Prove 2^(-n) converges

  • Thread starter jrsweet
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  • #1
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Homework Statement


Using the definition of convergence to prove that the sequence {2^(-n)} converges


Homework Equations





The Attempt at a Solution


So, I just don't think I am thinking straight or something. Here is what I got so far:

Chose e>0. Let N be any positive integer greater than ______.

How do I chose the N? Do I need to compare 2^(-n) to something larger to be able to find an N? Any help would be much appreciated. Thanks!
 

Answers and Replies

  • #2
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Homework Statement


Using the definition of convergence to prove that the sequence {2^(-n)} converges


Homework Equations





The Attempt at a Solution


So, I just don't think I am thinking straight or something. Here is what I got so far:

Chose e>0. Let N be any positive integer greater than ______.

How do I chose the N? Do I need to compare 2^(-n) to something larger to be able to find an N? Any help would be much appreciated. Thanks!
First off, if you're going to prove that the sequence {2-n} converges to some value, you have to know what that value is. In other words, what is
[tex]\lim_{n \to \infty} 2^{-n}?[/tex]

I don't see any evidence that you know what this value is; at least you didn't include that information in your post.

Second, show us that you know the definition of a sequence converging to some value.
 
  • #3
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Chose e>0. Let N be any positive integer greater than 1/e. Then, for n>=N we have

|1/(2n)-0| < |1/n|= (1/n) <= (1/N) < (1/(1/e)) = e

Thus, the sequence converges to 0??

Does that look right?
 
  • #4
Dick
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Yes, that looks ok. But you really didn't need the 1/n part. You could have just tried to find an N such that 1/2^N<e.
 
  • #5
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I was confused how to find it that way. I tried this:

1/2N < e
1/e < 2N
ln(1/e) < Nln2
ln(1/e)/ln2 < N

does that work?
 
  • #6
Dick
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I was confused how to find it that way. I tried this:

1/2N < e
1/e < 2N
ln(1/e) < Nln2
ln(1/e)/ln2 < N

does that work?

Yes. Pick N>ln(1/e)/ln(2).
 
  • #7
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I used the Ratio Test. I am not sure I did it correctly but here it goes.

lim n->inf {[2^(-n+1)]/[2^(-n)]}
= lim n-> inf {[2^(-n)*(1/2)]/[2^(-n)]
=lim n-> inf (1/2) = 1/2 < 1 Therefore by the ratio test the series converges.

Does this look right?
 
  • #8
Dick
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I used the Ratio Test. I am not sure I did it correctly but here it goes.

lim n->inf {[2^(-n+1)]/[2^(-n)]}
= lim n-> inf {[2^(-n)*(1/2)]/[2^(-n)]
=lim n-> inf (1/2) = 1/2 < 1 Therefore by the ratio test the series converges.

Does this look right?

That's fine. But the original problem was to show the SEQUENCE converge by the DEFINITION (using epsilons etc). You proved the SERIES converges using the RATIO test. That does imply the sequence converges, but it's not what's being asked for.
 

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