Prove closed form for $\sum_{n\in\Bbb N}\frac1{5n(5n-1)}$

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: $\frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) \simeq 0.07756$

Start with $$f(x) = \sum_{n=2}^\infty \frac{x^n}{n(n-1)} = x + (1-x) \log (1-x)$$ If $w$ is a primitive $5$-th root of unity then $$\frac{1}{5}\sum_{k=0}^4 w^{nk} = 1$$ if $5 | n$ and $0$ otherwise. Thus

$$\frac{1}{5}\sum_{k=0}^4 f(w^k) =\sum_{n> 2,\, 5 | n} \frac{1}{n(n-1)} = \sum_{n=1}^\infty \frac{1}{5n(5n-1)}$$

(use the limiting value $2$ for $f(1)$.)