Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. Let $P$ be a polynomial with real coefficients. Show that $$( 1 - x ) \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } P \left( x ^ { n } \right) \underset { \substack { x \rightarrow 1 \\ x < 1 } } { \longrightarrow } \int _ { 0 } ^ { 1 } P ( t ) \mathrm { d } t$$ We will first consider the special case $P ( x ) = x ^ { k }$, where $k \in \mathbb { N }$.
Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$.
Let $P$ be a polynomial with real coefficients. Show that
$$( 1 - x ) \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } P \left( x ^ { n } \right) \underset { \substack { x \rightarrow 1 \\ x < 1 } } { \longrightarrow } \int _ { 0 } ^ { 1 } P ( t ) \mathrm { d } t$$
We will first consider the special case $P ( x ) = x ^ { k }$, where $k \in \mathbb { N }$.