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Converge of infinite series

  1. Nov 16, 2006 #1
    Show that Σ[tex]_{n=1}^\infty ln(1 -1/n^2) = -ln 2[/tex]

    I'm not sure how to do this. Should I use telescopic sums or should I make a function [tex]y = ln(1 - 1/n^2)[/tex]? Is it possible to use telescopic sums here?
     
    Last edited: Nov 16, 2006
  2. jcsd
  3. Nov 16, 2006 #2
    When I make [tex]y = ln(1 - 1/n^2)[/tex]..then the limit of y = 0 which yields no conclusion about the series.
     
  4. Nov 16, 2006 #3

    quasar987

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    The sum cannot start at n=1 because then the first term blows up. I get the correct answer if n starts at 2. Two hints:

    1) Yes it's a telescoping series problem.

    2) ln(a+b) is not a friendly expressiom but ln(a*b) is because ln(a*b)=lna + lnb. And notice that [tex]1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{(n-1)(n+1)}{n^2}[/tex]
     
    Last edited: Nov 16, 2006
  5. Nov 16, 2006 #4

    quasar987

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    Incidentally, this problem proves indirectly that

    [tex]\prod_{i=2}^{\infty}\frac{i^2-1}{i^2}=\left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)\left(\frac{24}{25}\right)...=\frac{1}{2}[/tex]

    Nice!
     
  6. Nov 16, 2006 #5
    Thank you!
     
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