Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2008 / A / 2

For $\mathrm{Re}s>1$, the Euler product theorem isThe local series converge absolutely, and the product is absolutely convergent in the sense that the sum over primes of the absolute deviations of the local factors from one is finite. Indeed this sum is bounded by ${\sum }_{n\ge 2}|f(n)|n^{-\mathrm{Re}s}$. To prove the identity, expand the product over primes at most $y$. Absolute convergence permits all those local expansions, and multiplicativity identifies the coefficients with $f(n)$ for integers whose prime factors are at most $y$. Every omitted integer exceeds $y$, so the difference from the full Dirichlet series is bounded by its absolute tail beyond $y$, which tends to zero. The same tail bound on a closed right half-plane inside $\mathrm{Re}s>1$ proves local uniform convergence. With signed coefficients a local factor can be zero; the identity does not require claiming every such general Euler product is nonzero.

For $d(n^2)$, multiplicativity gives local coefficients $d(p^{2e})=2e+1$. With $z=p^{-s}$,The divisor bound from (1), applied with a sufficiently small exponent, proves absolute convergence when $\mathrm{Re}s>1$; divergence at one follows from $d(n^2)\ge 1$. Multiplying the local identities therefore givesFor $n\le x$, the inequality $n^{\delta }\le x^{\delta }$ gives $d(n^2)/n\le x^{\delta }d(n^2)/n^{1+\delta }$. Sum those nonnegative terms and enlarge to the whole infinite series to get the required comparison.

For real $1<s<2$, integral comparison gives $\zeta (s)\le 1+(s-1{)}^{-1}\le 2/(s-1)$, while $\zeta (2s)\ge 1$. Hence $\zeta (s{)}^3/\zeta (2s)=O((s-1{)}^{-3})$ uniformly in that interval. Choosing $\delta =1/\log x$ for $x\ge 3$ makes $0<\delta <1$ and $x^{\delta }=e$. Thus