A few procrastinal computations motivated by [Four infinite series involving Riemann zeta function](https://mathoverflow.net/questions/401468/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.)



A expression for the tangent function involving $\zeta(n),n=2,3,\ldots$
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 $x$)
$$\sum\_{n=1}^\infty\frac{x^n-1}{(x+1)^n}\zeta(n+1)=
\frac{\Gamma'(1-\frac{1}{x+1})}{\Gamma(1-\frac{1}{x+1})}+\gamma
-\frac{\Gamma'(1-\frac{x}{x+1})}{\Gamma(1-\frac{x}{x+1})}-\gamma$$
$$=\frac{\Gamma'(\frac{x}{x+1})}{\Gamma(\frac{x}{x+1})}-\frac{\Gamma'(\frac{1}{x+1})}{\Gamma(\frac{1}{x+1})}=\frac{\Gamma'(y)}{\Gamma(y)}-\frac{\Gamma'(1-y)}{\Gamma(1-y)}=\frac{d}{dy}\log(\Gamma(y)\Gamma(1-y))=\frac{d}{dy}\log\frac{\pi}{\sin\pi y}$$
$$=-\frac{\pi\cos\pi y}{\sin\pi y}=-\pi\cot\pi y=\pi\cot\frac{\pi x}{x+1}$$
$$=-\pi\tan\Bigl(\frac{\pi}{2}-\frac{\pi x}{x+1}\Bigr)=-\pi\tan\Bigl(\frac{\pi}{2}\frac{1-x}{1+x}\Bigr)=\pi\tan\Bigl(\frac{\pi}{2}\frac{x-1}{x+1}\Bigr).$$


