How to show $\sum\limits_{k\ge 1} \ln\left(\frac{k^2+k+1}{k^2+k-1}\right)=\ln\left(\cosh(\frac{\sqrt3}2\pi)\right)-\ln(\sin(\frac{\sqrt5-1}2\pi))$ I am interested to show
$$\sum_{k\ge 1} \ln\left(\frac{k^2+k+1}{k^2+k-1}\right)=\ln\left(\cosh\left(\frac{\sqrt3}2\pi\right)\right)-\ln\left(\sin\left(\frac{\sqrt5-1}2\pi\right)\right)$$
This series converges.
We can write  $\dfrac{n^2+n+1}{n^2+n-1}$ as $\dfrac{(n-\alpha)(n+1+\alpha)}{(n-\beta)(n+1+\beta)}$ with $\alpha=\dfrac{-1+i\sqrt3}2 $ and $\beta =\dfrac{-1+\sqrt5}2 $, but i can't use the Chamberland & Straub formula: if ( $a+b=c+d$)
$$\prod_{k\ge 0}\frac {(k+a)(k+b)}{(k+c)(k+d)}=\frac {\Gamma(c)\Gamma(d) }{\Gamma(a)\Gamma(b)}$$

Hint
Do not go to fast to infinity (too far) and consider the partial product
$$P_n=\prod_{k= 0}^n\frac {(k+a)(k+b)}{(k+c)(k+d)}=\frac{a b (a+1)_n (b+1)_n}{c d (c+1)_n (d+1)_n}$$ which is also
$$P_n=\frac{\Gamma (c)\, \Gamma (d)\, \Gamma (a+n+1) \,\Gamma (b+n+1)}{\Gamma
   (a) \,\Gamma (b)\, \Gamma (c+n+1) \,\Gamma (d+n+1)}$$ which is valid for any case.
