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precal

  • 2 years ago

sequence problem _ Calculus

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  1. precal
    • 2 years ago
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    |dw:1351476871325:dw|

  2. precal
    • 2 years ago
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    don't I need to use some formula?

  3. zepdrix
    • 2 years ago
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    So is the question, "Does the series converge?" ? :o

  4. precal
    • 2 years ago
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    yes, I know it converges to 9/4

  5. precal
    • 2 years ago
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    doesn't help that I don't know why?

  6. precal
    • 2 years ago
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    |dw:1351477345222:dw|

  7. precal
    • 2 years ago
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    |dw:1351477381056:dw|

  8. zepdrix
    • 2 years ago
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    |dw:1351477186152:dw| Hmm can't you do some type of comparison test? Sorry I'm a little rusty with these :D We know that the p-series 1/n^2 converses, and our series is smaller than that one :O something like that..

  9. precal
    • 2 years ago
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    |dw:1351477424396:dw|

  10. precal
    • 2 years ago
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    I am as rusty as you

  11. zepdrix
    • 2 years ago
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    Hmm :[

  12. precal
    • 2 years ago
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    how about if I do this as a telescoping series? as if I know what I am talking about

  13. zepdrix
    • 2 years ago
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    Yah that's prolly the way to approach it c: try to recognize a pattern coming out of it.

  14. precal
    • 2 years ago
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    I know if I factor out the 9 and then I just need to create the (1/4)

  15. zepdrix
    • 2 years ago
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    Oh oh oh i think it's coming back to me a little bit... I think in order to write it as a telescoping series, we need to rewrite it as partial fractions, then we'll get terms being subtracted in each n, allowing for some cancellations probably.

  16. colorful
    • 2 years ago
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    *

  17. zepdrix
    • 2 years ago
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    |dw:1351478870156:dw|

  18. zepdrix
    • 2 years ago
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    |dw:1351479010862:dw|

  19. colorful
    • 2 years ago
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    |dw:1351479486392:dw|

  20. zepdrix
    • 2 years ago
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    oh sorry XD hehe

  21. precal
    • 2 years ago
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    sorry, I ran off to watch some youtube videos on this

  22. zepdrix
    • 2 years ago
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    Mmmm that's prolly a good idea :) I don't think I'm getting anywhere with this one lol

  23. precal
    • 2 years ago
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    |dw:1351479960238:dw|got this as well

  24. precal
    • 2 years ago
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    |dw:1351480090695:dw|

  25. colorful
    • 2 years ago
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    the next few terms are something like (1/8-1/5+1/12)+(1/10-1/6+1/14)

  26. precal
    • 2 years ago
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    I thought somehow we would be left with 1/4 as the first term then it would make sense that they solution is 9/4

  27. precal
    • 2 years ago
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    wonder if we messed up on the partial fraction

  28. colorful
    • 2 years ago
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    I don't think so

  29. zepdrix
    • 2 years ago
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    Ummm I doubt it, it's likely that the terms just cancel out in a strange way :D

  30. precal
    • 2 years ago
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    well that is good to know

  31. precal
    • 2 years ago
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    yes, wonder if we have to do more than the first three terms to see it

  32. zepdrix
    • 2 years ago
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    |dw:1351480321431:dw|

  33. precal
    • 2 years ago
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    |dw:1351480632232:dw|

  34. zepdrix
    • 2 years ago
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    Oh hmm :D

  35. precal
    • 2 years ago
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    ok that is all I see right now

  36. zepdrix
    • 2 years ago
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    nasty little problem :3

  37. precal
    • 2 years ago
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    yes, well I think I will sleep on it, would not be the first time I went to bed with an unsolved problem....thanks..........

  38. precal
    • 2 years ago
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    |dw:1351531637965:dw|ok I think I solved it @zepdrix

  39. zepdrix
    • 2 years ago
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    Hmmmmmm that doesn't quite look right <:o you can't split up the base of a fraction. \[\frac{ 1 }{ 2n+4 }=\frac{ 1 }{ \infty+4 }=0\]

  40. zepdrix
    • 2 years ago
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    |dw:1351531992905:dw|