementary binomial theorem. Write the binomial coefficient "n choose k" as $C(n,k)$ and write $a > 1$ as $1+p$, with $p > 0 .$ Now the binomial expansion says that
$$(1+p)^n = 1 + n p + \underbrace{\sum_{k=2}^n C(n,k) p^k}_{>0},$$
so $a^n = (1+p)^n > 1 + np.$

As an exercise, you might like to use a similar argument to show that for any positive integer $k$, the sequence $a^n/n^k$ is unbounded (so that $a^n$ increases faster than any finite power of $n$). You can do all that without using logarithms anywhere.

 

[Delta2]

Becuase, the sequence has to be greater than an arbitrary real number, not just positive real numbers, I think. Also, that still doesn't solve the issue that I haven't formally defined logarithms yet and so can't use them.

If the sequence is greater than an arbitrary positive real number then it is greater than an arbitrary negative real number as well, because all negative real numbers are smaller than positive real numbers and ">" has transitive property. But yes if you haven't defined logarithms yet, your approach is not valid.

 

[StoneTemplePython]

Now the binomial expansion says that
$$(1+p)^n = 1 + n p + \underbrace{\sum_{k=2}^n C(n,k) p^k}_{>0},$$
so $a^n = (1+p)^n > 1 + np.$

Just to add a couple points: This is sometimes known as the Bernoulli inequality and is valid for any $p \geq -1$ (though technically I would not write the Bernouli inequality as strict in this case -- but the end result is the same though).

Trying to get a linear lower bound (which then has easy monotonic properties) is an effective strategy for OP to reach for... As a side note, I'd mention that for $p \in (-1,0)$ there is an interesting probability interpretation/proof via use of the union bound.

 

[Mark44]

We need to show that $\forall M \in \mathbb{R}~\exists N \in \mathbb{N},~~ a^n > M$.

You've omitted n (lower case) from you definition above.
All you need is $\forall M > 0~\exists N \in \mathbb{N}$ such that $n \ge N \Rightarrow a^n > M$

Becuase, the sequence has to be greater than an arbitrary real number, not just positive real numbers,

I don't understand your concern. You're given that a > 1, and your limit is as n grows large, so $a^n$ will be positive and large. So you don't need to be concerned about M being negative in your definition above..