• Support PF! Buy your school textbooks, materials and every day products Here!

Show that a^n is unbounded if a>1

  • #1
1,456
44

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?
 

Answers and Replies

  • #2
803
37
One, ##M## could be negative.
Why don't you just put absolute values around ##M##, then, when defining ##N##?
 
  • #3
1,456
44
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.
 
  • #4
803
37
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##.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

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.
 
Last edited:
  • #6
Delta2
Homework Helper
Insights Author
Gold Member
2,423
683
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.
 
  • #7
StoneTemplePython
Science Advisor
Gold Member
2019 Award
1,145
546
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.
 
Last edited:
  • #8
33,262
4,962
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##


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..
 

Related Threads for: Show that a^n is unbounded if a>1

  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
4
Views
727
Replies
12
Views
5K
  • Last Post
Replies
16
Views
10K
  • Last Post
Replies
4
Views
1K
Top