Series with cyclotomic polynomials
Series with cyclotomic polynomials
Let $\Phi$ denote the $n$ -th cyclotomic polynomial. Prove that:
$$\sum_{n=1}^\infty\frac{1}{n^4}\frac{\Phi'_n(e^{2\pi})}{\Phi_n(e^{2\pi})}=\frac{45\zeta(3)}{\pi^4e^{2\pi}}+\frac{7}{4\pi e^{2\pi}}$$
where $\Phi'$ stands for the derivative.
$$\sum_{n=1}^\infty\frac{1}{n^4}\frac{\Phi'_n(e^{2\pi})}{\Phi_n(e^{2\pi})}=\frac{45\zeta(3)}{\pi^4e^{2\pi}}+\frac{7}{4\pi e^{2\pi}}$$
where $\Phi'$ stands for the derivative.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Series with cyclotomic polynomials
Start with the well-known formulaRiemann wrote: ↑Tue Sep 13, 2016 8:02 pm Let $\Phi$ denote the $n$ -th cyclotomic polynomial. Prove that:
$$\sum_{n=1}^\infty\frac{1}{n^4}\frac{\Phi'_n(e^{2\pi})}{\Phi_n(e^{2\pi})}=\frac{45\zeta(3)}{\pi^4e^{2\pi}}+\frac{7}{4\pi e^{2\pi}}$$
where $\Phi'$ stands for the derivative.
$$\Phi_n(x)=\prod_{d\mid n}(x^d-1)^{\mu(n/d)},$$
where $\mu$ is the Mobius function. Then
$$\frac{\Phi_n'(x)}{\Phi_n(x)}=\sum_{d \mid n}\mu(n/d)\frac{dx^{d-1}}{x^d-1}$$
and so
$$f(x):=\sum_{n \ge 1}\frac{1}{n^4}\frac{\Phi_n'(x)}{\Phi_n(x)}=\sum_{n \ge 1}\frac{1}{n^4}\sum_{d \mid n}d \mu(n/d)\frac{x^{d-1}}{x^d-1}=\sum_{k \ge 1}\frac{1}{k^4d^4}\sum_{d \ge 1}d\mu(k)\frac{x^{d-1}}{x^d-1}=\frac{1}{x}\sum_{k \ge 1}\frac{\mu(k)}{k^4}\sum_{d \ge 1}\frac{1}{d^3}\frac{x^d}{x^d-1}$$
$$=\frac{1}{x\zeta(4)}\sum_{d \ge 1}\frac{1}{d^3}\frac{x^d}{x^d-1}.$$
So, using the identity $\frac{e^{2\pi d}}{e^{2\pi d}-1}=\frac{1+\coth(d\pi)}{2},$ we have
$$f(e^{2\pi})=\frac{1}{2e^{2\pi}\zeta(4)}\sum_{d \ge 1}\frac{1+\coth(d\pi)}{d^3}=\frac{1}{2e^{2\pi}\zeta(4)}\left(\zeta(3)+\sum_{d \ge 1}\frac{\coth(d\pi)}{d^3}\right). \ \ \ \ \ \ \ \ \ \ (1)$$
Now,
$$\sum_{d \ge 1}\frac{\coth(d\pi)}{d^3}=\sum_{d \ge 1}\frac{1}{d^3}\left(\frac{2d}{\pi}\sum_{m \ge 1}\frac{1}{m^2+d^2}+\frac{1}{d\pi}\right)=\frac{2}{\pi}\sum_{d,m \ge 1}\frac{1}{d^2(m^2+d^2)}+\frac{1}{\pi}\zeta(4). \ \ \ \ \ \ \ \ \ \ (2)$$
Finally,
$$S:=\sum_{d,m \ge 1}\frac{1}{d^2(m^2+d^2)}=\sum_{d,m \ge 1}\frac{1}{m^2}\left(\frac{1}{d^2}-\frac{1}{m^2+d^2}\right)=\sum_{d \ge 1}\frac{1}{d^2} \sum_{m \ge 1}\frac{1}{m^2}-\sum_{d,m \ge 1}\frac{1}{m^2(m^2+d^2)}$$
$$=(\zeta(2))^2-S,$$
which gives
$$S=\frac{1}{2}(\zeta(2))^2. \ \ \ \ \ \ \ \ (3)$$
The result now follows from $(1),(2),(3).$
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