I used Wolfram-Alpha to try to find the sum $$\sum_{n=0}^\infty (-1)^n\ln\frac{n+2}{n+1}.$$
While it only gave a numerical result, it showed the partial sum formula included $\text{LerchPhi}^{(0,1,0)}(-1,0,m+2)$, where $\Phi(x,s,a)$ is the Lerch Transcendent. I'm not sure how to relate the two. Is $(x,s,a)$ the superscripted parenthesis? Or the other, and why is there two?
Thanks.
By the unlimited power of Frullani's theorem
$$\sum_{n\geq 0}(-1)^n\log\frac{n+2}{n+1}=\int_{0}^{+\infty}\frac{dx}{x}\sum_{n\geq 0}(-1)^n\left(e^{-(n+1)x}-e^{-(n+2)x}\right)=\int_{0}^{+\infty}\frac{1-e^{-x}}{x(1+e^x)}\,dx$$ where the last integral can be computed from the Laplace transform of the $\tanh $ function, that depends on $\psi$. Or, simply, notice that your series is the logarithm of Wallis' product, hence its value is $\color{red}{\large{\log\frac{\pi}{2}}}$.