We set for every integer $n \geqslant 0$, $$B _ { n } = \sum _ { k = 0 } ^ { n } S ( n , k )$$ where $S(n,k)$ denotes the number of partitions of $\llbracket 1, n \rrbracket$ into $k$ parts. Let $R$ be the radius of convergence of the power series $\sum _ { n \geqslant 0 } \frac { B _ { n } } { n ! } z ^ { n }$, and for $x \in ] - R , R [$, set $f ( x ) = \sum _ { n = 0 } ^ { + \infty } \frac { B _ { n } } { n ! } x ^ { n }$. Show that for all $x \in ] - R , R [ , f ^ { \prime } ( x ) = \mathrm { e } ^ { x } f ( x )$.
We set for every integer $n \geqslant 0$,
$$B _ { n } = \sum _ { k = 0 } ^ { n } S ( n , k )$$
where $S(n,k)$ denotes the number of partitions of $\llbracket 1, n \rrbracket$ into $k$ parts. Let $R$ be the radius of convergence of the power series $\sum _ { n \geqslant 0 } \frac { B _ { n } } { n ! } z ^ { n }$, and for $x \in ] - R , R [$, set $f ( x ) = \sum _ { n = 0 } ^ { + \infty } \frac { B _ { n } } { n ! } x ^ { n }$.
Show that for all $x \in ] - R , R [ , f ^ { \prime } ( x ) = \mathrm { e } ^ { x } f ( x )$.