A natural series

$\displaystyle \sum_{2}^{\infty} \ln \left( 1-\frac{1}{n^{2}}\right)=\displaystyle \sum_{2}^{\infty} \ln \left( \frac{n^{2} -1}{n^{2}}\right)$

$\displaystyle \sum_{2}^{\infty} \ln \frac{\left(n -1\right)\left(n+1\right)}{n^{2}}$

$\ln\left( \displaystyle\prod^{\infty}_{2} \frac{\left(n -1\right)\left(n+1\right)}{n^{2}}\right)$

$\displaystyle\prod^{m}_{2} \frac{\left(n -1\right)\left(n+1\right)}{n^{2}} =\frac{\left(m -1\right)!\left(m+1\right)!}{2 \left(m!\right)^{2}}$

$\frac{\left(m -1\right)!\left(m+1\right)!}{2 \left(m!\right)^{2}} =\frac{\left(m \right)!\left(m\right)!\times \left(m+1\right)}{2 \times m \left(m! \times m!\right)}$

$\frac{m+1}{2m}$

sum of series is natural log of this limiting m to $\infty$ :

$\displaystyle\lim_{m\to \infty}\ln\left(\frac{m+1}{2m}\right)=\ln\left(\frac{1}{2m} +\frac{1}{2}\right)=\ln\left(\frac{1}{2}\right)$

$\begin{aligned} \sum_{n=2}^\infty \ln \left(1-\frac 1{n^2} \right) & = \sum_{n=2}^\infty \ln \left(\frac {n^2-1}{n^2} \right) \\ & = \sum_{n=2}^\infty \ln \left(\frac {(n-1)(n+1)}{n^2} \right) \\ & = \sum_{n=2}^\infty \Big(\ln(n-1) + \ln (n+1) - 2\ln n \Big) \\ & = \sum_{n=2}^\infty \Big(\ln(n-1) - \ln n - \ln n + \ln (n+1) \Big) \\ & = \sum_{n=2}^\infty \Big(\ln(n-1) - \ln n \Big) - \sum_{n=2}^\infty \Big(\ln n - \ln (n+1) \Big) \\ & = \ln 1 - \ln 2 = - \ln 2 = \ln \frac 12 \end{aligned}$

Therefore, $A+B = 1+2=\boxed 3$ .