Riemann zeta

  • #1
LagrangeEuler
699
19
[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.
 

Answers and Replies

  • #3
LagrangeEuler
699
19
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?
 
  • #4
mathman
Science Advisor
8,059
540

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