Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 2 / 13F / Solution

Because $|u_n(x)|\le M_n$ and the numerical series converges, its tails can be made arbitrarily small independently of $x$:Thus the partial sums satisfy the uniform Cauchy criterion; completeness of the real numbers supplies the pointwise limit, and the same tail bound proves uniform convergence. This is the Weierstrass M-test.

For $E_R$ with $R>0$, choose an integer $N>2R$. If $n>N$ and $|x|<R$, then $|x\pm n|\ge n-R>n/2$, so $|u_n(x)|\le 8/n^2$. The tail is therefore uniformly convergent by the M-test; adding its finitely many earlier terms preserves uniform convergence, even though those earlier terms can be unbounded near excluded integers. For $R\le 0$ the set is empty and the claim is vacuous.

Each summand is continuous on $E=\mathbb{R}\setminus \mathbb{Z}$. Around any $x\in E$, choose an interval containing $x$, avoiding integers and lying in some $E_R$. The uniform-limit continuity theorem then proves $\boxed{f\text{ is continuous on }E}$.

The series is not uniformly convergent on all of $E$. Uniform convergence would require $u_n\to 0$ uniformly, but $u_n(n+1/2)\ge 4$ for every $n>0$. These evaluation points belong to $E$, contradicting that necessary condition.