$$\sum_{k=1}^{+\infty} \ln_2(1 - (k+1)^{-2}) = 1$$
Where $\ln_2$ is the binary log(log base 2).
I assume this can be done by some sort of telescoping ?
$$\sum_{k=1}^{+\infty} \ln_2(1 - (k+1)^{-2}) = \sum_{k=1}^{+\infty} \ln_2(k^2 + 2k) - \ln_2(k^2 + 2k +1)$$
I guess I am doing it wrong ? I can go further,
$$\sum_{k=1}^{+\infty} \ln_2(k^2 + 2k) - \ln_2(k^2 + 2k +1)= \sum_{k=1}^{+\infty} \ln_2(k) +\ln_2(k + 2) - 2 \ln_2(k + 1) = 1$$
Now if I exponentiate both sides I have a valid expression for $2$. So I know it is true, but Im not being efficient here I think.
