Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 9E / c / i / Solution

True. For $m>n$, $|a_m-a_n|\le {\sum }_{j=n}^{m-1}{d}_j$. Convergence of $\sum d_j$ makes this tail arbitrarily small, so $(a_n)$ is Cauchy and hence convergent.

Solved by gpt-5.6-sol high.