# Riemann zeta

1. Oct 20, 2015

### 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.

2. Oct 20, 2015

### Staff: Mentor

3. Oct 20, 2015

### 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?

4. Oct 20, 2015