We denote $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$. We denote $g$ the function defined on $[ - 1 , + \infty [$ by $$g ( x ) = \sum _ { n = 2 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$$ Show that for every integer $n$ and for every $x$ in $] - 1,1 [$ $$\left| g ( x ) - \sum _ { k = 0 } ^ { n } \frac { g ^ { ( k ) } ( 0 ) } { k ! } x ^ { k } \right| \leqslant \zeta ( 2 ) | x | ^ { n + 1 }$$ Show that $g$ is expandable as a power series on $] - 1,1 [$.
We denote $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$. We denote $g$ the function defined on $[ - 1 , + \infty [$ by
$$g ( x ) = \sum _ { n = 2 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$$
Show that for every integer $n$ and for every $x$ in $] - 1,1 [$
$$\left| g ( x ) - \sum _ { k = 0 } ^ { n } \frac { g ^ { ( k ) } ( 0 ) } { k ! } x ^ { k } \right| \leqslant \zeta ( 2 ) | x | ^ { n + 1 }$$
Show that $g$ is expandable as a power series on $] - 1,1 [$.