<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] Look at finite sums first (such that we do not subtract two diverging series):
$$
\sum_{n=1}^N\frac1{5n(5n-1)} =\\
\sum_{n=1}^N\frac{-1}{5n} + \frac{1}{5n-1}  = \\
- \frac15 \sum_{n=1}^N(\frac{1}{n} - \frac{1}{n-1/5}) = \\
- \frac15 (\sum_{n=1}^N\frac{1}{n} - \sum_{n=0}^{N-1} \frac{1}{n+4/5} )= \\
- \frac15 (\sum_{n=1}^N\frac{1}{n} - \sum_{n=1}^N \frac{1}{n+4/5} - \frac54 + \frac{1}{N+4/5}) =\\
\frac{1}{4} - \frac{1}{5N+4} - \frac15 \sum_{n=1}^N(\frac{1}{n} - \frac{1}{n+4/5})
$$
The sum converges, so we can take the limit $N\to \infty$ to obtain 
$$
\sum_{n=1}^\infty \frac1{5n(5n-1)} =\\
\frac{1}{4}  - \frac15 \sum_{n=1}^\infty (\frac{1}{n} - \frac{1}{n+4/5})
$$
Now you can use a result (with proof by Achille Hui) from here which says 
$$
\mathcal{S}_{k/p} \stackrel{def}{=}
\sum_{n=1}^\infty\left(\frac{1}{n} - \frac{1}{n+\frac{k}{p}}\right)
 \\=
\frac{p}{k} - \log(2p) -\frac{\pi}{2}\cot\left(\frac{k\pi}{p}\right) + 
\sum_{l=1}^{p-1} 
\cos\left(\frac{2\pi k\ell}{p}\right) \log\sin\left(\frac{\ell\pi}{p}\right) \\ 
= \psi\left(\frac{k}{p}+1\right) + \gamma
$$
and conclude with $k=4$, $p=5$ that $\sum_{n=1}^\infty\frac1{5n(5n-1)} = \frac{1}{4}  - \frac15 (\psi\left(\frac{9}{5}\right) + \gamma)$ or, without using the Digamma function, 
$$
\sum_{n=1}^\infty\frac1{5n(5n-1)} =\\
\frac{1}{4}-\frac15 (\frac{5}{4} - \log(10) -\frac{\pi}{2}\cot\left(\frac{4\pi}{5}\right) + 
\sum_{l=1}^{4} 
\cos\left(\frac{2\pi 4\ell}{5}\right) \log\sin\left(\frac{\ell\pi}{5}\right)) =\\
 \frac{\log 10}{5} + \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) -\frac15 
\sum_{l=1}^{4} 
\cos\left(\frac{2\pi 4\ell}{5}\right) \log\sin\left(\frac{\ell\pi}{5}\right) = \\
 \frac{\log 10}{5} + \frac{\pi}{10}\cot\left(\frac{4\pi}{5}\right) -\frac1{20}  (\log(16/5) + \sqrt 5 (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)))\\
\simeq 0.07756
$$
which is exactly as given in the question: $\frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) \simeq 0.07756$
</s>