Uniform convergence of series

Remarks

Equivalently, the series converges at each point of $S$ and $\sup_{x \in S} \abs{\sum_{n > N} f_n(x)} \to 0$ as $N \to \infty$. A Cauchy form also holds: the series converges uniformly on $S$ if and only if for every $\varepsilon > 0$ there is $N_0$ with $\abs{\sum_{n=M+1}^{N} f_n(x)} < \varepsilon$ for all $N > M \ge N_0$ and all $x \in S$. For a subset of a larger domain, this is uniform convergence on a set of the partial sums.

Example: let $0 < r < 1$ and $f_n(x) = x^n$ on $S = [0, r]$, with $f(x) = 1/(1 - x)$. The tail is $\sum_{n > N} x^n = x^{N+1}/(1 - x) \le r^{N+1}/(1 - r)$ for all $x \in S$, and this bound tends to $0$. So the series converges uniformly on $[0, r]$.

Non-example: the same series on $S = [0, 1)$ converges at each point to $1/(1 - x)$, but not uniformly. For each $N$, the tail $x^{N+1}/(1 - x)$ tends to $\infty$ as $x \to 1^-$, so its supremum over $S$ is infinite.

The Weierstrass M-test is the usual way to prove uniform convergence.