= Solution
Because $|u_n(x)|\le M_n$ and the numerical series converges, its tails can be made arbitrarily small independently of $x$:
$$
\sup_{x\in E}\left|\sum_{n=N+1}^M u_n(x)\right|\le\sum_{n>N}M_n\longrightarrow0.
$$
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 $|u_n(x)|\le8/n^2$. 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 $R\le0$ 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 \b[not uniformly convergent on all of $E$]. <Uniform convergence> would require $u_n\to0$ uniformly, but $u_n(n+1/2)\ge4$ for every $n>0$. These evaluation points belong to $E$, contradicting that necessary condition.
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