Illustration of Proof of Theorem 11.6

Let $x=5$.

$$\begin{array}{rl}P(5)& =\prod _{p\le 5}\{1+f(p)+f(p^2)+\dots \}\\ & =\begin{array}{rl}(1+f(2)+& f(2^2)+f(2^3)+\dots )\\ (1+f(3)+& f(3^2)+f(3^3)+\dots )\\ (1+f(5)+& f(5^2)+f(5^3)+\dots ).\end{array}\end{array}$$

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$$\begin{array}{rl}A& =\{2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,\dots \},\\ B& =\{7,11,13,14,17,19,21,22,23,\dots \},\\ \{n:n>x\}& =\{6,7,8,9,10,11,12,13,14,15,16,\dots \}.\end{array}$$

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$$\sum _{n=1}^{\mathrm{\infty }}f(n)-\sum _{n\in A}f(n)=\sum _{n\in B}f(n).$$

$$\begin{array}{rl}(f(1)& +f(2)+f(3)+f(5)+f(6)+f(7)+\dots )\\ -(f(2)& +f(3)+f(4)+f(5)+f(6)+f(8)+\dots )\\ & =f(7)+f(11)+\dots .\end{array}$$

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$$\sum _{n\in B}|f(n)|\le \sum _{n>x}|f(n)|.$$

$$\begin{array}{rl}|f(7)|+|f(11)|+|f(13)|& +\dots \\ & \le |f(6)|+|f(7)|+|f(8)|+\dots \end{array}$$