How does zeta of zero equal to negative one half rather than to infinity?

959 Views Asked by At

$$\zeta(0)=(1/1^0)+(1/2^0)+(1/3^0)+(1/4^0)+(1/5^0)...$$ Am I right?

Anything raised to the power of zero is one. One to the power of zero is one. One divided by one is one. $$1/1^0=1$$ Am I right?

$$2^0=1$$ $$1/1=1$$ $$1/2^0=1$$ Am I right?

$$3^0=1$$ $$1/1=1$$ $$1/3^0=1$$ Am I right?

$$4^0=1$$ $$1/1=1$$ $$1/4^0=1$$ Am I right?

$$5^0=1$$ $$1/1=1$$ $$1/5^0=1$$ Am I right?

$$1+1+1+1+1...=\infty$$ Am I right?

Is $\zeta(0)$ not equal to $(1/1^0)+(1/2^0)+(1/3^0)+(1/4^0)+(1/5^0)...$?

My question is, just how does one compute $\zeta(0)$ to get $-1/2$?

1

There are 1 best solutions below

0
On BEST ANSWER

Here is a calculation. The Riemann formula is $$\zeta(s)=2^s\pi^{s-1}\sin\big(\frac{\pi s}{2}\big)\Gamma(1-s)\zeta(1-s)$$ We take the limit of this as $s\to 0$ and we use the fact that $s\zeta(1-s)\to -1$ as $s\to 0$, along with the fact that $\Gamma(1) = 1$. Furthermore, $\sin(\frac{\pi s}{2}) = \frac{\sin(\pi s/2)}{\pi s/2}\frac{\pi s}{2}$, and $\frac{\sin (\pi s/2)}{\pi s / 2} \to 1$ as $s \to 0$. Putting it all together gives you $$\zeta(0) = \lim_{s\to 0}\, 2^{s-1} \pi^s \cdot \frac{\sin(\pi s/2)}{\pi s/2} \cdot \Gamma(1-s) \cdot s\zeta(1-s) = 2^{-1} \pi^0 \cdot 1 \cdot \Gamma(1) \cdot (-1) = -\frac{1}{2}.$$