# Are these two Abel's criteria for uniform convergence different?

I wonder what differences are between the folowwing two versions of Abel's criteria for uniform convergence:

Abel's Test. Let $A \subset R^n$ and $\phi_n: A \rightarrow R$ be a sequence of functions which are decreasing; that is, $\phi_{n+1}(x) \leq \phi_n(x)$ for each $x \in A$. Suppose there is a constant $M$ such that$|\phi_n(x)| \leq M$ for all $x \in A$ and all $n$. If $\displaystyle\sum_{n=1}^\infty f_n(x)$ converges uniformly on $A$, then so does $\displaystyle \sum_{n=1}^\infty \phi_n(x)f_n(x)$.

From Wikipedia:

Abel's uniform convergence test. Let $\{g_n\}$ be a uniformly bounded sequence of real-valued continuous functions on a set $E$ such that $g_{n+1}(x) \leq g_n(x)$ for all $x ∈ E$ and positive integers $n$, and let $\{f_n\}$ be a sequence of real-valued functions such that the series $\displaystyle\sum f_n(x)$ converges uniformly on $E$. Then $\displaystyle\sum f_n(x)g_n(x)$converges uniformly on $E$.

1. Is the additional requirement of continuity for a sequence of functions in Wikipedia the only difference? If not, what else?
2. Is this continuity unnecessary and can be ignored as in Marsden's? If yes, is Marsden's a more general version? Or do you have a different one?

Thanks and regards!

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Doesn't Marsden specify what $f_n$ is? The assumption of continuity in Wiki's version is superfluous, as well as the assumption that $f_n$ be real-valued, they could be complex-valued as well. I see no other difference except that assumption. There are many of results of this flavor, see e.g. Dirichlet's criterion and Dedekind's criterion for two of them. –  t.b. Aug 23 '11 at 16:34
@Theo: Thanks! Does Marsden? I cannot find it. –  Tim Aug 23 '11 at 16:40
I'm asking. I don't know. Probably he has a universal assumption what $f$ and $f_n$ is supposed to mean. Google doesn't let me look and I never held that book in my hands. –  t.b. Aug 23 '11 at 16:42
Fix $\varepsilon>0$. The uniform convergence of the series implies that there exists $n$ such that $$\left|\sum_{k=n}^m f_k(x)\right|<\varepsilon \ \ \ \text{ for all } x\in A, m>n.$$ The key to the proof is "summation by parts": we have $$\sum_{k=n}^m\phi_k(x)f_k(x)=\phi_m(x)\sum_{k=n}^mf_k(x)-\sum_{k=n}^m(\phi_{k+1}(x)-\phi_k(x))\sum_{j=n}^kf_j(x).$$ Then