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 $-1/2$?
 A: 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
$$\zeta(0) = \lim_{s\to 0}\, 2^{s-1} \pi^s \cdot \frac{\sin(\pi s/2)}{\pi s/2} \cdot \Gamma(1-s) \cdot s\zeta(1-s) = 2^{-1} \pi^0 \cdot 1 \cdot \Gamma(1) \cdot (-1) = -\frac{1}{2}.$$ 
A: When you have a divergent series, there are ways to give it a "meaningful" finite answer anyways. One way is by using the Ramanujan summation. All of the ways of assigning values to divergent series don't tell you that they don't diverge; they simply do a partial sum (or other method). This example is using the Ramanujan summation:
Zander 6-13-19 (don't mind this, I just like to label my stuff with the date)
$ \begin{array}{l}
\zeta ( 0) =\sum ^{\infty }_{n=1}( 1) =\infty ,\ but\ with\ the\ ramanujan\ summation\\
\\
Ramanujan\ summation\\
\boxed{\sum\limits ^{\infty }_{n=1} f( n)\overset{\Re }{=} -\frac{f( 0)}{2} +i\int\limits ^{\infty }_{0}\frac{f( ix) -f( -ix)}{e^{2\pi x} -1} dx} \Longrightarrow \\
f( n) =1\\
\\
\zeta ( 0) =\sum ^{\infty }_{n=1}( 1)\overset{\Re }{=} -\frac{1}{2} +i\int\limits ^{\infty }_{0}\frac{1-1}{e^{2\pi x} -1} dx=-\frac{1}{2} +i\int\limits ^{\infty }_{0} 0dx=-\frac{1}{2}\\
\therefore \zeta ( 0) =-\frac{1}{2} \Re 
\end{array}$
The $\displaystyle \Re $ means that it is only equal to that value, because the Ramanujan summation assigned it
that value. The Ramanujan summation should not normally (if ever, I'm not sure) be used on a
convergent series, it is a method of analytic continuation if the sumagrand (function in the infinite series) has an appropriate growth pattern. Also, just because this (and other stuff too) says it is $-1/2$, it still equals +infinity. But we define it differently with
analytic continuation.
It's more like saying that zeta of zero equals $-1/2$ rather than the infinite sum.
