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anonymous

  • 5 years ago

summation series question: for the following equation do i use geometric series theorem, telescoping or harmonic methods? the problem is: (sorry idk how to illustrate summation LOL) infinity ----- `. { (1/2^k)-1/(2^k+1) } / ______ k=1

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  1. anonymous
    • 5 years ago
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    Is your summation,\[\sum_{k=1}^{\infty}\left( \frac{1}{2^k}-\frac{1}{2^{k+1}} \right)\]?

  2. anonymous
    • 5 years ago
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    Because if it is, you should look at expanding some terms and seeing what happens:\[(\frac{1}{2}-\frac{1}{2^2})+(\frac{1}{2^2}-\frac{1}{2^3})+(\frac{1}{2^3}-\frac{1}{2^4})\]\[=\frac{1}{2}+(-\frac{1}{2^2}+\frac{1}{2^2})+(-\frac{1}{2^3}+\frac{1}{2^3})-\frac{1}{2^4}\]\[=\frac{1}{2}-\frac{1}{2^4}\]

  3. anonymous
    • 5 years ago
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    So for the first N summations, you would have,\[\sum_{k=1}^{N}:=\frac{1}{2}-\frac{1}{2^{N+1}}\]

  4. anonymous
    • 5 years ago
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    This is a partial sum, and the limit of the sequence of partial sums equals the limit of the series. So all you have to do is take N to infinity now.

  5. anonymous
    • 5 years ago
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    The series sums to 1/2.

  6. anonymous
    • 5 years ago
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    thanks! so i this would be geometric, a = 1/2 and r= 1/2 ?

  7. anonymous
    • 5 years ago
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    Technically, you should use mathematical induction to show for the N'th partial sum, but I don't think people are too precious about it.

  8. anonymous
    • 5 years ago
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    No, this is telescoping.

  9. anonymous
    • 5 years ago
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    alright, so the first few terms will cancell out and then i'll just use the initial term and subtract the function thing.. is there a way i can just look at the problem and know what it is? or am i going to have to use trial and error?

  10. anonymous
    • 5 years ago
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    Well, when I saw what you'd written, because of the minus sign between the two fractions, I thought to look at the first few terms to get a feel. You can then see a pattern of one term between the outer ones killing off another term. This is the definition of telescoping. You can't technically take the limit until you've set up a partial sum. Then you send N to infinity. For a geometric, you would see in the summand, or upon expanding a few terms, a common growth factor. You can test for geometric by dividing the (n+1)th term by the nth. If what you get is something independent of n, you have a common ratio, and your series is geometric.

  11. anonymous
    • 5 years ago
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    wonderful. thanks for your help! i'm still not looking forward to this exam tomorrow...

  12. anonymous
    • 5 years ago
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    Good luck :)

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