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I need the following to proof a particular case of the Abel's general convergence criterion which goes as follows:

Let $\Omega$ be a non-empty subset of $\mathbb{C}$ and A a non-empty subset of $\Omega$. Let $f_n$ and $g_n$ be two sequences of functions from $\Omega$ to $\mathbb{C}$ with the following properties:

1.1) $\sum f_n$ converges uniformly in A.

1.2) The series $\sum (|g_1|+ \sum |g_n-g_{n+1}|)$ has bouded partial sums in A.

Then the series $\sum f_ng_n$ converges uniformly in A.

The particular criterion I want to proof goes as follows:

Let $\Omega$ be a non-empty subset of $\mathbb{C}$ and A a non-empty subset of $\Omega$. Let $f_n$ be a sequence of functions from $\Omega$ to $\mathbb{C}$ and $a_n$ a sequence of complex numbers with the following properties:

1.3) $\sum a_n$ is convergent

1.4) For each $z \in A, {f_n(z)}$ is a monotonic sequence of real numbers and ${f_n}$ is uniformly bounded in A.

then $\sum a_nf_n$ converges uniformly in A.

The theorem I should be using to proof it is:

If $b_n$ is monotonic and bounded then $\sum |b_n - b_{n+1}|$ is convergent.

It is clear for me that 1.4 is a particular case of 1.1 but my problem is that I don't really see how to mix the uniformly bounded condition in 1.3 with the bounded condition of partial sums in 1.2. It should begin like $\forall z \in A,\{f_n(z)\}$ is monotone and bounded therefore...

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  • $\begingroup$ I don't understand the $\Omega,A$ business. Why two sets? Also it seems the result is false as stated. Consider $A=[0,1], a_n = 1/n^2, f_n(x)= nx.$ $\endgroup$ – zhw. Apr 11 '16 at 5:43
  • $\begingroup$ @zhw I just edited my question. First, I forgot in 1.4) that $f_n$ should be uniformly bounded in A. Second, $\Omega$ stands for the definition set of $f_n$ and $A$ stands for the set of uniform convergence. $\endgroup$ – Javier Apr 12 '16 at 0:55

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