= Solution
\b[The implication is false]. Take the thin rectangles
$$
A_\varepsilon=\{x+iy:-1\leq x\leq1,\ 0<y\leq\varepsilon\},\qquad 0<\varepsilon\leq1.
$$
Each is a <compact H-hull>, and $\operatorname{diam}(A_\varepsilon)=\sqrt{4+\varepsilon^2}\to2$. We verify that its <half-plane capacity> nevertheless tends to zero.
Let $K=\{z\in\mathbb H:|z|\leq2\}$, which contains all these rectangles. At the <Brownian exit time> $\tau$ from $\mathbb H\setminus A_\varepsilon$, the exit height is at most $\varepsilon$, and a positive exit height requires hitting $K$ before $\mathbb R$. Therefore
$$
\mathbb E_{iy}[\operatorname{Im}B_\tau]\leq\varepsilon\,\mathbb P_{iy}(B\text{ hits }K\text{ before }\mathbb R).
$$
For $y>2$, <conformal invariance of planar Brownian motion> and $g_K(z)=z+4/z$ identify the probability on the right with the <harmonic measure> of $[-4,4]$ from $i(y-4/y)$ in $\mathbb H$. Integrating the <Poisson kernel for the upper half-plane> gives
$$
\mathbb P_{iy}(B\text{ hits }K\text{ before }\mathbb R)=\frac2\pi\arctan\frac4{y-4/y},\qquad \lim_{y\to\infty}y\,\mathbb P_{iy}(B\text{ hits }K\text{ before }\mathbb R)=\frac8\pi.
$$
The <Brownian representation of half-plane capacity> now proves the explicit estimate
$$
\boxed{0\leq\operatorname{hcap}(A_\varepsilon)\leq\frac8\pi\varepsilon\longrightarrow0,\qquad \operatorname{diam}(A_\varepsilon)\longrightarrow2.}
$$
Choose $\varepsilon=1/n$ to obtain the required sequence. This is an instance of <half-plane capacity of a low rectangle>: shrinking height can make <half-plane capacity> small while horizontal extent stays fixed.
\Image[/media/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-203-capacity-examples.png]
{description=Two explicit <half-plane capacities> and a thin <compact H-hull> with fixed width and vanishing <half-plane capacity>.}
{height=360}
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