Weierstrass M-test
The M-test is the most painless way to prove a series of functions converges uniformly. The trick is to forget the variable entirely: bound each term by a constant that does not depend on x, then check that those constants sum to a finite number. If a single convergent series of numbers dominates your whole series of functions, uniformity comes for free.
Statement: suppose f_n are functions on a set E and there exist constants M_n with |f_n(x)| <= M_n for all x in E, and the numeric series sum M_n converges. Then the series sum f_n(x) converges uniformly (and absolutely) on E. The M_n are upper bounds — typically M_n = sup over x of |f_n(x)| is the sharpest choice, but any valid bound works.
The proof is the uniform Cauchy criterion in disguise: the tail |sum_{k=n+1}^m f_k(x)| <= sum_{k=n+1}^m M_k, and the right side is independent of x and tends to 0 because sum M_n converges. An honest limitation: the M-test only detects ABSOLUTE uniform convergence, so it cannot certify series that converge uniformly only by delicate cancellation (for those, use Dirichlet's or Abel's test instead).
The series sum_{n>=1} sin(nx)/n^2 on the whole real line: |sin(nx)/n^2| <= 1/n^2 = M_n, and sum 1/n^2 = pi^2/6 < infinity. By the M-test the series converges uniformly on R; since each term is continuous, the sum is a continuous function on R.
A convergent numeric series dominates and forces uniform convergence.