Let $f$ be the function defined on $\mathbb { R } _ { + } ^ { * }$ by $f ( x ) = \frac { 1 } { ( 1 - \alpha ) x ^ { \alpha - 1 } }$. By applying to $f$ the Taylor formula with integral remainder at order 2, show that for every $k \in \mathbb { N } ^ { * } , f ( k + 1 ) - f ( k ) = \frac { 1 } { k ^ { \alpha } } - \frac { \alpha } { 2 } \frac { 1 } { k ^ { \alpha + 1 } } + A _ { k }$ where $A _ { k }$ is a real number satisfying $0 \leqslant A _ { k } \leqslant \frac { \alpha ( \alpha + 1 ) } { 2 k ^ { \alpha + 2 } }$.
Let $f$ be the function defined on $\mathbb { R } _ { + } ^ { * }$ by $f ( x ) = \frac { 1 } { ( 1 - \alpha ) x ^ { \alpha - 1 } }$. By applying to $f$ the Taylor formula with integral remainder at order 2, show that for every $k \in \mathbb { N } ^ { * } , f ( k + 1 ) - f ( k ) = \frac { 1 } { k ^ { \alpha } } - \frac { \alpha } { 2 } \frac { 1 } { k ^ { \alpha + 1 } } + A _ { k }$ where $A _ { k }$ is a real number satisfying $0 \leqslant A _ { k } \leqslant \frac { \alpha ( \alpha + 1 ) } { 2 k ^ { \alpha + 2 } }$.