Unbounded Seq.: 3 is an Upper Bound

• fishturtle1
In summary, the conversation discusses an unbounded sequence for which 3 is an upper bound and no number less than 3 can also be an upper bound. An example of this type of unbounded sequence is {k}3k=-∞ where k starts at -∞ and ends at 3, and k∈ℤ. The first five terms of this sequence are -∞, -∞+1, -∞+2, -∞+3, -∞+4. The notation used in the conversation is a bit confusing, but the answer in the back of the book is fn=3-n which is similar to ak=4-k, k=1...∞. The bold
fishturtle1

Homework Statement

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.

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.

fishturtle1 said:

Homework Statement

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.
How can an unbounded sequence have an upper bound?
fishturtle1 said:

The Attempt at a Solution

example: {k}3k=-∞
Can you list the first, say, five terms in this sequence?
fishturtle1 said:
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.

fishturtle1 said:

Homework Statement

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.

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.

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

fishturtle1 said:

Homework Statement

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.

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.

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

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, -∞+4I 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.

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?

fishturtle1 said:
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
No, these aren't numbers.
fishturtle1 said:
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.

fishturtle1 said:
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...∞.

They are the same, assuming ##n## starts at ##0## in ##3-n##.
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?

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##?

fishturtle1 said:
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...∞.
This sequence is bounded since all of the terms are less than or equal to 3.
fishturtle1 said:
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?
I think the problem is poorly worded.

Mark44 said:
This sequence is bounded since all of the terms are less than or equal to 3.
I think the problem is poorly worded.

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

LCKurtz said:
Aren't you confusing "bounded above" with "bounded"?
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

LCKurtz said:
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##?
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?

Mark44 said:
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
Then why do you say in post #9 that ##a_k = 4-k,~k=1..\infty## is bounded?

Last edited:
fishturtle1 said:
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?

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

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

LCKurtz said:
Then why do you say in post #9 that ##a_k = 4-k,~k=1..\infty## is bounded?
I should have said "bounded above."

Mark44 said:
I should have said "bounded above."
Right. That's what I pointed out in post #10 in the first place.

1. What does it mean for 3 to be an upper bound in an unbounded sequence?

When 3 is an upper bound in an unbounded sequence, it means that all the terms in the sequence are less than or equal to 3. In other words, 3 is the highest possible value that can appear in the sequence.

2. How can we prove that 3 is an upper bound in an unbounded sequence?

To prove that 3 is an upper bound in an unbounded sequence, we need to show that for every term in the sequence, the value is less than or equal to 3. This can be done by using mathematical induction or by directly evaluating the terms in the sequence.

3. Is 3 the only possible upper bound in an unbounded sequence?

No, 3 is not the only possible upper bound in an unbounded sequence. Depending on the values and rules of the sequence, there can be multiple upper bounds. However, 3 is the minimum upper bound in an unbounded sequence where the terms are all positive integers.

4. Can an unbounded sequence have both an upper bound and a lower bound?

No, an unbounded sequence cannot have both an upper bound and a lower bound. By definition, an unbounded sequence has terms that continue infinitely without any limit or boundary. Therefore, it cannot have a lower bound.

5. How does knowing that 3 is an upper bound in an unbounded sequence help in understanding the sequence?

Knowing that 3 is an upper bound in an unbounded sequence can help in understanding the behavior and limitations of the sequence. It tells us that the terms in the sequence will never exceed the value of 3, which can provide insights into the patterns and trends in the sequence. Additionally, it can also help in determining the convergence or divergence of the sequence.

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