# Showing $\sum c_n = \sum a_n \sum b_n$ if $\sum c_n x^n=\sum a_n x^n \sum b_n x^n$ on $[0, 1)$ [closed]

Given that $$\sum c_n x^n=\sum a_n x^n \sum b_n x^n$$ on $$[0, 1)$$ and all partial sums $$\sum c_n x^n, \sum a_n x^n, \sum b_n x^n$$ converges uniformly on $$[0, 1]$$, is it true that $$\sum c_n = \sum a_n \sum b_n$$?

## closed as off-topic by Carl Mummert, Nosrati, Scientifica, Delta-u, MicahSep 26 '18 at 16:38

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Yes. In the conditions you mentioned the sum $$\sum c_n$$ is equal to the Abel sum which is $$\lim_{x\to{1^{-}}}\sum c_n x^n$$, and same can be said about the other two sums. So then:
$$\sum c_n=\lim_{x\to{1^{-}}}\sum c_n x^n=\lim_{x\to{1^{-}}}\sum a_n x^n\sum b_n x^n=\lim_{x\to{1^{-}}}\sum a_n x^n\lim_{x\to{1^{-}}}\sum b_n x^n=$$
$$=\sum a_n\sum b_n$$