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Determine whether the series converges

  1. Nov 10, 2005 #1
    Hello. The question says: “determine whether the series converges conditionally or absolutely, or diverges. The series they gave me is an alternating series so I used the alternating series test. The alternating series test has 2 parts. The 1st part says the limit as an goes to infinity must be 0 and an+1 <= an, for all n. I know how to do the 1st part but I don’t know how to find out if an+1 <= an. The answer in the back of the book says that it converges absolutely.
    [img=http://img392.imageshack.us/img392/9695/calchw46zm.th.jpg]
    Thanks
     
  2. jcsd
  3. Nov 10, 2005 #2

    quasar987

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    Are you a girl?

    to show that [itex]a_{n+1}\leq a_n[/itex], just...

    1. Write out the equality you want to prove:

    [tex]\frac{n+1}{(n+1)^3-1} \leq \frac{n}{n^3-1}[/tex]

    2. Manipulate the hostile inequality as you would an equation to reduce it to a simpler form for which you know the "value of thruth" (i.e. wheter it is true of not). For exemple, you would write

    [tex]\frac{n+1}{(n+1)^3-1} \leq \frac{n}{n^3-1} \Leftrightarrow ... \Leftrightarrow 0\leq n[/tex]

    which is true, hence the inequality of origin is true.
     
  4. Nov 10, 2005 #3
    Yes I'm a girl:smile: . How did you know?
    Can I plug in numbers and see if it is less than a_n because that equation gets really messy?
     
  5. Nov 10, 2005 #4
    I tried to simplify it and I got:

    (n-1)/[(n+1)(n+2)] <= 1

    If I simplified correctly then where do I go from here?
     
  6. Nov 10, 2005 #5

    Pyrrhus

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    why don't you try d'Alambert's method or the ratio test?
     
  7. Nov 10, 2005 #6
    What's d'Alambert's method ?
     
  8. Nov 10, 2005 #7

    Pyrrhus

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    If you're dealing with absolute convergence of a series, you should start with

    [tex] \sum_{n=2}^{\infty} |\frac{(-1)^{n} n}{n^{3} - 1}| [/tex]

    If this series of absolute values converge then the series absolutely converges.

    If that doesn't work you should apply the ratio test

    [tex] \lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}| < 1 [/tex]

    then the series absolutely converges.
     
  9. Nov 10, 2005 #8
    can you find out if it conditionally converges by the ratio test?
     
  10. Nov 10, 2005 #9

    Pyrrhus

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    Conditional convergence basicly means the series converges but its series of absolute values diverges. With the ratio test you can find absolute convergence and divergence.
     
  11. Nov 10, 2005 #10
    If I did the ratio test would I have to include the (-1)^n?
     
  12. Nov 10, 2005 #11

    Pyrrhus

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    of course.
     
  13. Nov 10, 2005 #12
    If it is not an alternating series then can it have conditional convergence?
     
  14. Nov 10, 2005 #13

    Pyrrhus

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    I said a series, i didn't especifically meant Alternating series, so what i said above applies to all.
     
  15. Nov 10, 2005 #14
    ok thanks. I think I understand it now.:smile:
     
  16. Nov 10, 2005 #15

    quasar987

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    I wanted to test my "girl handwritting recognizing" skills. :cool:

    You can't plug in the numbers. It has to be perfectly general (i.e. true for every n, or at least for all n greater than a certain N). The inequality may seem messy at first but once you get rid of the denominators, things start canceling out very quickly from side to side of the inequality and you're left with 4n >= -1, which is true for all n.
     
  17. Nov 10, 2005 #16
    and n can only be positive right?
     
  18. Nov 11, 2005 #17

    quasar987

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    Look at your sum: n goes from 2 to infinity. So yeah, n is always positive. Sometimes, n will run from 0 to infinity. In that case, n is not always positive. But the important thing is that there exists an N such that for n>N, the inequality [itex]a_{n+1} \leq a_n[/itex] holds. Do you see why only this "weaker" condition is necessary, and not "for all n"?
     
    Last edited: Nov 11, 2005
  19. Nov 11, 2005 #18
    If you have a rational expression for a_n where degree(numerator) < degree(denominator) IMo it's generally easier to show the result by dividing top and bottom by the polynomial which dominates in the denominator for large values of n.

    [tex]
    a_n = \frac{n}{{n^3 - 1}} = \frac{1}{{n^2 - \frac{1}{n}}}
    [/tex]

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

    So you want to show that a_n >= a_(n+1).

    The numerators are equal if you write them as I did above. If you can show that the denominator of a_(n+1) is greater than the denominator of a_n then you've shown the result. Considering that n is natural number it's not hard to do so. You should be able to finish it off.

    Hmm...reminds me of how I forgot to finish off a question on my exam with [tex]...\forall n > \max \left\{ {N,N'} \right\}[/tex].:grumpy:
     
    Last edited: Nov 11, 2005
  20. Nov 11, 2005 #19

    quasar987

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    sorry 'bout that :p
     
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