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

Mark Fischler
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  • Looks like this question may be helpful, or at least the methods, with $p=5$ – Dietrich Burde Mar 11 '19 at 17:21
  • Replacing $5$ by an arbitrary parameter $k,$ $$\sum_{n=1}^m \frac{1}{kn(kn-1)}$$ can be expressed in terms of zeroth Digamma functions. Other values of $k$ yield similar answers, so I'm guessing this relates heavily to Digamma identities – Brevan Ellefsen Mar 11 '19 at 17:24
  • Checking, this indeed follows quickly from Gauss' Diagam Theorem using my comment above. Using this theorem, we can express the finite sum in terms of logarithms and basic trig functions and then take limits and choose a specific $k$. – Brevan Ellefsen Mar 11 '19 at 17:31
  • I think it is simply$$\frac {-1}{5}\left(\psi\left(\frac 45\right)+\gamma\right)$$ Where $\gamma$ is the Euler-Mascheroni Constant – Rohan Shinde Mar 11 '19 at 17:41
  • @Darkrai What remains is to prove that the expression is equal to the claimed one. – user Mar 11 '19 at 17:46
  • @user For that, as stated in above comments, Gauss Digamma Theorem is here to help. – Rohan Shinde Mar 11 '19 at 17:52
  • As $\cot(\pi/5) = \sqrt{1 + 2/\sqrt5 }$ one may actually change the $\cot$ term for roots. – Andreas Mar 11 '19 at 22:04

4 Answers4

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

Andreas
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  • @OP: The identity cited from Achille Hui is known as Gauss' Diagamma Theorem, so this answer more or less uses the same logic I outlined in my comments above, and indeed shows that this generalizes to the sum $$\sum_{n=1}^\infty \frac{1}{kn(kn-1)}$$ – Brevan Ellefsen Mar 11 '19 at 17:33
  • @BrevanEllefsen Thanks. I put that in the main text as an alternative. – Andreas Mar 11 '19 at 17:53
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    Interesting, though I am not happy with writing $\sum_{n=1}^\infty\frac{1}{n} - \frac{1}{n-1/5} = \sum_{n=1}^\infty\frac{1}{n} - \sum_{n=0}^\infty \frac{1}{n+4/5}$ – Math-fun Mar 11 '19 at 18:01
  • I'm puzzled here: The expression you get works out numerically to be negative. In fact, it is exactly $\frac9{100}$ lower than the correct sum. But I can't find the error in your work. If you replace the $\frac54$ in the third from last expression with $\frac45$ it does get the right answer; yet from the page by Hui, it sure looks like the $\frac54$ belongs there. – Mark Fischler Mar 11 '19 at 18:31
  • @MarkFischler The result by Hui appears ok. I corrected a silly mistake in the $n=0$ term in the fifth line. Should be fine now. – Andreas Mar 11 '19 at 21:46
  • Are you a physicist, Andreas? I only ask because, well, $\infty-\infty$... – TonyK Mar 11 '19 at 22:03
  • @TonyK I know the danger. It works here, however. For rigor, I put it in the main text. – Andreas Mar 11 '19 at 22:18
  • That's better! +1 for rigour. – TonyK Mar 11 '19 at 22:29
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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 $f(1)$.)

Jair Taylor
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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: $f(1)$

Angel Moreno
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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: $$ -\frac1{2} (\log(5 - \sqrt 5) - \log(5 + \sqrt 5)) = \log(\rho) $$ $$ \log(5 + \sqrt 5) - \log(5 - \sqrt 5) = 2\log(\rho) $$ $$ \log \frac {5 + \sqrt 5} {5 - \sqrt 5} = \log(\rho^2) $$ $$ \frac {5 + \sqrt 5} {5 - \sqrt 5} = \rho^2 $$ $$ 5 + \sqrt 5 = (5 - \sqrt 5)\rho^2 $$ $$ 5 + \sqrt 5 = (5 - \sqrt 5)\frac{3+\sqrt 5}{2} $$ $$ 5 + \sqrt 5 = 5 + \sqrt 5 $$

Angel Moreno
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