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Riemann zeta

  1. Oct 20, 2015 #1
    [tex]\sum^{\infty}_{n=1}\frac{1}{n^{\alpha}}=\zeta(\alpha) [/tex]
    For ##\alpha=-1##

    ##\zeta(-1)=-\frac{1}{12}##
    I do not see any difference between sum
    ##1+2+3+4+5+...##
    and ##\zeta(-1)##. How the second one is finite and how we get negative result when all numbers which we add are positive. Thanks for the answer.
     
  2. jcsd
  3. Oct 20, 2015 #2

    jedishrfu

    Staff: Mentor

  4. Oct 20, 2015 #3
    I do not understand this so well. So
    Series
    ##\sum^{\infty}_{n=1}\frac{1}{n^{\alpha}}## converges for ##\alpha>1##. Why in complex plane ##\zeta(-1)## makes sence?
     
  5. Oct 20, 2015 #4

    mathman

    User Avatar
    Science Advisor

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