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anonymous

  • 5 years ago

how do you determine the final terms in a telescoping series when the cancelling terms are not consecutive?

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  1. anonymous
    • 5 years ago
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    Foolproof method: If you can't figure out a smart way, just write out the first 2/3/4 terms and check

  2. anonymous
    • 5 years ago
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    There will be better ways to figure it out, but that may be the most general/hardest to get wrong one.

  3. anonymous
    • 5 years ago
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    ok, I will try that now... stuck on a problem

  4. anonymous
    • 5 years ago
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    You could post it, if you wanted ¬_¬

  5. anonymous
    • 5 years ago
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    ok, that might help... \[\sum_{n=2}^{\infty} 2/n ^{2}-1\]

  6. anonymous
    • 5 years ago
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    I understand how to do the partial fraction decomposition

  7. anonymous
    • 5 years ago
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    I am getting stuck on the final terms when I sub in the series and cancel out

  8. anonymous
    • 5 years ago
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    how do I arrive to the following answer....

  9. anonymous
    • 5 years ago
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    \[s _{n}=1 + 1/2- (1/n-1) - (1/n)\]

  10. anonymous
    • 5 years ago
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    Write it out something like this (I would) 1/1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 ..... +1/(n-4) - 1/(n-2) +1/(n-3) - 1/(n-1) +1/(n-2) - 1/(n) If you start cancelling in pairs, you see that from the start only 1 + 1/2 don't cancel ... you could 'guess' that the second half of the last two will stay, but if you write out the last three too you can show this.

  11. anonymous
    • 5 years ago
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    so the last two terms in the last two parentheses stay?

  12. anonymous
    • 5 years ago
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    or the last two iterations

  13. anonymous
    • 5 years ago
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    not parentheses

  14. anonymous
    • 5 years ago
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    I don't see how you are coming up with the last 3 terms with the n...??

  15. anonymous
    • 5 years ago
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    See cancelling in image

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  16. anonymous
    • 5 years ago
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    As for the last terms, just sub in n-2, n-1 and n to the partial fraction 1/(n-1) - 1/(n+1)

  17. anonymous
    • 5 years ago
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    I guess that is what I don't get... why am i subbing in n-2, n-1, and n? is that always?

  18. anonymous
    • 5 years ago
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    Well, if it's a sum from 2 to N, it will include 2, 3, 4 .... n-3, n-2, n-1 and n etc. Why do you ONLY go back to n-2? The difference in the partial fractions n-1 and n+1 is 2, so you can 'guess' that that will cancel back with the one two behind, so you need the last two at the end (but I used three so you can 'check' that they cancel how you think they will (see diagram).

  19. anonymous
    • 5 years ago
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    You could sub in the first 5 and the last 5 if you have enough time, just to make sure. Or none, if you can guess which will stay. Just depends on how confident you are.

  20. anonymous
    • 5 years ago
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    ok, I think I understand.... thank you, Mr. Newton :)

  21. anonymous
    • 5 years ago
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    No problem

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