Es fácil probar el último paso de Andreas: $$ \sum_{n=1}^\infty\frac1{5n(5n-1)} =\\ \frac{\log 10}{5} + \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) -\frac1{20} (\log(16/5) + \sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)))\\ \simeq 0.07756 $$ que es exactamente como se indica en la pregunta: $\frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) \simeq 0.07756$
Sólo tienes que comprobar estos tres anuncios:
a) $$ \frac{\log 10}{5} -\frac1{20} \log(16/5) = \frac14\log(5) $$ pasos:
$$ \frac{\log 10}{5} -\frac1{20} \log(16/5) =\\ \frac{\log 2}{5}+\frac{\log 5}{5}-\frac{\log 16}{20}+\frac{\log 5}{20} =\\ \frac{\log 2}{5}+\frac{\log 5}{5}-\frac{4\log 2}{20}+\frac{\log 5}{20} =\\ (\frac{1}{5}-\frac{4}{20})\log2+(\frac{1}{5}+\frac{1}{20})\log5 =\\ \frac14\log(5) $$
b) es trivial: $$ \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) = -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) $$
c) $$ -\frac1{20} (\sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5))) = \frac{\sqrt{5}}{10}\log(\rho) $$ pasos: $$ -\frac1{2} (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)) = \log(\rho) $$ $$ \log(5 + \sqrt 5) - \log(5 - \sqrt 5) = 2\log(\rho) $$ $$ \log \frac {5 + \sqrt 5} {5 - \sqrt 5} = \log(\rho^2) $$ $$ \frac {5 + \sqrt 5} {5 - \sqrt 5} = \rho^2 $$ $$ 5 + \sqrt 5 = (5 - \sqrt 5)\rho^2 $$ $$ 5 + \sqrt 5 = (5 - \sqrt 5)\frac{3+\sqrt 5}{2} $$ $$ 5 + \sqrt 5 = 5 + \sqrt 5 $$