Understanding the Finite Value of ##\zeta(-1)##

[LagrangeEuler]

[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.

 

[jedishrfu]

There's an article here on how its computed:

http://blogs.scientificamerican.com/roots-of-unity/does-123-really-equal-112/

 

[LagrangeEuler]

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?

 

[mathman]

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?

The basic concept is analytic continuation.
https://en.wikipedia.org/wiki/Analytic_continuation
http://math.columbia.edu/~nsnyder/tutorial/lecture4.pdf

The second is specific for analytic continuation of zeta function.