Basic question about the zeros of Riemann Zeta Function I'm starting to learn about RZF. 
$$\zeta:\mathbb{C}\rightarrow\mathbb{C}, \,\,\,\,\zeta(z)=\sum_{n=1}^{\infty}\frac{1}{n^{z}} $$
I read that for $n\in\mathbb{N}$, $\zeta(-2n)=0$. But this is not making any sense to me, because $$\zeta(-2n)=\sum_{k=1}^{\infty}\frac{1}{k^{-2n}}=\frac{1}{1^{-2n}}+\frac{1}{2^{-2n}}+\frac{1}{3^{-2n}}+\cdots $$
How is a sum of positive terms not equal to zero? Maybe I'm making some stupid mistake, but I'm not understanding.
 A: You are correct that the initial definition of the Riemann zeta function does not help us make sense of the expression $\zeta(-2n)$. The core idea here is the idea of analytic continuation, which allows us to uniquely extend certain analytic functions on a given region to a holomorphic function on the entire complex plane. I will sketch below one approach how this can be done.
One can, rather easily, extend the region of convergence of the Riemann Zeta function to $\operatorname{Re} s > 0$. 
Using the Dirichlet Eta function, one easily sees that we have:
\begin{align}
\eta(s) = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s} = (1-2^{1-s})\zeta(s)
\end{align}
The Dirichlet Eta function, however, converges for $\operatorname{Re} s > 0$ (as long as $s \neq 1$), so we can define the Riemann Zeta function for any $s$ with $\operatorname{Re} s > 0$ as
$$\zeta(s) = \frac{1}{1-2^{1-s}}\eta(s)$$
This is equal to the initial series definition for any complex $s$ whenever both definitions exist, so it seems like a natural extension to the Riemann Zeta function.
Now, one can observe, that in the region $\{s\in\mathbb{C}:0<\operatorname{Re} s<1\}$, that the thus defined Riemann Zeta function obeys the functional equation
$$\zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s) \tag1$$
Now, for any value of $s$ where $\operatorname{Re} s < 0$, we have that $\operatorname{Re}(1-s) > 1$ - therefore, for any $s$, at least one of the two Zeta functions appearing in $(1)$ exists, and can thus be used to define the other. Therefore, ensuring that this functional equation holds everywhere (which it must, for the continuation of the zeta function to be holomorphic) gives as a way to define the Riemann zeta function on the entirety of the complex plane.
Now, however, if you substitute $s = -2n$ for some $n\in\mathbb{N}$ into $(1)$, we have:
$$\zeta(-2n) = -2^{-2n}\pi^{-2n-1}\sin(n\pi)\Gamma(2n + 1)\zeta(2n+1)$$
which can be seen to be zero due to the $\sin(n\pi) = 0$ for any integer $n$ and all the other terms remaining finite.
A: Your series definition of the Riemann zeta function is only valid for values of $z \in \mathbb C$ with ${\rm Re}(z) > 1$. That is because the condition ${\rm Re}(z) > 1$ is required to ensure that the series converges. You can't use this series definition for $z \in -2\mathbb N$, because, as you have observed, the series diverges.
Instead, you should use the analytic continuation of the Riemann zeta function. This is a function, holomorphic on the whole of $\mathbb C \setminus \{ 1 \}$, that agrees with the your series expression $z \mapsto \sum_{k=1}^\infty \frac 1 {n^z}$ on the region $\{ z \in \mathbb C : {\rm Re}(z) > 1\}$.
[If such an analytic continuation exists, then it is unique by the identity theorem in complex analysis.]
One way to construct such an analytic continuation is described in Baker's "Comprehensive Course in Number Theory". There are two steps in this construction:
(1) Extend $\zeta(z)$ to the larger region $\{ z \in \mathbb C : {\rm Re}(z) > -1, z \neq 1 \}$ using the formula
$$ \zeta(z) = z \int_1^\infty \frac{\lfloor x \rfloor - x + \tfrac 1 2 }{x^{z+1}} dx + \frac{1}{z - 1} + \frac 1 2.$$
(That this formula agrees with your series expression when ${\rm Re}(z) > 1$ is proved in the book.)
(2) Show that the version of $\zeta(z)$ defined in (1) obeys the reflection formula
$$ \zeta(z) = 2^z \pi^{z - 1} \sin(\tfrac \pi 2 z) \Gamma(1 - z) \zeta(1-z)$$
when $-1 < {\rm Re}(z) < 2$, and use this reflection formula to define $\zeta(z)$ for ${\rm Re}(z) \leq -1$ based on the values of $\zeta(z)$ for ${\rm Re}(z) \geq 2$ which we already know from your series expression.
This reflection formula is what you would use to verify that $\zeta(-2n) = 0$ for $n \in \mathbb N$, which originally prompted your question. Indeed, we have
$$ \zeta(-2n) = 2^{-2n}\pi^{-2n-1} \sin(\pi n) \Gamma(1  + 2n) \zeta(1 + 2n),$$
and $\sin(\pi n)$ is zero, while $\Gamma(1+2n)$ and $\zeta(1+2n)$ are finite, for all $n \in \mathbb N$.
In fact, the reflection formula tells you a lot more about the analytic structure of the zeta function. Observe that:


*

*$\sin(\frac \pi 2 z)$ has simple zeroes at $z \in 2 \mathbb Z$.

*$\Gamma(1-z)$ has simple poles when $z \in \mathbb N$.

*$\zeta(z)$ has a simple pole at $z = 1$, and no other poles for ${\rm Re}(z) > -1$. This follows from the integral expression from step 1.

*$\zeta(z)$ has no zeroes when ${\rm Re}(z) \geq 1$. This follows easily from the product expansion of $\zeta(z)$ in the case when ${\rm Re}(z) > 1$ though it requires more work to prove this when ${\rm Re}(z) = 1$.
Combining these observation with the reflection formula, you learn that:


*

*$\zeta(z)$ has a simple pole at $z = 1$, and no other poles.

*$\zeta(z)$ has simple zeroes at $z \in -2\mathbb N$. Excluding the strip $0 < {\rm Re}(z) < 1$, $\zeta(z)$ has no other zeroes. (What happens on the strip $0 < {\rm Re}(z) < 1$ is the subject of the Riemann hypothesis.)
