# Show that a^n is unbounded if a>1

• Mr Davis 97
In summary: 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. In summary, the sequence ##a^n## is unbounded.
Mr Davis 97

## Homework Statement

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

## 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?

Delta2
Mr Davis 97 said:
One, ##M## could be negative.

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

Delta2
Eclair_de_XII said:
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.

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

Mr Davis 97 said:
##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##.

Mr Davis 97 said:

## Homework Statement

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

## 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.

Last edited:
StoneTemplePython and Delta2
Mr Davis 97 said:
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.

Ray Vickson said:
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.

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Mr Davis 97 said:
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##
Mr Davis 97 said:
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..

Delta2

## 1. What does it mean for a^n to be unbounded?

When a^n is unbounded, it means that as n increases without bound, the value of a^n also increases without bound. In other words, there is no limit to how large a^n can get.

## 2. What is the significance of a>1 in this statement?

The condition a>1 is important because it guarantees that a^n will always be a positive value. If a is less than 1, then a^n would decrease as n increases and would not be unbounded.

## 3. Can you provide an example to demonstrate this statement?

Sure, let's take a=2. When n=1, a^n=2^1=2. As n increases, a^n will also increase: a^2=2^2=4, a^3=2^3=8, and so on. As n approaches infinity, a^n will continue to increase without bound.

## 4. How does this concept relate to exponential growth?

Exponential growth is a type of unbounded growth, where the value of a function increases rapidly as the input increases. In this case, a^n represents the exponential growth function and a>1 ensures that the growth is unbounded.

## 5. Is there a way to prove that a^n is unbounded for all values of a>1?

Yes, we can use the limit definition of a^n to prove that it is unbounded. We can show that as n approaches infinity, a^n will also approach infinity, which is the definition of an unbounded function. This proof can be done using mathematical induction or the epsilon-delta definition of limits.

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