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My simple problem relates with my question about sign-alternating sums of numbers, which opposite to coprime with $m$.

$$1-\frac{1}{3}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}-\frac{1}{13}+\cdots=\frac{\pi}{10}\left[\cot\left(\frac{\pi}{10}\right)-\cot\left(\frac{3\pi}{10}\right)\right]=\frac{\pi}{5^{3/4}\sqrt{\varphi}}=A$$

Here $\varphi=\frac{1+\sqrt{5}}{2}$.

$$A=\prod\limits_{p|10}\left(1+\frac{1}{p}\right)\prod\limits_{p}^{\infty}\left(1+\frac{1}{p}\right)^{-1}\prod\limits_{p[10n+3]}^{\infty}\frac{p+1}{p-1}\prod\limits_{p[10n+9]}^{\infty}\frac{p+1}{p-1}$$

$$A=\prod\limits_{p|10}\left(1-\frac{1}{p}\right)\prod\limits_{p}^{\infty}\left(1-\frac{1}{p}\right)^{-1}\prod\limits_{p[10n+1]}^{\infty}\frac{p-1}{p+1}\prod\limits_{p[10n+7]}^{\infty}\frac{p-1}{p+1}$$

$$\prod\limits_{p[10n+1]}^{\infty}\frac{p-1}{p+1}\prod\limits_{p[10n+3]}^{\infty}\frac{p+1}{p-1}\prod\limits_{p[10n+7]}^{\infty}\frac{p-1}{p+1}\prod\limits_{p[10n+9]}^{\infty}\frac{p+1}{p-1}=\frac{A^2}{\zeta(2)}\prod\limits_{p|10}\left(1-\frac{1}{p^2}\right)^{-1}=B$$

$$B=\frac{\pi^2}{5\sqrt{5}\varphi}\cdot\frac{6}{\pi^2}\cdot\frac{4\cdot25}{3\cdot24}=\frac{\sqrt{5}}{3\varphi}$$

But by computing we have $$B\approx\sqrt{\frac{\varphi^5}{5}}$$ I use $\approx$, because I not sure. But next I find result, which I sure is true.

$$1+\frac{1}{3}-\frac{1}{7}-\frac{1}{9}+\frac{1}{11}+\frac{1}{13}-\cdots=\frac{\pi}{10}\left[\cot\left(\frac{\pi}{10}\right)+\cot\left(\frac{3\pi}{10}\right)\right]=\frac{\pi\sqrt{\varphi}}{5^{3/4}}=C$$

$$\prod\limits_{p[10n+1]}^{\infty}\frac{p-1}{p+1}\prod\limits_{p[10n+3]}^{\infty}\frac{p-1}{p+1}\prod\limits_{p[10n+7]}^{\infty}\frac{p+1}{p-1}\prod\limits_{p[10n+9]}^{\infty}\frac{p+1}{p-1}=\frac{B^2}{\zeta(2)}\prod\limits_{p|10}\left(1-\frac{1}{p^2}\right)^{-1}=D$$

So it must be equal to $\frac{\sqrt{5}\varphi}{3}$, but by computing we have $$D\approx\frac{4}{\sqrt{5\varphi^5}}$$

Then we can be sure, that

$$\prod\limits_{p[10n+1]}^{\infty}\frac{p-1}{p+1}\prod\limits_{p[10n+9]}^{\infty}\frac{p+1}{p-1}=\sqrt{BD}=\frac{2}{\sqrt{5}}$$

Why we have discrepancy of $A$ with $B$, and $C$ with $D$?

user514787
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2 Answers2

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$$\chi(n)=\left\{\begin{array}{rcl}1&\text{if}&n\pmod{10}\in\{1,7\}\\-1&\text{if}&n\pmod{10}\in\{3,9\}\\0&&\text{otherwise}\end{array}\right.$$ is not a multiplicative function (for instance, $1=\chi(3)\chi(13)\neq\chi(39)=-1$), so how do you associate $\sum_{n\geq 1}\frac{\chi(n)}{n}$ to an Euler product?

Jack D'Aurizio
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Now I understand my mistake: $$f_{1}(10,I[+])=1-\frac{1}{3}+\frac{1}{7}-\frac{1}{9}+\cdots=\frac{\pi}{5^{3/4}\sqrt{\varphi}}=A$$ $$f_{1}(10,I[++])=1+\frac{1}{3}-\frac{1}{7}-\frac{1}{9}+\cdots=\frac{\pi\sqrt{\varphi}}{5^{3/4}}=a$$ $$a=A\varphi$$ $$f_{1}(30,I[+++-])=1+\frac{1}{7}+\frac{1}{11}-\frac{1}{13}+\cdots=B=A+\frac{1}{3}a=A\left(1+\frac{\varphi}{3}\right)$$ $$f_{1}(30,I[+-++])=1-\frac{1}{7}+\frac{1}{11}+\frac{1}{13}-\cdots=b=a-\frac{1}{3}A=A\left(\varphi-\frac{1}{3}\right) $$ $$b=Bx$$ $$f_{1}(210,I[++-+---+++-+--+++----++-])=$$ $$1+\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\cdots=C=B-\frac{1}{7}b=B\left(1-\frac{x}{7}\right)$$ $$f_{1}(210,I[+++--+-+-++-+-+-++-+--++])=$$ $$1+\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\cdots=c=b+\frac{1}{7}B=B\left(x+\frac{1}{7}\right)$$ $$c=Cy$$ And so on: $$D=C-\frac{1}{11}c=C\left(1-\frac{y}{11}\right)$$ $$d=c-\frac{1}{11}C=C\left(y-\frac{1}{11}\right)$$ $$d=Dz$$ Of course it's never give us Euler product in standard form.

user514787
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