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Let $\displaystyle J(x) = \sum_{n=1}^{\infty} H_n x^n = -\frac{\ln(1-x)}{1-x}$ for $|x|<1$

We have $\displaystyle J'(\frac{1}{2}) = \sum_{n=1}^{\infty} \frac{nH_n}{2^{n-1}} = \frac{1}{(1-\frac{1}{2})^2} - \frac{\ln(1-\frac{1}{2})}{(1-\frac{1}{2})^2}$

Thus , $\displaystyle \sum_{n=1}^{\infty} \frac{nH_n}{2^{n-1}} = \color{#D61F06}{4}+\ln\color{#3D99F6}{16}$ making the answer $\boxed{4+16=20}$

Note : There's a typo , It should be $A+\ln B$

$\begin{aligned} S & = \sum_{n=1}^\infty \frac{n H_n}{2^{n-1}} \\ & = \color{#3D99F6}{1} + \sum_{n=\color{#3D99F6}{2}}^\infty \frac{n H_n}{2^{n-1}} \\ & = 1 + \sum_{n=2}^\infty \frac{n \left(H_{n-1}+\frac1n\right)}{2^{n-1}} \\ & = 1 + \sum_{n=2}^\infty \frac{n H_{n-1}+1}{2^{n-1}} \\ & = 1 + \sum_{n=\color{#3D99F6}{1}}^\infty \frac{\color{#3D99F6}{(n+1)}H_\color{#3D99F6}{n}+1}{2^\color{#3D99F6}{n}} \\ & = \color{#D61F06}{1} + \frac{1}{\color{#3D99F6}{2}} \sum_{n=1}^\infty \frac{n H_n}{2^\color{#3D99F6}{n-1}} + \sum_{n=1}^\infty \frac{H_n}{2^n} + \color{#D61F06}{\sum_{n=1}^\infty \frac1{2^n}} \\ \implies \frac12 S & = \sum_{n=1}^\infty \frac{H_n}{2^n} + \color{#D61F06}{\sum_{n=0}^\infty \frac1{2^n}} \\ & = \frac{-\ln \left(1-\frac12\right)}{1-\frac12} + \frac1 {1-\frac12} \\ & = 2\ln 2 + 2 \\ \implies S & = 4\ln 2 + 4 \\ & = 4 + \ln 16 \end{aligned}$

$\implies A+B = 4+16 = \boxed{20}$ .