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Limit of a sequence

  • Thread starter srfriggen
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  • #1
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Homework Statement



Show that if sn[itex]\leq[/itex]b for all but finitely many n, then lim sn[itex]\leq[/itex]b.




Homework Equations





The Attempt at a Solution



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

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

l sn-s l < [itex]\epsilon[/itex].

Choose [itex]\epsilon[/itex]=s-b.

Then l sn-s l < s-b.

-(sn-s)<s-b

sn>b, contradiction to problem statement.



My question is this: The proof only seems to work if we assume sn-s is negative. But why couldn't sn 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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
SammyS
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Homework Statement



Show that if sn[itex]\leq[/itex]b for all but finitely many n, then lim sn[itex]\leq[/itex]b.

Homework Equations



The Attempt at a Solution



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

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

l sn-s l < [itex]\epsilon[/itex].

Choose [itex]\epsilon[/itex]=s-b.

Then l sn-s l < s-b.

-(sn-s)<s-b

sn>b, contradiction to problem statement.

My question is this: The proof only seems to work if we assume sn-s is negative. But why couldn't sn 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 sn - s l < s - b ,

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

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

Add s to all : b < sn .

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 sn > b, one of those n's is largest, call it N. What does that say about sn if n > N?
 
Last edited:
  • #3
288
3
If l sn - s l < s - b ,

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

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

Add s to all : b < sn .

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 sn > b, one of those n's is largest, call it N. What does that say about sn 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.
 
  • #4
288
3
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?
 
  • #5
SammyS
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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 sn > b . Right.

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

I.e. if sn ≥ b, then n ≤ N.

So, if n > N, then sn < b.
 
  • #6
Ray Vickson
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Homework Statement



Show that if sn[itex]\leq[/itex]b for all but finitely many n, then lim sn[itex]\leq[/itex]b.




Homework Equations





The Attempt at a Solution



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

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

l sn-s l < [itex]\epsilon[/itex].

Choose [itex]\epsilon[/itex]=s-b.

Then l sn-s l < s-b.

-(sn-s)<s-b

sn>b, contradiction to problem statement.



My question is this: The proof only seems to work if we assume sn-s is negative. But why couldn't sn 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

Homework Statement





Homework Equations





The Attempt at a Solution

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

RGV
 
  • #7
288
3
Maybe I could have stated it better.

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

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

I.e. if sn ≥ b, then n ≤ N.

So, if n > N, then sn < 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!
 

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