<s>[INST] Prove closed form for $\sum_{n\in\Bbb N}\frac1{5n(5n-1)}$ While looking for solutions to a difficult geometric problem, I encountered this sum:
$$
\sum_{n\in\Bbb N}\frac1{5n(5n-1)} = \frac1{4\cdot 5} + \frac1{9\cdot 10} + \frac1{14\cdot 15} + \ldots
$$
A bit of numerical exploring has convinced me that the answer is
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right)
$$
where $\rho$ is the "Golden Ratio" $\rho = \frac{1+\sqrt{5}}{2}$.
But I can't find a way to prove it.

Prove 
  $$
\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right)
$$

 [/INST] Is easy to prove the last step of Andreas:
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} =\\
 \frac{\log 10}{5} + \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) -\frac1{20}  (\log(16/5) + \sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)))\\
\simeq 0.07756
$$
which is exactly as given in the question: $\frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) \simeq 0.07756$
You just have to check these three adds:
a) $$ \frac{\log 10}{5} -\frac1{20} \log(16/5) = \frac14\log(5) $$
steps:
$$ \frac{\log 10}{5} -\frac1{20} \log(16/5) =\\ \frac{\log 2}{5}+\frac{\log 5}{5}-\frac{\log 16}{20}+\frac{\log 5}{20} =\\
\frac{\log 2}{5}+\frac{\log 5}{5}-\frac{4\log 2}{20}+\frac{\log 5}{20} =\\
(\frac{1}{5}-\frac{4}{20})\log2+(\frac{1}{5}+\frac{1}{20})\log5 =\\
\frac14\log(5)
$$
b) is trivial: $$  \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) = -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) $$
c) $$ -\frac1{20}  (\sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5))) = \frac{\sqrt{5}}{10}\log(\rho) $$
steps:
$$  -\frac1{2} (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)) = \log(\rho) $$
$$  \log(5 + \sqrt 5) - \log(5 - \sqrt 5) = 2\log(\rho) $$
$$  \log \frac {5 + \sqrt 5} {5 - \sqrt 5} = \log(\rho^2) $$
$$  \frac {5 + \sqrt 5} {5 - \sqrt 5} = \rho^2 $$
$$  5 + \sqrt 5 = (5 - \sqrt 5)\rho^2 $$
$$  5 + \sqrt 5 = (5 - \sqrt 5)\frac{3+\sqrt 5}{2} $$
$$  5 + \sqrt 5 = 5 + \sqrt 5 $$
</s>