Infinite Series of Functions is Uniformly Convergent iff Sequence of Partial Sums is Uniformly Cauchy

Theorem
Let $S \subseteq \R$.

Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$.

Then the infinite series:


 * $\ds \sum_{n \mathop = 1}^\infty f_n$

converges uniformly on $S$ for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that:


 * $\ds \size {\sum_{k \mathop = m + 1}^n \map {f_k} x} < \varepsilon$

for all $x \in S$ and $n > m > N$.

Proof
Let $\sequence {s_n}$ be a sequence of real functions $S \to \R$ with:


 * $\displaystyle \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_k} x$

for each $n \in \N$ and $x \in S$.

By the definition of uniform convergence of an infinite series:


 * $\ds \sum_{n \mathop = 1}^\infty f_n$ is uniformly convergent $\sequence {s_n}$ is uniformly convergent.

By Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent:


 * $\sequence {s_n}$ is uniformly convergent it is uniformly Cauchy.

That is for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that for all $m, n > N$ we have:


 * $\size {\map {s_n} x - \map {s_m} x} < \varepsilon$ for all $x \in S$.

Note that if $n = m$:


 * $\size {\map {s_n} x - \map {s_m} x} = 0 < \varepsilon$

, we can therefore take $n > m$.

If $n > m$, then:

So $\sequence {s_n}$ is uniformly Cauchy for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that:


 * $\ds \size {\sum_{k \mathop = m + 1}^n \map {f_k} x} < \varepsilon$

for all $n > m > N$ and $x \in S$.