Non-vanishing of the zeta function on the line Re s = 1

Remarks

For $\operatorname{Re} s > 1$ this follows from the Euler product formula, since a convergent product of nonzero factors is nonzero. For the line $\operatorname{Re} s = 1$, the classical argument uses the inequality $3 + 4\cos\theta + \cos 2\theta = 2(1 + \cos\theta)^2 \ge 0$. With the Euler product it gives $\abs{\zeta(\sigma)^3 \, \zeta(\sigma + it)^4 \, \zeta(\sigma + 2it)} \ge 1$ for $\sigma > 1$. If $\zeta(1 + it) = 0$ with $t \ne 0$, then as $\sigma \to 1^+$ the left side behaves like $(\sigma - 1)^{-3} (\sigma - 1)^{4} \to 0$, because $\zeta$ has a simple pole at $1$ and $\zeta(1 + 2it)$ is finite; this is a contradiction.

Going deeper

For a student

The prime number theorem $\pi(x) \sim x / \log x$ follows from this non-vanishing by a Tauberian theorem, such as the Wiener–Ikehara theorem, applied to $-\zeta'(s)/\zeta(s) = \sum_n \Lambda(n) n^{-s}$. Hadamard and de la Vallée Poussin proved the prime number theorem this way in 1896. Quantitative zero-free regions, such as $\sigma > 1 - c/\log(\abs{t} + 2)$, give the prime number theorem with an error term. The same method shows $L(1 + it, \chi) \ne 0$ for Dirichlet L-functions.