Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 11I (Number Theory)

For $\sigma =\mathrm{Re}s>1$, the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ is absolutely convergent. A finite product over primes $p\le P$ can be expanded into absolutely convergent geometric series. By unique prime factorization,The omitted integers are all greater than $P$, so the difference from $\zeta (s)$ has modulus at most ${\sum }_{n>P}{n}^{-\sigma }\to 0$. ThusThis defines the infinite Euler product as the limit of finite products. Its absolute convergence also follows from ${\sum }_p{p}^{-\sigma }<\infty$, with $\log (1-p^{-s}{)}^{-1}=p^{-s}+O(p^{-2\sigma })$ for sufficiently large $p$.

Since $n^{-s}-(n+1{)}^{-s}=s{\int }_n^{n+1}{x}^{-s-1}dx$, the assumed identity becomesfor $\mathrm{Re}s>1$, where $\{x\}$ is the fractional part. ConsequentlyTo justify the analytic continuation, on any compact subset of this half-plane choose $ϵ>0$ with $\mathrm{Re}s\ge ϵ$. The integrand is bounded by $x^{-1-ϵ}$, an integrable function. Each derivative with respect to $s$ is bounded by a constant times $(1+\log x{)}^j{x}^{-1-ϵ}$, also integrable. Differentiation under the integral is valid, so the right-hand side is holomorphic throughout the half-plane, including $s=1$.