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A few procrastinal computations motivated by Four infinite series involving Riemann zeta function suggest the identity $$\tan\left(\frac{\kappa-1}{\kappa+1}\frac{\pi}{2}\right)=\frac{1}{\pi}\sum_{n=1}^\infty \frac{\kappa^n-1}{(\kappa+1)^n}\zeta(n+1)$$ for all $\kappa>0$. (Both sides change sign when replacing $\kappa$ by $1/\kappa$. It is thus enough to consider $\kappa\geq 1$.)

I checked it on $100$ random values up to 400 digits which convinced me that it must hold.

Is there an easy explanation?

(Remark: This identity, if true, provides of course easy answers to question M0401468 mentionned above.)

(Computational remark: For high precision computation, it is perhaps best to cut the sum at some not very large value $N$ (I worked with $N=400$) and to approximate the omitted terms by summing the first few terms contributing to $\zeta(N+1),\ldots$, by considering the associated geometric progressions.)

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When I was writing my answer, Dan Romik answer appeared. Mine is the same but with more detail.

We have $$\log\Gamma(1-x)=\gamma x+\sum_{n=2}^\infty\zeta(n)\frac{x^n}{n}$$ (this is known and is also an exercise in complex analysis). Hence by differentiation $$-\frac{\Gamma'(1-x)}{\Gamma(1-x)}-\gamma=\sum_{n=2}^\infty \zeta(n) x^{n-1}$$ It follows that (I change $\kappa$ into $x$) $$\sum_{n=1}^\infty\frac{x^n-1}{(x+1)^n}\zeta(n+1)= \frac{\Gamma'(1-\frac{1}{x+1})}{\Gamma(1-\frac{1}{x+1})}+\gamma -\frac{\Gamma'(1-\frac{x}{x+1})}{\Gamma(1-\frac{x}{x+1})}-\gamma$$ $$=\frac{\Gamma'(\frac{x}{x+1})}{\Gamma(\frac{x}{x+1})}-\frac{\Gamma'(\frac{1}{x+1})}{\Gamma(\frac{1}{x+1})}=\frac{\Gamma'(y)}{\Gamma(y)}-\frac{\Gamma'(1-y)}{\Gamma(1-y)}=\frac{d}{dy}\log(\Gamma(y)\Gamma(1-y))=\frac{d}{dy}\log\frac{\pi}{\sin\pi y}$$ $$=-\frac{\pi\cos\pi y}{\sin\pi y}=-\pi\cot\pi y=\pi\cot\frac{\pi x}{x+1}$$ $$=-\pi\tan\Bigl(\frac{\pi}{2}-\frac{\pi x}{x+1}\Bigr)=-\pi\tan\Bigl(\frac{\pi}{2}\frac{1-x}{1+x}\Bigr)=\pi\tan\Bigl(\frac{\pi}{2}\frac{x-1}{x+1}\Bigr).$$

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  • $\begingroup$ Thank you for this detailled answer. (I have already accepted Dan Romiks answer but you beat him for clarity.) $\endgroup$ Sep 23, 2021 at 20:25
  • $\begingroup$ @RolandBacher If you think this is a better answer (I'm not going to judge either way) then you can change your accepted answer to this one. $\endgroup$ Sep 23, 2021 at 20:45
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I haven't checked the details, but this should follow in a straightforward way from a standard relation for the digamma function $\psi(z) = \Gamma'(z)/\Gamma(z)$, namely $$ \psi(z+1) = -\gamma - \sum_{n=1}^{\infty} (-1)^n \zeta(n+1) z^n, $$ together with the reflection formula $$ \psi(1-z)-\psi(z) = \pi \cot(\pi z). $$ Mathematica knows about these formulas, and confirms that your identity is correct, as seen in this screenshot: enter image description here

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  • $\begingroup$ Sorry for "disaccepting" your answer (which I like) but juan's answer puts all the dots on the i's. $\endgroup$ Sep 24, 2021 at 8:27
  • $\begingroup$ @RolandBacher no problem. $\endgroup$
    – Dan Romik
    Sep 24, 2021 at 14:22

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