How Are Series Formulas and Analytic Continuation Different for the Riemann Zeta Function?

Drafted with GPT-5.5 Pro.

The most common confusion around the Riemann zeta function is treating a function and one formula for representing that function as if they were the same thing.

The Riemann zeta function first appears through the series

$$\zeta (s)=\sum _{n=1}^{\infty }\frac{1}{n^s}$$

but this series does not converge for every complex number $s$. The direct series representation is valid only when

$$\mathrm{Re}(s)>1$$

Here $\mathrm{Re}(s)$ means the real part of the complex number $s$. If

$$s=a+bi$$

then

$$\mathrm{Re}(s)=a,\mathrm{Im}(s)=b$$

So $\mathrm{Re}(s)>1$ does not mean that the complex number $s$ is greater than $1$. It means that the real-coordinate part of $s$ in the complex plane is greater than $1$.

The Euler Product Has the Same Domain Limit

Another formula often confused with the whole zeta function is the Euler product:

$$\zeta (s)=\prod _{p\in P}\frac{1}{1-p^{-s}}$$

This formula connects the zeta function directly to prime numbers. It is important, but it also does not hold for every $s$. It is valid when

$$\mathrm{Re}(s)>1$$

If a problem writes only

$$\zeta (s)=\prod _{p\in P}\frac{1}{1-p^{-s}}$$

without the condition, the intended reference to the Euler product may be understandable, but the mathematical statement is incomplete.

For example, if $s=0$, the right side becomes

$$\prod _{p\in P}\frac{1}{1-p^0}=\prod _{p\in P}\frac{1}{1-1}$$

Each factor has the form $\frac{1}{0}$. So the Euler product cannot be treated as a formula valid across the whole complex plane.

Formula Failure Is Not Function Nonexistence

This naturally raises the question:

If $\mathrm{Re}(s)\le 1$, does the Riemann zeta function itself fail to exist?

No. What fails there is the basic series representation, not the function itself.

This is where analytic continuation matters.

Analytic continuation extends a function from a smaller domain to a larger domain while preserving the original function on the region where both definitions overlap. It is not the act of forcing arbitrary values into a broken formula. It is the act of finding a wider analytic function that agrees with the old one where the old one worked.

Geometric Series Example

The simplest example is the geometric series:

$$1+z+z^2+z^3+\cdots$$

This series converges only when

$$|z|<1$$

Inside that range,

$$1+z+z^2+z^3+\cdots =\frac{1}{1-z}$$

But the function

$$\frac{1}{1-z}$$

is defined on a wider domain, except at $z=1$. So

$$F(z)=\frac{1}{1-z}$$

can be understood as the analytic continuation of the original geometric-series function.

This does not mean that

$$1+2+4+8+\cdots =-1$$

as an ordinary convergent series. At $z=2$, the series still diverges. It only means that the extended function that agrees with the series on $|z|<1$ gives

$$\frac{1}{1-2}=-1$$

at $z=2$.

The Same Distinction Applies to Zeta

The basic zeta series

$$\sum _{n=1}^{\infty }\frac{1}{n^s}$$

converges only when $\mathrm{Re}(s)>1$. But there is a function that agrees with this series in that domain and continues analytically to a wider region. We keep writing that continued function as

$$\zeta (s)$$

More precisely, the Riemann zeta function has an analytic continuation to the complex plane except at $s=1$, where it has a simple pole.

What Happens on the Critical Line

On the critical line

$$s=\frac{1}{2}+bi$$

the basic series

$$1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots$$

does not converge, because $\mathrm{Re}(s)=\frac{1}{2}$ is outside the direct convergence domain.

One useful related function is the Dirichlet eta function:

$$\eta (s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\cdots$$

In the appropriate domain,

$$\eta (s)=(1-2^{1-s})\zeta (s)$$

so

$$\zeta (s)=\frac{\eta (s)}{1-2^{1-s}}$$

This gives a way to understand zeta values in regions where the original zeta series is not the right formula.

When we write

$$\zeta \left(\frac{1}{2}+bi\right)$$

we are not directly substituting $s=\frac{1}{2}+bi$ into the original series. We are referring to the analytically continued zeta function.

How to Judge Three Common Statements

P1. The Riemann zeta function can be expressed as an Euler product.

$$\zeta (s)=\prod _{p\in P}\frac{1}{1-p^{-s}}$$

This is true with the condition $\mathrm{Re}(s)>1$. Without that condition, it is misleading. The Euler product is a domain-limited representation, just like the original zeta series.

P2. There are infinitely many zeros of the zeta function on the critical line.

$$\zeta \left(\frac{1}{2}+bi\right)=0$$

for infinitely many values of $b$ is a statement about the analytically continued zeta function, not the original series. This is a known theorem. The critical line lies outside the direct convergence domain of the basic series, so this statement must be about the continued function.

P3. Every nontrivial zero has real part $1/2$.

This is the Riemann hypothesis. It claims that every nontrivial zero lies on the critical line. It is not a proven theorem.

The key is to separate the object from its formulas:

Object or Formula	Meaning	Valid Domain
Basic series	$\sum 1/n^s$	$\mathrm{Re}(s)>1$
Euler product	${\prod }_p(1-p^{-s}{)}^{-1}$	$\mathrm{Re}(s)>1$
Continued zeta function	$\zeta (s)$ obtained by analytic continuation	Complex plane except $s=1$
Riemann hypothesis target	Nontrivial zeros of the continued $\zeta (s)$	Mainly $0<\mathrm{Re}(s)<1$

Core Takeaway

The Riemann zeta function is represented by the series $\sum 1/n^s$ when $\mathrm{Re}(s)>1$. Outside that region, the series no longer defines the function directly. The zeta function is then understood through analytic continuation, and the Riemann hypothesis concerns the zeros of that continued function.