$$\sum_{n=1}^\infty\frac{1}{n^4}\frac{\Phi'_n(e^{2\pi})}{\Phi_n(e^{2\pi})}=\frac{45\zeta(3)}{\pi^4e^{2\pi}}+\frac{7}{4\pi e^{2\pi}}$$
where $\Phi'$ stands for the derivative.
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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