For $r\geq1$, let $R_r=[-r,r]\times(0,1]$. The <Brownian representation of half-plane capacity> gives
$$
\operatorname{hcap}(R_r)=\lim_{y\to\infty}y\,\mathbb E_{iy}[\operatorname{Im}B_\tau].
$$
On hitting $R_r$ the imaginary part is at most one, while the <harmonic measure> estimate supplied in the question shows that the probability of reaching a disc of radius $O(r)$ containing $R_r$ is $O(r/y)$. Hence $\operatorname{hcap}(R_r)\leq Cr$, the <half-plane capacity of a low rectangle> bound.

Set
$$
A_n=n^{-1}R_n=[-1,1]\times(0,n^{-1}].
$$
The <scaling and translation of half-plane capacity> gives
$$
\operatorname{hcap}(A_n)=n^{-2}\operatorname{hcap}(R_n)\leq Cn^{-1}\longrightarrow0,
$$
whereas $\operatorname{diam}(A_n)=\sqrt{4+n^{-2}}\to2$. This supplies the required sequence.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 / 1 / b / ii / Solution

For $R_r=[-r,r]\times(0,1]$ with $r\geq1$,
$$
\operatorname{hcap}(R_r)\leq Cr.
$$
Indeed, the imaginary part at the Brownian exit point is at most one and the probability of reaching the radius-$O(r)$ neighbourhood containing the rectangle from $iy$ is $O(r/y)$. Consequently $r^{-1}R_r=[-1,1]\times(0,r^{-1}]$ has capacity $O(r^{-1})$ although its diameter tends to two.


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