Based upon Oloa's question here Evaluating $\sum_{n \geq 1}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)$ I was thinking if we possibly can get a nice way to evaluate the series
$$\sum_{n=1}^{\infty}\ln^2 \!\left(1+\frac1{2n}\right) \!\ln^2\!\left(1+\frac1{2n+1}\right).$$
Maybe using the same telescoping idea or it simply doesn't work? What then?
It can be shown that: \begin{align} \sum_{n=1}^{\infty} \ln^{2}\left(1+\frac{1}{2n}\right) \, \ln^{2}\left(1 + \frac{1}{2n+1}\right) = - \frac{\ln^{4} 2}{2} + 2 \, \sum_{n=1}^{\infty} \ln^{2}\left(1 + \frac{1}{n}\right) \, \ln\left(1+\frac{1}{2n}\right) \, \ln\left(1 + \frac{1}{2n+1}\right). \end{align} The remaining summation may be presentable in a form not based on numerical estimation. In the event a few decimal place accuracy is sought the following identity is presented. Accurate to 9 decimal places the series is given by \begin{align} \sum_{n=1}^{\infty} \ln^{2}\left(1+\frac{1}{2n}\right) \, \ln^{2}\left(1 + \frac{1}{2n+1}\right) = \left( \frac{1}{2} - \frac{1}{5^{\frac{32}{37}} \, \gamma} \right) \, \ln^{4}2 \approx 0.01600318562\cdots \end{align} where $\gamma$ is the Euler-Mascheroni constant.
Proof: Let $$f(x) = \ln\left(1 + \frac{1}{x}\right) \tag{1}$$ then $$f^{2}(n) = ( f(2n) + f(2n+1) )^{2} = f^{2}(2n) + f^{2}(2n+1) + 2 \, f(2n) \, f(2n+1) \tag{2}$$ which yields $$f(2n) \, f(2n+1) = \frac{1}{2} \, [ f^{2}(n) - f^{2}(2n) - f^{2}(2n+1) ]. \tag{3}$$ Squaring both sides provides \begin{align} & f^{2}(2n) \, f^{2}(2n+1) \\ & \hspace{5mm} = \frac{1}{4} \, [ f^{4}(n) + f^{4}(2n) + f^{4}(2n+1) - 2 \, f^{2}(n) \, ( f^{2}(2n) + f^{2}(2n+1) ) + 2 \, f^{2}(2n) \, f^{2}(2n+1) ] \end{align} or $$f^{2}(2n) \, f^{2}(2n+1) = \frac{1}{2} \, [ f^{4}(n) + f^{4}(2n) + f^{4}(2n+1) - 2 \, f^{2}(n) \, ( f^{2}(2n) + f^{2}(2n+1) ) ] \tag{4}$$ By making use of (1) this reduces to $$f^{2}(2n) \, f^{2}(2n+1) = - \frac{1}{2} \, [f^{4}(n) - f^{4}(2n) - f^{4}(2n+1) ] + 2 \, f^{2}(n) \, f(2n) \, f(2n+1). \tag{5}$$
Now, summing over the index then it is seen that: \begin{align} S &= \sum_{n=1}^{\infty} f^{2}(2n) \, f^{2}(2n+1) \\ &= - \frac{1}{2} \, \sum_{n=1}^{\infty} \left[ f^{4}(n) - f^{4}(2n) - f^{4}(2n+1) \right] + 2 \, \sum_{n=1}^{\infty} f^{2}(n) \, f(2n) \, f(2n+1) \\ &= - \frac{1}{2} \, f^{4}(1) + + 2 \, \sum_{n=1}^{\infty} f^{2}(n) \, f(2n) \, f(2n+1) \tag{6} \end{align} From (6) the first statement presented is obtained. As to the remaining summation it is conjectured that it is also of the form $A_{0} \, \ln^{4}2$.
