How to prove $$\sum _{k=0}^{\infty } \frac{L_{2 k+1}}{(2 k+1)^2 \binom{2 k}{k}}=\frac{8}{5} \left(C-\frac{1}{8} \pi \log \left(\frac{\sqrt{50-22 \sqrt{5}}+10}{10-\sqrt{50-22 \sqrt{5}}}\right)\right)$$ Where $L_k$ denotes Lucas number and $C$ Catalan? Any help will be appreciated.

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1 Answers1


This is more of an extended comment at this point until I link to the final result, but numerically the above relation is plausible.

There are two observations to make to take the main step forward:

$$L_n = \phi^n + (1-\phi)^n $$

where $\phi = (\sqrt{5}+1)/2$ is the golden ratio, and

$$\sum_{k=0}^{\infty} \frac{x^{2 k+1}}{(2 k+1) \binom{2 k}{k}} = \frac{2 \arcsin{(x/2)}}{\sqrt{1-(x/2)^2}} $$

Then after some manipulation, one may show that the above sum is equal to the following definite integral:

$$2 \int_{\arcsin{(\phi-1)/2}}^{\arcsin{(\phi/2)}} d\theta \, \theta \, \csc{\theta} $$

Note that the lower limit is equal to $\pi/10$ and the upper limit is $3 \pi/10$. This may be integrated by parts and manipulated a little further to produce

$$\frac{\pi}{5} \log{\left (\frac{\tan^3{\left ( \frac{3 \pi}{20} \right )}}{\tan{\left ( \frac{\pi}{20} \right )}} \right )} - 4 \int_{\tan{(\pi/20)}}^{\tan{(3 \pi/20)}} du \frac{\log{u}}{1+u^2}$$

What remains for us to prove is that

$$\int_{\tan{(\pi/20)}}^{\tan{(3 \pi/20)}} du \frac{\log{u}}{1+u^2} = -\frac25 C$$

which amazingly checks out numerically, and that the ratio

$$\frac{\tan^3{\left ( \frac{3 \pi}{20} \right )}}{\tan{\left ( \frac{\pi}{20} \right )}} = \frac{10-\sqrt{50-22 \sqrt{5}}}{10+\sqrt{50-22 \sqrt{5}}}$$

which also checks out numerically (and should be simple algebra).

Ron Gordon
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    Also the upper bound is $\frac{3\pi}{10}$. – Zacky Oct 12 '19 at 13:12
  • @LeBlanc: thanks for that, should have mentioned it. That said, even though you are right and we should be able to avoid dilogs, the truth is that the path to avoidance is very complex (unless there is a simple way of which I am not aware) and we still may be stuck with simplifying them. – Ron Gordon Oct 12 '19 at 13:17
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    $$2\int_{\arcsin{(\phi-1)/2}}^{\arcsin{(\phi/2)}} \theta \csc{\theta}d\theta=2\int_{\frac{\pi}{10}}^\frac{3\pi}{10}\theta \left(\ln\left(\tan \frac{\theta }{2}\right)\right)'d\theta $$ $$=\frac{3\pi}{5}\ln\left(\sqrt 5-1-\sqrt{5-2\sqrt 5}\right)-\frac{\pi}{5}\ln\left(\sqrt 5+1-\sqrt{5+2\sqrt5 } \right)-2\int_\frac{\pi}{10}^\frac{3\pi}{10}\ln\left(\tan \frac{\theta }{2}\right)d\theta$$ And the last integral equals $-\frac45 G$ as shown [here](https://math.stackexchange.com/q/1847894/515527), which completes the proof. – Zacky Oct 12 '19 at 13:49
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    @LeBlanc: Amazing! And it looks like we completely avoided dilogs! – Ron Gordon Oct 12 '19 at 13:59