# Riemann zeta

LagrangeEuler
$$\sum^{\infty}_{n=1}\frac{1}{n^{\alpha}}=\zeta(\alpha)$$
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.