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Telescoping Series

  1. Sep 30, 2006 #1
    I want to determine whether [tex] \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} [/tex] is convergent or divergent. I did the following:

    [tex] \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} + \frac{1}{j+1}) [/tex]. Writing down some terms, I got:

    [tex] (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5}) + (\frac{1}{4}-\frac{1}{6})... [/tex]. It seemed like all the terms were canceling except [tex] (1+\frac{1}{2}) [/tex].

    But when I looked at the answer, it said the sum was [tex] s_{n}= 1+\frac{1}{2}-\frac{1}{n-1}-\frac{1}{n} [/tex].

    Can anybody tell me how they got this?


    Thanks
     
  2. jcsd
  3. Sep 30, 2006 #2

    quasar987

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    S_n is the partial sum, not the whole sum. to get the whole sum, take the limit of S_n as n-->oo and you do get 1+1/2.
     
  4. Sep 30, 2006 #3

    quasar987

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    btw, it should be

    [tex] \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} - \frac{1}{j+1}) [/tex]

    Oh, but you developped the sum correctly, so never mind.
     
  5. Sep 30, 2006 #4
    but how did they get the partial sum?
     
  6. Sep 30, 2006 #5

    quasar987

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    but hum... I do calculate something different for S_n... I get

    [tex]S_n = 1+\frac{1}{2}-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n+1}[/tex]
     
  7. Sep 30, 2006 #6

    shmoe

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    It looks like they want:

    [tex]s_n=\sum_{j=2}^{n-1}\frac{1}{j-1}-\frac{1}{j+1}[/tex]

    i.e. s_4 is really the first two terms of your sum, that is s_4=(1-1/3)+(1/2-1/4). Then:

    [tex]s_n=\sum_{j=2}^{n-1}\frac{1}{j-1}-\sum_{j=2}^{n-1}\frac{1}{j+1}=\sum_{j=1}^{n-2}\frac{1}{j}-\sum_{j=3}^{n}\frac{1}{j}[/tex]
     
  8. Oct 1, 2006 #7
    So lets say I have [tex] \sum_{n=1}^{\infty} (\frac{3}{n(n+3)}) = (\frac{1}{n}-\frac{1}{n+3}) [/tex]. I said that this equals:

    [tex] (1-\frac{1}{4})+(\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+...+(\frac{1}{n-1}-\frac{1}{n+2})+(\frac{1}{n}-\frac{1}{n+3}) [/tex]. So how do I simplify this sequence of partial sums so that I can take the limit to get the value of the sum


    Also if the nth partial sum of a series [tex] \sum_{n=1}^{\infty} a_{n} [/tex] is [tex] s_{n} = 3-\frac{n}{2^{n}} [/tex] find [tex] a_{n} [/tex] and [tex] \sum_{n=1}^{\infty} a_{n} [/tex]. So would [tex] a_{n} = s_{n}-s_{n-1} = (\frac{-n}{2^{n}}+\frac{n-1}{2^{n-1}}) [/tex]? The sum would be [tex] \lim_{n\rightarrow \infty} s_{n} = 3 [/tex]?


    Thanks
     
    Last edited: Oct 1, 2006
  9. Oct 1, 2006 #8

    shmoe

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    The same way I showed you above, break the partial sum into two sums and adjust their variables of summation so the things inside the sums match and you can subtract them (except a few terms on the ends).

    Looks fine.
     
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