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5

We have: $$ \zeta(s)=\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1},\qquad \zeta(s)^2=\sum_{n\geq 1}\frac{d(n)}{n^s} $$ hence, assuming that the sequence $\{a(n)\}_{n\in\mathbb{N}^*}$ is the sequence of coefficients of the Dirichlet series associated with $f(s)$: $$ f(s)=\zeta(s)^2\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)=\sum_{n\geq ...


4

Approach, which uses Fourier series. Denote $$ S = \sum_{k=1}^\infty \left[\frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right]. $$ Consider function $$ f(x) = \dfrac{\pi x(\pi-x)}{8}, \qquad x\in[0,\pi];\tag{1} $$ construct the odd extension of $f(x)$ to the interval $[−\pi, \pi]$: $f(-x)=-f(x), x\in [0,\pi]$; and make it $2\pi$-periodic: copy to each segment ...


2

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13

Notice$\color{blue}{^{[1]}}$ $$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right) = \sum_{\substack{k=-\infty\\ k\ne 0}}^\infty \frac{1}{(6k-1)^3} = 1 - \frac{1}{6^3}\sum_{k=-\infty}^\infty \frac{1}{(\frac16-k)^3}$$ Recall the infinite product expansion of $\sin x$ $$\sin x = x \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2\pi^2}\right)$$ ...



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