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Euler sum

  1. May 23, 2007 #1
    This may be a stupid question, but I do not understand why the Euler sum is infinite for zeta=1. Why is "1+1/2+1/3+1/4+... " infinite, but zeta=2 (1+1/4+1/9+...) not?
     
  2. jcsd
  3. May 23, 2007 #2

    mjsd

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    1+1/2+1/3+... is [tex]\sum_n \frac{1}{n}[/tex] is a harmonic series and there are various tests to show that this diverges while
    1+1/4+1/9+.... is really [tex]\sum_n \frac{1}{n^2}[/tex] and this series converges
     
  4. May 23, 2007 #3

    Gib Z

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    The first series is a p series where p is equal to 1. We know for convergence of p series that a condition is that p>1.

    The second series converges to [itex]\pi^2/6[/itex] but is somewhat harder to show. Just to show it converges at all though, p=2 which is more than 1. Look up p series.
     
  5. May 23, 2007 #4

    mathman

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    1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+.....>
    1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+.....=
    1+1/2+1/2+ 1/2+.... which diverges.
     
  6. May 23, 2007 #5
    Brilliant!
     
  7. May 24, 2007 #6
    You can use the integral test: [tex] 1+1/2+1/3+++1/n> \int_N\frac{1}{x}dx =ln (N )\rightarrow \infty. [/tex]

    And: [tex] 1/4+1/9+1/16 +(1/(N+1)^2 < \int_1^N\frac{1}{x^2}dx =1-1/N. [/tex]
     
    Last edited: May 25, 2007
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