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I'm quite lost at this, I've tried to express the $csch(nx)$ as a sum like this: $$\frac{csch(nx)}{2}= \frac{1}{e^{nx}-e^{-nx}}=\frac{1}{2nx}\sum_{k=0}^{\infty} \frac{(2nx)^k B_k(1/2)}{k!}$$ Where $B_k(x)$ are the Bernoulli polynomials. Then I put these sums together: $$\sum_{n=1}^\infty \frac{\mu(n)}{\sinh(nx)}=\sum_{n=1}^\infty \frac{\mu(n)}{x}\sum_{k=0}^\infty \frac{(2x)^kB_k(1/2)}{n^{1-k} k!}$$ And then (without worrying too much whether I can swap the sums or not): $$\sum_{n=1}^\infty \frac{\mu(n)}{x}\sum_{k=0}^\infty \frac{(2x)^kB_k(1/2)}{n^{1-k} k!}= \frac{1}{x} \sum_{k=0}^\infty \frac{(2x)^kB_k(1/2)}{k!} \sum_{n=1}^\infty \frac{\mu(n)}{n^{1-k}}= \frac{1}{x} \sum_{k=0}^\infty \frac{(2x)^kB_k(1/2)}{k!} \cdot \frac{1}{\zeta(1-k)}$$

After this I tried to express the zeta function as a function of the Bernoulli numbers and the other way around, but to no good result. I hope someone can help me with this. Thanks in advanced.

Rafa
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1 Answers1

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Let $$f(x)=2e^{-2x}(e^x-1)=2(e^{-x}-e^{-2x}).$$ Then, $$\sum^\infty_{n=1}f(nx)=2\left(\frac1{e^x-1}-\frac1{e^{2x}-1}\right)=\frac{2\,e^x}{e^{2x}-1}=\frac1{\sinh x}.$$ By the Moebius inversion formula, that implies $$f(x)=\sum^\infty_{n=1}\frac{\mu(n)}{\sinh nx}.$$