We denote $\mathcal { D }$ the open unit disk of $\mathbb { C } : \mathcal { D } = \{ z \in \mathbb { C } ; | z | < 1 \}$. The sequence $(a_n)$ is defined by $$\left\{ \begin{array} { l } a _ { 0 } = 1 \\ a _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { ( - 1 ) ^ { n - k } } { ( n - k ) ! } H _ { k } \left( \frac { n + k } { 2 } - 1 \right) \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$ Show that, for all $z \in \mathcal { D }$, the power series $\sum _ { n \geqslant 0 } a _ { n } z ^ { n }$ converges.
We denote $\mathcal { D }$ the open unit disk of $\mathbb { C } : \mathcal { D } = \{ z \in \mathbb { C } ; | z | < 1 \}$. The sequence $(a_n)$ is defined by
$$\left\{ \begin{array} { l } a _ { 0 } = 1 \\ a _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { ( - 1 ) ^ { n - k } } { ( n - k ) ! } H _ { k } \left( \frac { n + k } { 2 } - 1 \right) \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Show that, for all $z \in \mathcal { D }$, the power series $\sum _ { n \geqslant 0 } a _ { n } z ^ { n }$ converges.