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The Proposition states if $(f_n)_{n\in\mathbb{N}}$ is a secuence of functions $f_n:D\rightarrow \mathbb{C}$ with $\sum_{n=1}^{\infty} ||f_n||_D < +\infty$ then the series $\sum_{n=1}^{\infty}f_n$ converges absolutely uniform and pointwise.

The series converges uniformly (or uniformous?)if and only if the secuence of partial sums do so. The series of partialsums is $(\sum_{i=1}^{n}f_n)_{n\in\mathbb{N}}$

A sequence of functions converge if and only if

$\exists_{f:D\rightarrow\mathbb{C}}\forall_{\epsilon>0}\exists_{n(\epsilon)\in\mathbb{N}}\forall_{n\geq n(\epsilon)}\forall_{z\in D}:|f_n(z)-f(z)|<\epsilon$

Which would mean

$\exists_{f:D\rightarrow\mathbb{C}}\forall_{\epsilon>0}\exists_{n(\epsilon)\in\mathbb{N}}\forall_{n\geq n(\epsilon)}\forall_{z\in D}:|\sum_{i=1}^{n}f_n(z)-f(z)|<\epsilon$

My Question what would be the $f$ in this Proposition?

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    $\begingroup$ Perhaps you are having a translation difficulty (English is an immensely irregular language). "Uniformous" is not a word in English, although logically it could be. The correct word is "uniformly". $\endgroup$ – DanielWainfleet Jan 10 at 9:28
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For each $z$ the series $ \sum\limits_{k=1}^{\infty} f_k(z)$ is absolutely convergent. $f(z)$ is the sum of this series. Note that $| \sum\limits_{k=1}^{n} f_k(z) - \sum\limits_{k=1}^{m} f_k(z) |<\epsilon$ for all $z \in D$ for all $n,m \geq n_0$ implies $| \sum\limits_{k=1}^{n} f_k(z) -f(z) |\leq \epsilon$ for all $z \in D$ for all $n \geq n_0$.

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