# Prove $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\ln{(2n+1)}}{2n+1}=\pi/4(\gamma-\ln{\pi})+\pi\ln{(\Gamma(3/4))}$

In the title, $$\gamma$$ is the Euler-Mascheroni constant and $$\Gamma(3/4)$$ represents the extension of the factorial function.

This isn't a homework question or something, someone left it on a board in one of the buildings in my university and I'm just really surprised by it. The only thing I've tried is writing out the first few terms and trying to manipulate them into some sort of pattern, but I don't see where to go from there.

$$\frac{\ln{3}}{3}-\frac{\ln{5}}{5}+\frac{\ln{7}}{7}-\frac{\ln{9}}{9}+...$$ $$\ln{(3^{1/3})}-\ln{(5^{1/5})}+\ln{(7^{1/7})}-\ln{(9^{1/9})}+...$$ $$\ln{\Bigg(\frac{3^{1/3}}{5^{1/5}}\Bigg)}+\ln{\Bigg(\frac{7^{1/7}}{9^{1/9}}\Bigg)}+...$$

From here, I know that I could combine log terms even more, multiplying the numerators/denominators, but I don't think that's the right path to follow for this.

• haha, sounds like something out of Good Will Hunting. – zoidberg Dec 9 '18 at 19:51
• I have verified using software that the left side appears to converge somewhere close to the right side. – Calvin Godfrey Dec 9 '18 at 19:52
• A straightforward consequence of differentiation of the Dirichlet Beta function, or of Kummer-Malmstein Fourier series of $\log\Gamma$. – Jack D'Aurizio Dec 9 '18 at 20:14
• Thanks for that, Jack! I found the paper by Malmsten where he proved this, looking through it now. – Calvin Godfrey Dec 9 '18 at 20:25
• The original paper by Malmsten is here, with the statement in question on page 24 of the pdf, and the proof preceding it. – Calvin Godfrey Dec 9 '18 at 20:41