Riemann zeta

  • #1
596
10
[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
596
10
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
7,876
452

Related Threads on Riemann zeta

  • Last Post
Replies
1
Views
1K
Replies
2
Views
649
Replies
2
Views
684
Replies
7
Views
932
  • Last Post
Replies
2
Views
675
  • Last Post
Replies
2
Views
2K
Replies
11
Views
1K
Replies
5
Views
4K
Replies
2
Views
3K
Replies
1
Views
535
Top