TITLE: Evaluating $\sum_{n \geq 1}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)$
QUESTION [20 upvotes]: Is there a direct way to evaluate the following series?

$$
\sum_{n=1}^{\infty}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)=\frac12\ln^2 2. \tag1
$$ 

I've tried telescoping sums unsuccessfully. The convergence is clear. Given the simplicity of the result, I'm inclined to think it might exist an elegant way to get $(1)$.

REPLY [36 votes]: Use:
$$\left(\ln\left(1+\frac{1}{2n}\right)+\ln\left(1+\frac{1}{2n+1}\right)\right)^2=\ln^2\left(1+\frac{1}{2n}\right)+\ln^2\left(1+\frac{1}{2n+1}\right)+2\ln\left(1+\frac{1}{2n}\right)\ln\left(1+\frac{1}{2n+1}\right)$$
$$\Rightarrow \ln\left(1+\frac{1}{2n}\right)\ln\left(1+\frac{1}{2n+1}\right)=\frac{1}{2}\left(\ln^2\left(1+\frac{1}{n}\right)-\ln^2\left(1+\frac{1}{2n}\right)-\ln^2\left(1+\frac{1}{2n+1}\right)\right)$$
Therefore, the sum in the given problem is:
$$\frac{1}{2}\left(\sum_{n=1}^{\infty}\ln^2\left(1+\frac{1}{n}\right)-\sum_{n=1}^{\infty}\ln^2\left(1+\frac{1}{2n}\right)-\sum_{n=1}^{\infty}\ln^2\left(1+\frac{1}{2n+1}\right)\right)$$
Notice that all the terms of the first sum are cancelled by the other two sums except $n=1$, hence our answer is:
$$\boxed{\dfrac{1}{2}\ln^22}$$