Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 10G (Analysis II)

Uniform convergence means: for every $\epsilon >0$ there is $N$ such that $n\ge N$ implies $|f_n(x)-f(x)|<\epsilon$ for every $x\in E$.

If $|f_n(x)|\le M_n$ and $\sum M_n<\infty$, then every tail is bounded uniformly by the corresponding numerical tail. The uniform Cauchy criterion therefore proves uniform convergence.

For each $a>1$, $n^{-x}\le n^{-a}$ on $[a,\infty )$ and the p-series converges. The M-test gives locally uniform convergence there, so the sum $\zeta$ is continuous on $(1,\infty )$.

The series for $S$ converges uniformly by the M-test. Therefore termwise integration is valid and every term integrates to $\sin (n^3\pi )/n^7=0$, so the integral is zero.

It is not uniform. For a small fixed $c>0$, set $x_N=c/N^3$. On a block $N\le n\le (1+\delta )N$, all $n^3{x}_N$ lie in a compact subinterval of $(0,\pi )$, so the corresponding tail has a positive lower bound independent of $N$; this violates the uniform Cauchy criterion.