Board index Computation of Series A Limits' question

## A Limits' question

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### A Limits' question

Fri Aug 26, 2016 8:25 am

Posts: 105
Define $A\left( x \right) = \sum\limits_{k = 1}^\infty {{a_k}{x^k}} ,{A_n}\left( x \right) = \sum\limits_{k = 1}^n {{a_k}{x^k}} ,B\left( x \right) = \sum\limits_{k = 1}^\infty {{b_k}{x^k}} ,{B_n}\left( x \right) = \sum\limits_{k = 1}^n {{b_k}{x^k}}$. When $x=1$, $A(1)$ and $B(1)$ are convergent, and we assume that the following identity is convergent
$\sum\limits_{n = 1}^\infty {\frac{{A\left( 1 \right){B_n}\left( 1 \right) - B\left( 1 \right){A_n}\left( 1 \right)}}{n}} = c,$
where $c$ is a constant. Then, the limit is whether or not holds?
$\mathop {\lim }\limits_{x \to 1} \left\{ {A\left( x \right)\sum\limits_{n = 1}^\infty {\frac{{{B_n}\left( 1 \right)}}{n}{x^n}} - B\left( x \right)\sum\limits_{n = 1}^\infty {\frac{{{A_n}\left( 1 \right)}}{n}{x^n}} } \right\}\mathop = \limits^? \sum\limits_{n = 1}^\infty {\frac{{A\left( 1 \right){B_n}\left( 1 \right) - B\left( 1 \right){A_n}\left( 1 \right)}}{n}} = c$