looks like double trouble is kicking in

$\begin{aligned} S & = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac 1{2^{n+m-1}(n+m-1)(n+m)} \\ & = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac 1{2^{n+m-1}}\left(\frac 1{n+m-1} - \frac 1{n+m} \right) \\ & = \sum_{n=1}^\infty \sum_{m=1}^\infty \left(\frac 1{(n+m-1)2^{n+m-1}} - \frac 2{(n+m)2^{n+m}} \right) \\ & = \sum_{n=1}^\infty \left(\frac 1{n(2^n)} - \sum_{m=1}^\infty \frac 1{(n+m)2^{n+m}} \right) & \small \blue{\text{Replace }\frac 12 \text{ by x}} \\ & = \sum_{\blue{n=1}}^\infty \int x^{\blue{n-1}} dx - \sum_{n=1}^\infty \sum_{\blue{m=1}}^\infty \int x^{n+\blue{m-1}} dx \\ & = \sum_{\red{n=0}}^\infty \int x^{\red n} dx - \sum_{n=1}^\infty \sum_{\red{m=0}}^\infty \int x^{n+\red m} dx & \small \blue{\text{Since }x^k \text{ converges absolutely for}} \\ & = \int \left(\sum_{n=0}^\infty x^n \right) dx - \int \left(\sum_{n=1}^\infty \sum_{m=0}^\infty x^{n+m} \right) dx & \small \blue{|x| < 1 \text{, we can swap} \sum \text{ with }\int.} \\ & = \int \frac {dx}{1-x} - \int \left(\sum_{n=1}^\infty x^n \sum_{m=0}^\infty x^m \right) dx \\ & = - \ln (1-x) - \blue{\int \frac x{(1-x)^2} dx} & \small \blue{\text{By integration by parts}} \\ & = - \ln(1-x) - \blue{\frac x{1-x} + \int \frac 1{1-x} dx} \\ & = - 2\ln(1-x) - \frac x{1-x} & \small \blue{\text{Put back }\frac 12 \text{ as }x} \\ & = \ln 4 - 1 \end{aligned}$

Therefore, $A+B= 4+1 = \boxed 5$ .

By expanding $\displaystyle \int_{0}^{x}\frac{1}{1+x} dx$ using integration by parts and not by directly finding the anti-derivative we can find an infinite series for $\ln\left(1+x\right)$ which converges for all $x$ in the domain of this function in the form of $\left(\displaystyle\sum_{m=1}^{\infty} \frac{x^{m}}{m\left(1+x\right)^{m}}\right)$ . We find the anti-derivative to this series which is $\displaystyle\int\ln\left(1+x\right) dx=\displaystyle\int \left(\displaystyle\sum_{m=1}^{\infty} \frac{x^{m}}{m\left(1+x\right)^{m}}\right) dx$

For each term the integral becomes $\displaystyle\sum_{i=1}^{\infty} \frac{x^{m+i}}{\left(m+i-1\right)\left(m+i\right)\left(1+x\right)^{m+i-1}}$

applying this to all terms the general form will become $\displaystyle\sum_{m=1}^{\infty} \left( \sum_{i=1}^{\infty} \frac{x^{m+i}}{\left(m+i-1\right)\left(m+i\right)\left(1+x\right)^{m+i-1}}\right) =\displaystyle\int \ln\left( 1+x\right) dx =\left(1+x\right)\ln\left( 1+x\right) -x$

for $x=1$ we have , $\left(2\right)\ln\left( 2\right) -1= \ln \left(4\right) -1$

This is also same as

$\displaystyle\sum_{m=1}^{\infty} \left( \sum_{i=1}^{\infty} \frac{1^{m+i}}{\left(m+i-1\right)\left(m+i\right)\left(2\right)^{m+i-1}}\right)$

$\displaystyle\sum_{m=1}^{\infty} \left( \sum_{i=1}^{\infty} \frac{1}{\left(m+i-1\right)\left(m+i\right)\left(2\right)^{m+i-1}}\right)$

We have, putting $N = m+n-1$ , $\begin{aligned} \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{2^{m+n-1}(m+n-1)(m+n)} & = \; \sum_{N=1}^\infty \sum_{n=1}^N \frac{1}{2^NN(N+1)} \; = \; \sum_{N=1}^\infty \frac{1}{2^N(N+1)} \\ & = \; 2\sum_{N=1}^\infty \frac{1}{2^NN} - 1 \; = \; -2\ln\tfrac12 - 1 \; = \; \ln4 - 1 \end{aligned}$ Assuming that $A,B$ are integers, this makes $A+B=4+1=\boxed{5}$ .

Of course, we could also have $A = 4e$ and $B=2$ , which is allowed by the wording of the question (but disallowed by the integer answer format).