On the closed form for $\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}$

We have $$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-1}=\frac{3}{2}+\frac{\ln(\sqrt[3]{2}-1)}{2^{7/3}}+\frac{\sqrt{3}}{2^{4/3}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}\tag1$$ $$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-2}=\frac{3}{2}-\frac{\ln(\sqrt[3]{2}-1)}{2^{5/3}}+\frac{\sqrt{3}}{2^{2/3}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}\tag2$$ with the first discussed in this post. (Update) Courtesy of the answers below, we also have, $$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}=\frac{4}{3}+\frac{\ln\big((4^{1/4}-3^{1/4})\sqrt{2+\sqrt{3}}\big)}{3^{5/4}\sqrt{2}}+\frac{\sqrt{2}}{3^{5/4}}\arctan\big(3^{1/4}\sqrt{2+\sqrt{3}}\big)\tag3$$ $$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-3}=\frac{4}{3}-\frac{\ln\big((4^{1/4}-3^{1/4})\sqrt{2+\sqrt{3}}\big)}{3^{3/4}\sqrt{2}}+\frac{\sqrt{2}}{3^{3/4}}\arctan\big(3^{1/4}\sqrt{2+\sqrt{3}}\big)\tag4$$Walpha gives a closed-form for $(3)$, but uses two log functions and two arctans. The two answers below show it can be simplified further similar to $(1),(2)$.

Solution 1:

Yes. The logarithms have the same denominator. So factoring this out gives a simple difference of logs: namely the simple log of a quotient, which itself can be simplified to $$2\ln\left(\frac12+\frac{\surd3}2-\sqrt{2\surd3}\right)$$(if my calculation is right). A similar method will simplify the inverse cotangents to a single inverse cotangent, by use of the formula $$\alpha+\beta=\operatorname{arcot}\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}$$(exactly or to within $\pm\pi$).

(Revised) According to my (corrected) calculation, Wolfram's expression for formula 3 can be written as $$\frac43+\frac4{3a}\arctan \frac{2a+a^3}4-\frac2{3a}\ln\frac{2+2a+a^2}4,$$where $a:=\sqrt{2\surd3}$.

Solution 2:

What you need is $\enspace\displaystyle \arctan x+\arctan y=\arctan\frac{x+y}{1-xy}\enspace$ because the values in front of the two $\arctan$ are equal of the results of wolframalpha. Also the values in front of the two $\ln$ are the same.