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Consider a convergent sum of the form $$ A(x) = \sum_{n=0}^\infty c\, a_n(x)\, x^n. $$ This sum is also bounded as $x \rightarrow \infty$, for example, it is a Taylor series for $A(x)$ at $0$.

I am interested in the product of $A(x)$ with $B(x)$, where $B(x)$ is bounded $\forall\, x$, $$ B(x)A(x) = B(x)\sum_{n=0}^\infty c_n\, a_n(x)\, x^n. $$ I want to take $B(x)$ into the sum to get $$ B(x)A(x) = \sum_{n=0}^\infty c_n\, B(x)\, a_n(x)\, x^n. $$ Under what conditions is this sum also guaranteed to converge, particularly as $x \rightarrow \infty$?

Note that I know the product $B(x)\, a_n(x)\, x^n$ is bounded $\forall\,x$.

I found this question and this question which seem similar. Apologies if this is a duplicate, but I'm not sure if I can apply their answers to my situation. Can I simply use the fact that $c \sum_n a_n = \sum_n c\, a_n$ for bounded $c$? Or are there any special considerations, due to the fact that $B$ depends on $x$ which also occurs in the sum?

Also, does anything change if $B(x)$ is complex?

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For any $x$ you can always take $B(x)$ inside the sum. This does follow from the fact that $c \sum a_n =\sum ca_n$. It is not clear what you mean by 'particularly as $ x \to \infty$'

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