The series $\sum_{k=1}^{+\infty} 2 \log(2k+1) - \log(2k) - \log(2k+2) = (2\log3-\log2-\log4) + (2\log5-\log4-\log6) + ...$ converges to a value approximately equal to 0.242 (obtained computing the partial sum with 10000 terms).
I have been playing around but I can't find an explicit formula or link it with a known constant. Do you have any thoughts or ideas?
It is a series of positive terms that tends to zeros. Interestingly, we cannot split the terms in the following way: $\sum_{k=1}^{+\infty} 2 \log(2k+1) - \log(2k) - \log(2k+2) = -\log2 + 2\log3 - 2\log4 + 2 \log5 - 2 \log6 + \ldots$ This last series does not converge since the terms are growing larger and larger.
The hint.
Use $$\frac{\pi}{2}=\prod_{n=1}^{+\infty}\frac{4n^2}{4n^2-1}.$$
I got the following answer: $$\ln\frac{4}{\pi}.$$
Indeed, since $\ln$ is a continuous function, we obtain: $$\sum_{k=1}^{+\infty}\left(2\ln(2k+1)-\ln2k-\ln(2k+2)\right)=\sum_{k=1}^{+\infty}\ln\frac{(2k+1)^2}{2k(2k+2)}=$$ $$=\ln\prod_{k=1}^{+\infty}\frac{(2k+1)^2}{2k(2k+2)}=\ln\frac{2}{\prod\limits_{k=1}^{+\infty}\frac{4k^2}{4k^2-1}}=\ln\frac{4}{\pi}.$$