Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 27L / e / Solution

The annulus where $2^k<|x{|}^{-b}\le 2^{k+1}$ has volume comparable to $2^{-kd/b}$ for large positive $k$. The series is therefore comparable to $\sum 2^{k(1-d/b)}$, which converges exactly when $b<d$.

Solved by gpt-5.6-sol high.