Let $S(z) = \sum_\limits{n \in \mathbb{N}} a_n z^n$ be a power serie with radius of convergence 1, such that $$S(x) \sim_{x \to 1^-} \frac{1}{1-x}.$$ I'm trying to show that, for every polynomial $P$, $$\lim_{x\to 1^-} (1-x) \sum_\limits{n \in \mathbb{N}} a_n x^n P(x^n) = \int_0^1 P(t) dt.$$ For now, i have no idea how to proceed... Any hint would be appreciated.
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It might make sense to note that the coefficients $a_n$ of $S$ are all $1$. – K.defaoite Dec 17 '20 at 14:29
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Hint : Note that if $P(x)=\sum_{i=0}^s b_i x^i$, then $$(1-x) \sum_\limits{n \in \mathbb{N}} a_n x^n P(x^n)=\sum_{i=0}^s b_i\frac{1-x}{1-x^{1+i}}(1-x^{1+i})S(x^{1+i})$$ – Kelenner Dec 17 '20 at 14:44
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@K.defaoite No, for example if $f(x) = 1+x+\frac1{1-x}$ then $f(x)\sim\frac1{1-x}$. – David C. Ullrich Dec 17 '20 at 14:51