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Series converge?

  1. Jun 2, 2007 #1
    1. The problem statement, all variables and given/known data

    series converge?

    DL the img, sry, can't type it.
    [​IMG]

    what's the sum if converge?

    2. Relevant equations

    none.

    3. The attempt at a solution

    alternating series, but it's (-2)
     

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    Last edited: Jun 2, 2007
  2. jcsd
  3. Jun 2, 2007 #2

    quasar987

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    Are you saying it converges, or diverge?
     
  4. Jun 2, 2007 #3
    I don't know where to start, I didn't know what test to use?

    Can u give me a hint of which test to use?
     
  5. Jun 2, 2007 #4

    quasar987

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    Based on what you said, I assume that you already understood that the series can be rewritten as

    [tex]\sum_{n=1}^{\infty}2(-1)^{n}[/tex]

    ??

    I don't see which test can be used on this. But you can fall back on the very definition of convergence: A series converge if the sequence of the partial sums converge. But if you find two subsequences that converge to different values, then the sequence itself diverges. Can you find those subsequences?
     
  6. Jun 2, 2007 #5
    thx, that helps alot, so the alternating series?

    Can u factor out the 2? r u sure that can be done?i don't understand
     
  7. Jun 2, 2007 #6
    I got the series diverge, because the 2 cancels and it left with [tex]\sum_{n=1}^{\infty}(-1)^{n}[/tex]

    is it right?

    thx for ur help
     
  8. Jun 2, 2007 #7

    quasar987

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    Ah, there is a test you can use! And it's the easiest. If [itex]\lim_{n\rightarrow \infty} a_n\neq 0[/itex], then the series [itex]\sum a_n[/itex] diverges.
     
  9. Jun 2, 2007 #8
    Thanks for your help. But I still don't understand how can did u rewrite the series... to this [tex]\sum_{n=1}^{\infty}2(-1)^{n}[/tex]

    Sorry, I am a bit of slow, Please Explain more, Thank You!
     
  10. Jun 2, 2007 #9

    quasar987

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    Ok, here is how.

    [tex]\frac{(-2)^{n+1}}{2^n}=\frac{(-2)(-2)^n}{2^n}=(-2)\left(\frac{(-2)}{2}\right)^n=(-2)(-1)^n=2(-1)^{n+1}[/tex]

    Thus,

    [tex]\sum_{n=0}^{\infty}\frac{(-2)^{n+1}}{2^n}=\sum_{n=0}^{\infty}2(-1)^{n+1}=\sum_{n=1}^{\infty}2(-1)^{n}[/tex]
     
  11. Jun 3, 2007 #10
    catching the bump up in the index is tricky. notice that the series now starts at n=1.
     
  12. Jun 3, 2007 #11
    Thank you very much, this helps alot.

    So the series converge, the sum is either 0 or -2, depends if it's even or odd, right?
     
  13. Jun 3, 2007 #12

    Dick

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    A series cannot converge to two limits. That sort of behavior is called 'divergent'.
     
  14. Jun 3, 2007 #13
    o yea, thanks.
     
  15. Jun 4, 2007 #14

    Gib Z

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    Ahh I think a more appropriate word would have been is oscillating =]
     
  16. Jun 4, 2007 #15

    Dick

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    Right. And the sum of i from i=0 to infinity could more accurately be said to 'increase without bound' rather than diverge. :smile:
     
  17. Jun 4, 2007 #16

    Gib Z

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    Taking out the factor of 2, partial sums gives : -1, 0, -1,0,-1.... So it is bounded...
     
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