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Number theory Difficulty 6.2 National olympiad Prove it

1. Let f(n)f(n) be a function defined on the set of positive integers, and let ss be a real number. The series ∑n=1∞f(n)n−s(1)\sum_{n=1}^{\infty} f(n) n^{-s(1)} converges absolutely and is not equal to zero. Prove:
(i) If f(mn)=f(m)f(n)f(m n)=f(m) f(n) whenever (m,n)=1(m, n)=1, then
∑n=1∞f(n)n−s=∏p(1+f(p)p−s+f(p2)p−2s+⋯ )\sum_{n=1}^{\infty} f(n) n^{-s}=\prod_{p}\left(1+f(p) p^{-s}+f\left(p^{2}\right) p^{-2 s}+\cdots\right)
(ii) If f(mn)=f(m)f(n)f(m n)=f(m) f(n) for all m,nm, n, then
∑n=1∞f(n)n−s=∏p(1−f(p)p−s)−1\sum_{n=1}^{\infty} f(n) n^{-s}=\prod_{p}\left(1-f(p) p^{-s}\right)^{-1}

Here, both infinite products on the right-hand side converge absolutely.

Solution

1. (i) For any given positive integer kk and x<2kx<2^{k}, by the fundamental theorem of arithmetic and the multiplicative condition, we have
∣∏p⩽x(1+f(p)p−s+f(p2)p−2s+⋯+f(pk)p−ks)−∑n⩽xf(n)n−s∣⩽∑x<n<x(x)k∣f(n)∣n−s.\begin{array}{c} \left|\prod_{p \leqslant x}\left(1+f(p) p^{-s}+f\left(p^{2}\right) p^{-2 s}+\cdots+f\left(p^{k}\right) p^{-k s}\right)-\sum_{n \leqslant x} f(n) n^{-s}\right| \\ \leqslant \sum_{x<n<x^{(x) k}}|f(n)| n^{-s} . \end{array}

From this and the absolute convergence of the series, we obtain the desired conclusion; (ii) follows from (i).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.