Value of Riemann zeta function at $-1$

Question

This claim is false $\sum_{n=1}^{\infty}n=\sum_{n=1}^{\infty}n^{-(-1)}= \zeta(-1)=-1/12$.

The error is that we should

$\sum_{n=1}^{\infty}n=\sum_{n=1}^{\infty}(1/n ^1)^{-1}=(0)^{-1}$.

Am I correct? It's difficult to say that an infinite sum like that don't diverge and that sum of positive numbers can give negative number.

Answer 1

Ralph Mellish said :

$\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$ is only true when Re(s)>1. To evaluate at s=−1 you would need to use the functional equation. Point being: the error you're making is with your second equality: ∑∞n=1n−(−1)≠ζ(−1). Also, no: ∑∞n=1(1/n1)−1≠(0)−1. It seems as though you're thinking that ∑(stuff)−1=(∑stuff)−1, which is not true, and you seem to be be thinking that ∑∞n=11/n=0, which is not true (perhaps you're not thinking these things, but it is the only way I can interpret what you wrote above "i'm i correct?"). –

but he doesnt know that this trick is often used in string theory which is the final theory of everything

(this was posted as an answer in the place of a comment because have small inductive reputation)

Answer 2

Use the functional equation: $$\zeta(s)=2^s\pi^{-1+s}\Gamma(1-s)\sin\frac{\pi s}2\zeta(1-s)\;,\;\;s\neq1\implies$$

$$\zeta(-1)=\frac1{2\pi^2}\cdot1\cdot(-1)\zeta(2)=-\frac1{12}$$