I have been given a set

S = {(-1)^n [(3n-1)/(n+1)] : n E N}

That is, n is an element of the Natural Numbers {1,2,3,...}

I am required to prove that 3 is an upper bound for S and justify whether I think it is a least upper bound.

Now I think I can do this but would appreciate some guidance.

I will start.

Intuitively, we can see that the sequence of numbers approaches 3 and -3 as n goes to infinity.

S = {-1, 1.7, -2, 2.2, -2.3, ... }

But for a precise solution we need: For every e > 0 there exists s E S such that s > 3 - e.

Now I think this last part is saying that there exists some upper bound for the set. Is this right?

So I will fix e > 0 and I want to find an n E N such that

|((3n-1)(-1)^n)/(n+1) - 3| > e

If I can find an n which is less than epsilon have I proven that my chosen upper bound is indeed an upper bound?

Now I am thinking of simplifying the above expression and I really want to get rid of the (-1)^n bit because it is very annoying.

I will assume that n is even. In this case I can eliminate (-1)^n. So my equation becomes

|(3n-1)/(n+1) - 3| < e for n is even.

|((3n-1)/(n+1)) - 3| = |-4/(n+1)|

|-4/(n+1)| < e

- 4/e < n+1

n > -1 - 4/e

Take N to be any integer bigger than -1 - 4/e then n >= N implies |((3n-1)/(n+1))| < e so 3 is an upper bound.


I have a feeling that this is completely and utterly wrong. Please feel free to tell me. I would appreciate it if someone could check and tell me where I went wrong or a better or correct way of doing this.

Thanks for the time.

Cheers.
Hence

|((3n-1)/(n+1))

I think I have done this incorrectly.

The question asks to show that 3 is an upper bound. To prove that a non-empty set has an upper bound there must be k such that k >= s for every s in the set S.

This definition says that no matter what element we choose in the set S there is a k which is bigger. This k is called the upper bound. And I want to prove that k = 3 for the set S = {(-1)^n (3n-1)/(n+1)}.

S = {-1, 1.7, -2, 2.2, -2.3, ...} unordered
S = {..., -2.3, -2, -1, 1.7, 2.2, ...} ordered

I would like to prove that k = 3 is an upper bound. An upper bound is defined as being a number in N such that for all s in S, k >= s. If k < s then k is not an upper bound.

How would I prove that k = 3 is an upper bound?

Would I have to prove by contradiction/contrapositive/ or by cases?

How about this...

Ignoring the (-1)^n bit we have

(3n-1)/(n+1) = 3((n-1)/(n+1))

We know that n-1 < n+1 for all n.

Hence (n-1)/(n+1) < 1

Therefore 1 is an upper bound for the set T = {(n-1)/(n+1):nEN}

In other words the elements of T do not get bigger than 1.

Multiply all elements of T by 3 such that 3T = S (since 3T = 3[(n-1)/(n+1)] = S) it makes sense that the upper bound of S is now 3.

Does this make sense??? How would I write this logically???

In order to show that 3 is an upper bound you only need to show that 3 is larger than any member of the set. Your intuitive notion that the sequence has the two limit points -3 and 3 is correct. You would still need to show that every member of the sequence is less than (or equal to) 3.

Clearly, if n is odd, S_n is negative and less than 3. Therefor, we can simply "throw" that case away. If n is even, S_n is positive and is equal to (3n-1)/(n+1). If, for any n, (3n-1)/(n+1)> 3, then 3n-1> 3(n+1)= 3n+ 3 (clearly n+1 is positive). That's the same as -1> 3 which is false. S_n can never be greater than 3 and so 3 is an upper bound.

Showing that 3 is the LEAST upper bound is harder. Since 3 is an upper limit, showing that 3 is the limit of the terms for even n would suffice. That would mean that no number less than 3 could be an upper bound since we could take "epsilon"= 3 minus that number and show that there must be a member of the sequence closer to 3 than that: i.e. larger than the presumed upper bound.

Excellent! Thanks HallsofIvy.

Showing that 3 is the LEAST upper bound is harder. Since 3 is an upper limit, showing that 3 is the limit of the terms for even n would suffice. That would mean that no number less than 3 could be an upper bound since we could take "epsilon"= 3 minus that number and show that there must be a member of the sequence closer to 3 than that: i.e. larger than the presumed upper bound.

HallsofIvy, I understand this method. Could you guide me on my calculations.



Prove that (3n-1)/(n+1) has the limit of 3 as n goes to infinity.

(epsilon is denoted "e")

Fix e > 0. We need to find an n E N such that n >_ N

|(3n - 1)/(n + 1) - 3| < e

<=> |(3n - 1 - 3n - 1)/(n + 1)| < e

<=> -2/(n + 1) < e

<=> -2 < e(n + 1)

<=> n > -2/e - 1

Now take N to be any integer bigger than -2/e - 1 then

n >_ N => |(3n-1)/(n+1) - 3| < e is true

Thus 3 is a least upper bound (or supremum)

<=> |(3n - 1 - 3n - 1)/(n + 1)| < e

<=> -2/(n + 1) < e


You have a mistake on the second line here, you didn't take the absoulute value into account. -2/(n + 1) is always negative, since n>0.

In the end you said to take an N that's larger than -2/e - 1, which is a negative number. That's usually a good sign that you made a mistake somewhere :).

You are right Shmoe. I tried to fix it up. How does it look now?

Fix e > 0. We need to find an n E N such that n >_ N

|(3n - 1)/(n + 1) - 3| < e

<=> |(3n - 1 - 3n - 1)/(n + 1)| < e

<=> |-2/(n + 1)| < e

Since n > 0 the term on the left is always negative

<=> 2/(n + 1) < e

<=> 2 < e(n + 1)

<=> n > 2/e - 1

Now take N to be any integer bigger than 2/e - 1 then

n >_ N => |(3n-1)/(n+1) - 3| < e is true

Thus 3 is a least upper bound (or supremum)

May I ask what the '<=>' is? Does it just mean 'from previous line' or does it have an extra meaning?

oxymoron- that looks correct. This may seem picky, but you've shown that the limit of the absolute value of your original sequence is 3. This also shows that the limit of the even terms of your original sequence is 3 (limit of a subsequence of a convergent sequence), which is what you wanted (see HallofIvy's post). Picky, but worth mentioning.

Now combine that with a proof that 3 is an upper bound, and you're done.


May I ask what the '<=>' is? Does it just mean 'from previous line' or does it have an extra meaning?

It means "if and only if" or "iff" for short, so both lines are equivalent.

Ah, thank you.

Now combine that with a proof that 3 is an upper bound, and you're done.

Okay, so I have my proof that 3 is definately an upper bound. Then I proved that 3 is also the least upper bound (by showing that 3 is an upper bound and then showing that there is no number smaller than 3 than is also an upper bound - which I have shown above).

Can I also show that 3 is a least upper bound by additionally proving that the sequence can be thought of as monotone increasing.

Ah, but the set is non-convergent isn't it? Or is it? Hold on, I am getting confused.

If it is monotone increasing, and has an upper bound (3), then the sequence has a least upper bound. This number (LUB) will be the number that the sequence converges to.

Is this logical? Or have I done enough already to prove that 3 is a Least Upper Bound?

Thanks for any input And I appreciate the help so far.

Cheers.

Can I also show that 3 is a least upper bound by additionally proving that the sequence can be thought of as monotone increasing.

No, having a monotone sequence that's bounded by 3 doesn't mean that it's limit is 3. It means it's limit is less than or equal to 3. You need the limit to be equal to 3 to show that 3 is the lub.


Ah, but the set is non-convergent isn't it? Or is it? Hold on, I am getting confused.

A "set" on it's own isn't convergent or divergent. If you're thinking of the set as a sequence (with the nice order they've given to you), then it's divergent. However, if you look at the subsequence of even terms, this sequence converges to 3.



Is this logical? Or have I done enough already to prove that 3 is a Least Upper Bound?

You have the pieces. 3 being an upper bound for your set and the limit of a sequence taken from your set shows it's an upper bound.