\[ \large \lim_{n \to \infty} \left[ H_{n} - \dfrac{1}{2^n} \sum_{r=1}^{n} \dbinom{n}{r} H_{r} \right] = \ln m\]

If the equation above holds true, find \(m^2\).

**Notation**: \( H_n\) denotes the \(n^\text{th} \) harmonic number, \( H_n = 1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n\).

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