Let $S(\alpha)=\sum_yQ(y)^\alpha$. Both $\log S(\alpha)$ and $1-\alpha$ vanish at one. By <L'Hopital rule>, with logarithms to base two,
$$
\lim_{\alpha\to1}H_\alpha(Y)
=-\left.\frac{d}{d\alpha}\log_2S(\alpha)\right|_{\alpha=1}
=-\sum_yQ(y)\log_2Q(y)
=H(Y).
$$
Terms with $Q(y)=0$ contribute zero by continuity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 224 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-224/2/a/solution