Too much pi

$\begin{aligned} \mathscr L & = \lim_{n \to \infty} \left[\frac n2 \left(2 \sqrt [n] {\frac \pi 2} - \sqrt [n] {\frac \pi 3} - \sqrt [n] {\frac \pi 4} \right) \right] \\ & = \lim_{n \to \infty} \frac {2 \left(\frac \pi 2 \right)^\frac 1n - \left(\frac \pi 3 \right)^\frac 1n - \left(\frac \pi 4 \right)^\frac 1n }{\frac 2n} & \small \color{#3D99F6}{\text{Let }x = \frac 1n.} \\ & = \lim_{x \to 0} \frac {2 \left(\frac \pi 2 \right)^x - \left(\frac \pi 3 \right)^x - \left(\frac \pi 4 \right)^x}{2x} & \small \color{#3D99F6}{\text{This is a 0/0 case and L'Hôpital's rule applies.}} \\ & = \lim_{x \to 0} \frac {2 \ln \left(\frac \pi 2 \right) \left(\frac \pi 2 \right)^x - \ln \left(\frac \pi 3 \right) \left(\frac \pi 3 \right)^x - \ln \left(\frac \pi 4 \right) \left(\frac \pi 4 \right)^x}{2} & \small \color{#3D99F6}{\text{Differentiate up and down w.r.t. }x.} \\ & = \frac 12 \left( 2 \ln \left(\frac \pi 2 \right) - \ln \left(\frac \pi 3 \right) - \ln \left(\frac \pi 4 \right) \right) \\ & = \frac 12 \left(\ln \left( \frac {\pi^2}4 \times \frac 3 \pi \times \frac 4 \pi \right) \right) \\ & = \frac {\ln 3}2 \end{aligned}$

$\implies A+B = 3+2 = \boxed{5}$