Limit of a Sequence: Why Do We Assume the Difference is Negative?

[srfriggen]

Homework Statement

Show that if s_n\leqb for all but finitely many n, then lim s_n\leqb.

Homework Equations

The Attempt at a Solution

My question is regarding the absolute value portion of the proof:

by contradiction: Call lim s_n s. Suppose s>b. Then

l s_n-s l < \epsilon.

Choose \epsilon=s-b.

Then l s_n-s l < s-b.

-(s_n-s)<s-b

s_n>b, contradiction to problem statement.



My question is this: The proof only seems to work if we assume s_n-s is negative. But why couldn't s_n be greater than s? Am I missing something important in the wording of the problem? Perhaps the "all but finitely man n"? I actually don't quite grasp what that means.

If the sequence was 1/n



Choose

 

[SammyS]

Homework Statement

Show that if s_n\leqb for all but finitely many n, then lim s_n\leqb.

Homework Equations

The Attempt at a Solution

My question is regarding the absolute value portion of the proof:

by contradiction: Call lim s_n s. Suppose s>b. Then

l s_n-s l < \epsilon.

Choose \epsilon=s-b.

Then l s_n-s l < s-b.

-(s_n-s)<s-b

s_n>b, contradiction to problem statement.

My question is this: The proof only seems to work if we assume s_n-s is negative. But why couldn't s_n be greater than s? Am I missing something important in the wording of the problem? Perhaps the "all but finitely man n"? I actually don't quite grasp what that means.

If the sequence was 1/n

Choose

If l s_n - s l < s - b ,

then -(s - b) < s_n - s < s - b .

That's -s + b < s_n - s < s - b .

Add s to all : b < s_n .

As for the "all but finitely many n":

That's going to be an important part of the proof.

Certainly, of those values of n, for which s_n > b, one of those n's is largest, call it N. What does that say about s_n if n > N?

 

[srfriggen]

If l s_n - s l < s - b ,

then -(s - b) < s_n - s < s - b .

That's -s + b < s_n - s < s - b .

Add s to all : b < s_n .

As for the "all but finitely many n":

That's going to be an important part of the proof.

Certainly, of those values of n, for which s_n > b, one of those n's is largest, call it N. What does that say about s_n if n > N?

Ok. I've looked over the problem for a while now and I understand the absolute value portion. You still considered when sn-s is positive but it didn't affect the contradiction.

The only conclusion I am able to make about your last question, and I don't know how this affects the proof, is that s(n) doesn't exist where n>N.

 

[srfriggen]

Wait, but that means there is no N in Naturals such that n>N implies ls(n)-sl<epsilon, for some epsilon >0. So that is the contradiction? That the limit doesn't exist?

 

[SammyS]

Ok. I've looked over the problem for a while now and I understand the absolute value portion. You still considered when sn-s is positive but it didn't affect the contradiction.

The only conclusion I am able to make about your last question, and I don't know how this affects the proof, is that s(n) doesn't exist where n>N.

Maybe I could have stated it better.

There are only a finite number of n values for which s_n > b . Right.

Let N be the index (subscript) of the last s_n for which s_n ≥ b.

I.e. if s_n ≥ b, then n ≤ N.

So, if n > N, then s_n < b.

 

[Ray Vickson]

Homework Statement

Show that if s_n\leqb for all but finitely many n, then lim s_n\leqb.

Homework Equations

The Attempt at a Solution

My question is regarding the absolute value portion of the proof:

by contradiction: Call lim s_n s. Suppose s>b. Then

l s_n-s l < \epsilon.

Choose \epsilon=s-b.

Then l s_n-s l < s-b.

-(s_n-s)<s-b

s_n>b, contradiction to problem statement.



My question is this: The proof only seems to work if we assume s_n-s is negative. But why couldn't s_n be greater than s? Am I missing something important in the wording of the problem? Perhaps the "all but finitely man n"? I actually don't quite grasp what that means.

If the sequence was 1/n



Choose

Suppose r = s-b > 0. Choose any ε > 0, ε < r. There exists N so that for all n ≥ N we have
|s-s_n| < ε, meaning that s - ε < s_n < s + ε, so s_n > s-ε > s-r = b, and this contradicts the original hypothesis.

RGV

 

[srfriggen]

Maybe I could have stated it better.

There are only a finite number of n values for which s_n > b . Right.

Let N be the index (subscript) of the last s_n for which s_n ≥ b.

I.e. if s_n ≥ b, then n ≤ N.

So, if n > N, then s_n < b.

aha! I see it now! Confusion between notation of s(n) and n's was throwing me off in the wrong direction (literally, on the number line, in the wrong direction!).

Thank you for your patience and guidance!