# Unbounded sequence

1. Jan 23, 2017

### fishturtle1

1. The problem statement, all variables and given/known data
Either give an example or show that no example exists.

An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound.

2. Relevant equations

3. The attempt at a solution

example: {k}3k=-∞

and by this notation i mean that k starts at -∞ and ends at 3, and k∈ℤ .

I chose -∞ as the lower bound because I wanted this sequence to be unbounded. Then I chose 3 as the upper bound because that is what the original statement asked for.

My questions:
Does this example fulfill the original statement because I am unsure of what the question means by "and no β < 3 is an upper bound." The book uses β as the upper bound in previous pages.

Also does the way I wrote the sequence mean what I want it to mean? Because I am also confused on the notation.

2. Jan 23, 2017

### Staff: Mentor

How can an unbounded sequence have an upper bound?
Can you list the first, say, five terms in this sequence?

3. Jan 23, 2017

### Ray Vickson

Perhaps an "unbounded sequence" means what everybody else calls an "infinite sequence". An infinite sequence can be bounded or unbounded.

4. Jan 23, 2017

### LCKurtz

If you mean what I think you do, a better way to write it would be $a_k = 4-k,~k = 1 ..\infty$.

5. Jan 23, 2017

### fishturtle1

This is the definition of an unbounded sequence I've been using from online,

a sequence is bounded if it is bounded above and below <=> if ∃k∈ℝ such that | xn | ≤ k ∀n∈ℕ.

The first five terms in this sequence would be
-∞, -∞+1, -∞+2, -∞+3, -∞+4

I was thinking that an unbounded sequence can have an upper bound if it goes to infinity in some direction but converges to a number as well.

6. Jan 23, 2017

### fishturtle1

I checked the answer in the back of the book, and the answer is fn=3-n which seems similar to ak=4-k, k=1...∞.

I think I would have gotten this same answer had the original statement been " An unbounded sequence for which 3 is an upper bound ".

I'm still confused by the bold part: "An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound. "
I think that it means no number less than 3 can be an upper bound, but isn't this redundant since we said 3 is an upper bound in the first part of the statement?

7. Jan 23, 2017

### Staff: Mentor

No, these aren't numbers.

8. Jan 23, 2017

### LCKurtz

They are the same, assuming $n$ starts at $0$ in $3-n$.
No, it isn't redundant. Since $3$ is a term of the sequence, no number $x$ less than $3$ can be an upper bound. How could it be if $x < 3$?

9. Jan 23, 2017

### Staff: Mentor

This sequence is bounded since all of the terms are less than or equal to 3.
I think the problem is poorly worded.

10. Jan 23, 2017

### LCKurtz

Aren't you confusing "bounded above" with "bounded"?

11. Jan 23, 2017

### Staff: Mentor

I don't think so. The definition of bounded sequence I am using is that it is a sequence that is bounded above and bounded below. That is, for some M > 0, |an| < M for all n in Z+ (or similar restriction on n).
See https://proofwiki.org/wiki/Definition:Bounded_Sequence

12. Jan 23, 2017

### fishturtle1

Ok I think I get it. So we're told 3 is an upper bound. That means that the greatest number in this sequence is less than or equal to 3.

we're also told that no number less than 3 is an upper bound.

Therefore 3 must be included in this sequence.

If we were not told that no number less than 3 is an upper bound, then our sequence could have been something like xn=-10-n, n=0 . . ∞

Am i understanding this correctly?

13. Jan 23, 2017

### LCKurtz

Then why do you say in post #9 that $a_k = 4-k,~k=1..\infty$ is bounded?

Last edited: Jan 23, 2017
14. Jan 23, 2017

### LCKurtz

Yes, almost. You could have it true if the sequence just got arbitrarily close to $3$ but less than $3$.

15. Jan 23, 2017

### fishturtle1

Ok, thank you both for your help, helped me a lot on this question .

16. Jan 23, 2017

### Staff: Mentor

I should have said "bounded above."

17. Jan 23, 2017

### LCKurtz

Right. That's what I pointed out in post #10 in the first place.