A tough series

[Tolaso J Kos]

Prove that:

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \sum_{k=0}^{2n} \frac{1}{2n+4k+3}= \frac{3\pi}{8} \log \left( \frac{1+\sqrt{5}}{2} \right) - \frac{\pi}{16} \log 5$$

[galactus]

I deleted what I posted because I was dissatisfied with it. There were convergence issues.

Using digamma and the arctan series, I found the sum is equal to the integral:

$$\int_{0}^{1}\frac{x(\tan^{-1}(x)-\tan^{-1}(x^{5}))}{x^{4}-1}dx=\int_{0}^{1}\frac{x\tan^{-1}\left(\frac{x(x^{2}-1)}{x^{4}-x^{2}+1}\right)}{x^{4}-1}dx$$

I think contours can finish it, but I have not...yet anyway .

Hey RD?.