For $s \in \mathbb{C}$ with $\textrm{Re } s > 1$, the zeta function is defined as

$$\begin{aligned} \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \end{aligned}$$

We can readily see $\zeta(s)$ converges absolutely in $\{\textrm{Re } s > 1\}$. What's interesting is the convergence is uniform in $\{\textrm{Re } s > r\}$ for any $r > 1$, which can be used to show that $\zeta(s)$ is analytic in $\{\textrm{Re } s > 1\}$.

Riemann was able to show that $\zeta(s)$ has an analytic continuation into $\mathbb{C} \setminus \{1\}$, and this continuation satisfies that $\zeta(s) \rightarrow \infty$ as $s \rightarrow 1$.

Some interesting facts:

• The only zeros of the zeta function (ie when $\zeta(s) = 0$) outside the strip $\{0 \le \textrm{Re } s \le 1\}$ are at the negative even integers, $-2, -4, -6, \cdots$, which are called the "trivial zeros".
• Zeta has no zeros on the line $\{ \textrm{Re } s = 1\}$, nor the line $\{ \textrm{Re } s = 0\}$.

Riemann Hypothesis

    In the strip $\{0 \le \textrm{Re } s \le 1\}$, all zeros of $\zeta$ are on the line $\displaystyle \{ \textrm{Re } s = \frac{1}{2} \}$

which has significant implication on the distribution of prime numbers and the growth of many important arithmetic functions, but remains unproved.

Amazingly,

$$\begin{aligned} \zeta(s) = \prod_p \frac{1}{1 - p^{-s}} \end{aligned}$$

where the product is over all primes ! Why ? Observe

$$\begin{aligned} \zeta(s) &= \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \\ &= \big(\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{8^s} + \cdots\big)\big(\frac{1}{1^s} + \frac{1}{3^s} + \frac{1}{9^s} + \cdots\big)\big(\frac{1}{1^s} + \frac{1}{5^s} + \frac{1}{25^s} + \cdots\big) \cdots \\ \end{aligned}$$

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 10:23 AM