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 $n \in \mathbb{N}$. We denote $\widetilde { a } _ { n } = \frac { A _ { n } } { n + 1 }$ where $A_n = \sum_{k=0}^n a_k$. Conclude that there exists a real $\nu > 0$ such that from a certain rank onwards we have $\widetilde { a } _ { n } \geqslant \nu$.
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 $n \in \mathbb{N}$. We denote $\widetilde { a } _ { n } = \frac { A _ { n } } { n + 1 }$ where $A_n = \sum_{k=0}^n a_k$.
Conclude that there exists a real $\nu > 0$ such that from a certain rank onwards we have $\widetilde { a } _ { n } \geqslant \nu$.