For $\sigma=\Re s>1$,
$$
\sum_{n=1}^\infty\left|\frac{f(n)}{n^s}\right|
\leq\sum_{n=1}^\infty\frac1{n^\sigma}
=\zeta(\sigma)<\infty.
$$
On every compact subset of $\Re s>1$, the terms are bounded by a convergent series $\sum n^{-1-\delta}$. The <Weierstrass M-test> gives locally uniform convergence, and the theorem on <locally uniform convergence of holomorphic functions> shows that the limit is an <analytic function>. Hence $D_f$ is analytic throughout $\Re s>1$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 3 / a / Solution

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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-150/3/a/solution