Evaluating $\sum_{n \geq 1}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)$

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 $(1)$.

• Your instinct seems good, but I couldn't make telescoping sums work either.... – Greg Martin Aug 12 '15 at 16:28
• @Chris'ssistheartist: Can you please provide the link to the source? Thank you! :) – Pranav Arora Aug 12 '15 at 18:12
• @PranavArora I think I mixed up the problems. – user 1357113 Aug 12 '15 at 18:15
• Another interesting one would be $$\sum_{n=1}^{\infty}\ln^2 \!\left(1+\frac1{2n}\right) \!\ln^2\!\left(1+\frac1{2n+1}\right).$$ – user 1357113 Aug 12 '15 at 18:25

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: $$\boxed{\dfrac{1}{2}\ln^22}$$
• I thought that maybe I could just somehow split the logs. Seeing the product, I immediately thought of $(a+b)^2=a^2+b^2+2ab$ and the fact that $(1+\frac{1}{2n})(1+\frac{1}{2n+1})=1+\frac{1}{n}$ also helped. :) – Pranav Arora Aug 12 '15 at 18:11