I was told this interesting question today, but I haven't managed to get very far:
Evaluate $$\sum_{n=1}^\infty \log \left(1+\frac{1}{n}\right)\log \left(1+\frac{1}{2n}\right)\log \left(1+\frac{1}{2n+1}\right).$$
I am interested in seeing at least a few solutions.
Here is a solution I just found. Notice that $$\log\left(1+\frac{1}{2n+1}\right)=\log\left(1+\frac{1}{n}\right)-\log\left(1+\frac{1}{2n}\right)$$ so that our series becomes $$\sum_{n=1}^{\infty}\left(\log\left(1+\frac{1}{n}\right)^{2}\log\left(1+\frac{1}{2n}\right)-\log\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{2n}\right)^{2}\right).$$ Since $$\log\left(1+\frac{1}{2n+1}\right)^{3}=\log\left(1+\frac{1}{n}\right)^{3}-3\log\left(1+\frac{1}{n}\right)^{2}\log\left(1+\frac{1}{2n}\right)+3\log\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{2n}\right)^{2}-\log\left(1+\frac{1}{2n}\right)^{3},$$ we see that our series equals $$\frac{1}{3}\left(\sum_{n=1}^{\infty}\log\left(1+\frac{1}{n}\right)^{3}-\log\left(1+\frac{1}{2n}\right)^{3}-\log\left(1+\frac{1}{2n+1}\right)^{3}\right),$$ and the above telescopes and equals $$\frac{\left(\log2\right)^{3}}{3}.$$