Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 3

For $\sigma =\Re s>1$,On every compact subset of $\Re s>1$, the terms are bounded by a convergent series $\sum n^{-1-\delta }$. The Weierstrass M-test gives locally uniform convergence, and the theorem on locally uniform convergence of holomorphic functions shows that the limit is an analytic function. Hence $D_f$ is analytic throughout $\Re s>1$.

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For arithmetic functions $f,g$ bounded in modulus by one, their pretentious distance up to $x$ is defined by

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Complete multiplicativity and absolute convergence give the Euler productSet $\sigma =1+1/\log x$. Taking logarithms of absolute values and expanding the local factors gives, uniformly in real $t$,The prime powers with exponent at least two contribute $O(1)$; changing $p^{-\sigma }$ to $p^{-1}$ below $x$ and estimating the tail above $x$ also cost $O(1)$. By Mertens theorem,Exponentiating yields

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For $\sigma >1$, take logarithms of Euler products. With $z=f(p)p^{-it}$, the inequalityat every prime power gives the three-four-one inequality for Euler products

Suppose $D_f(1+it)=0$ with multiplicity $m\ge 1$. The assumed analytic continuation giveswhile $D_{f^2}(\sigma +2it)$ remains bounded and $\zeta (\sigma {)}^3\asymp (\sigma -1{)}^{-3}$ as $\sigma \downarrow 1$. The left side of the inequality would then becontradicting its lower bound one. Hence $D_f$ has no zero on $\Re s=1$. Absolute convergence of its Euler product already excludes zeros for $\Re s>1$, so $D_f(s)\ne 0$ throughout $\Re s\ge 1$.

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