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Sum of an Infinite Series

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine the sum of the following series:

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

    Sorry for poor latex, that is supposed to say infinity.

    2. Relevant equations
    How might we turn this into an easier function to deal with?


    3. The attempt at a solution

    So far I've only proved convergence of the series. I'm not really sure where to begin. Any help is appreciated. I've ran it through maple and come out with the sum equaling -ln(2).

    I thought of trying to work backwards to see if this series was some sort of taylor expansion but I failed at that. I just don't see any elementary techniques. Obviously maple spotted something I haven't lol.
     
    Last edited: Mar 23, 2009
  2. jcsd
  3. Mar 23, 2009 #2
    Maybe it is a geometric series. If so, why not figure out the 1st and 2nd term and the find the r. You are given the first term then use the formula a/ (1-r) where a is the first term and r is the rate.
     
  4. Mar 23, 2009 #3
    Okay,

    So I worked out the first two terms:

    A1 = log(1-1/4) = log(3/4)

    A2 = log(1-1/9) = log(8/9)

    then r = log(8/9) / log(3/4)

    But then checking A3 = log(1-1/16),

    A3 does not equal r*A2.

    So this is not a geometric series?
     
  5. Mar 23, 2009 #4

    Dick

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    It's not geometric. Your last hope is that it is a telescoping series. Is it? Write 1-1/(n+1)^2 as a single ratio and factor it up cleverly.
     
  6. Mar 23, 2009 #5
    Defenitely not.

    Try this instead:


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

    [tex]=\sum ln\keft(\frac{n(n+2)}{(n+1)^2}\left)=...=\sum[ln(n)+ln(n+2)-2ln(n+1)]=[/tex]

    [tex] =\sum[ln(n+2)-ln(n+1)]+\sum [ln(n)-ln(n+1)][/tex]

    Now stuff will cancel out, i already did more than i was supposed to.

    Edit: Sorry Dick, i didn't know you were already on it!
     
  7. Mar 23, 2009 #6

    Dick

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    S'ok. How could you, I hadn't posted yet. But you should start with a hint. You shouldn't steal all the fun from what seems to be a pretty clever poster. I consider 'pretty clever' to be realizing and proving that a series is not geometric.
     
    Last edited: Mar 23, 2009
  8. Mar 23, 2009 #7
    Haha, well thanks alot guys :) I had thought of building it into a single ratio but I thought I was only further complicating it. Thanks again.
     
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