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

 A: 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: $\frac14\log(5) + \frac{\sqrt{5}}{10}\log(\rho) -\frac{\pi}{10}\cot\left(\frac{\pi}{5} \right) \simeq 0.07756$
A: 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 $f(1)$.)
A: Let: $$ f(x) = \sum_{n=1}^{\infty}{\frac{x^{5n}}{5n(5n-1)}} $$
$$ f''(x) = \sum_{n=1}^{\infty}x^{5n-2} = \frac{x^3}{1-x^5} = g(x) $$
And I do not know very well in what form, but I know it's related to hypergeometric functions...
https://en.wikipedia.org/wiki/Hypergeometric_function
And the constant value is: $f(1)$
A: 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:
$$  -\frac1{2} (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)) = \log(\rho) $$
$$  \log(5 + \sqrt 5) - \log(5 - \sqrt 5) = 2\log(\rho) $$
$$  \log \frac {5 + \sqrt 5} {5 - \sqrt 5} = \log(\rho^2) $$
$$  \frac {5 + \sqrt 5} {5 - \sqrt 5} = \rho^2 $$
$$  5 + \sqrt 5 = (5 - \sqrt 5)\rho^2 $$
$$  5 + \sqrt 5 = (5 - \sqrt 5)\frac{3+\sqrt 5}{2} $$
$$  5 + \sqrt 5 = 5 + \sqrt 5 $$
