How does zeta of zero equal to negative one half rather than to infinity? $$\zeta(0)=(1/1^0)+(1/2^0)+(1/3^0)+(1/4^0)+(1/5^0)...$$
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 $-1/2$?

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
$$\zeta(0) = \lim_{s\to 0}\, 2^{s-1} \pi^s \cdot \frac{\sin(\pi s/2)}{\pi s/2} \cdot \Gamma(1-s) \cdot s\zeta(1-s) = 2^{-1} \pi^0 \cdot 1 \cdot \Gamma(1) \cdot (-1) = -\frac{1}{2}.$$ 
