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Papers (176)
2025
centrale-maths1__official 40 centrale-maths2__official 36 mines-ponts-maths1__mp 17 mines-ponts-maths1__pc 21 mines-ponts-maths1__psi 21 mines-ponts-maths2__mp 28 mines-ponts-maths2__pc 23 mines-ponts-maths2__psi 25 polytechnique-maths-a__mp 35 polytechnique-maths__fui 9 polytechnique-maths__pc 27 x-ens-maths-a__fui 10 x-ens-maths-a__mp 18 x-ens-maths-b__mp 6 x-ens-maths-c__mp 6 x-ens-maths-d__mp 31 x-ens-maths__pc 27 x-ens-maths__psi 30
2024
centrale-maths1__official 21 centrale-maths2__official 28 geipi-polytech__maths 9 mines-ponts-maths1__mp 23 mines-ponts-maths1__psi 9 mines-ponts-maths2__mp 14 mines-ponts-maths2__pc 19 mines-ponts-maths2__psi 20 polytechnique-maths-a__mp 42 polytechnique-maths-b__mp 27 x-ens-maths-a__mp 43 x-ens-maths-b__mp 29 x-ens-maths-c__mp 22 x-ens-maths-d__mp 41 x-ens-maths__pc 20 x-ens-maths__psi 23
2023
centrale-maths1__official 37 centrale-maths2__official 32 e3a-polytech-maths__mp 4 mines-ponts-maths1__mp 14 mines-ponts-maths1__pc 21 mines-ponts-maths1__psi 21 mines-ponts-maths2__mp 21 mines-ponts-maths2__pc 13 mines-ponts-maths2__psi 22 polytechnique-maths__fui 3 x-ens-maths-a__mp 24 x-ens-maths-b__mp 10 x-ens-maths-c__mp 10 x-ens-maths-d__mp 10 x-ens-maths__pc 22
2022
centrale-maths1__mp 22 centrale-maths1__pc 33 centrale-maths1__psi 42 centrale-maths2__mp 26 centrale-maths2__pc 37 centrale-maths2__psi 40 mines-ponts-maths1__mp 26 mines-ponts-maths1__pc 20 mines-ponts-maths1__psi 23 mines-ponts-maths2__mp 22 mines-ponts-maths2__pc 9 mines-ponts-maths2__psi 18 x-ens-maths-a__mp 8 x-ens-maths-b__mp 19 x-ens-maths-c__mp 17 x-ens-maths-d__mp 47 x-ens-maths1__mp 13 x-ens-maths2__mp 26 x-ens-maths__pc 7 x-ens-maths__pc_cpge 14 x-ens-maths__psi 22 x-ens-maths__psi_cpge 26
2021
centrale-maths1__mp 34 centrale-maths1__pc 36 centrale-maths1__psi 28 centrale-maths2__mp 21 centrale-maths2__pc 38 centrale-maths2__psi 28 x-ens-maths2__mp 35 x-ens-maths__pc 29
2020
centrale-maths1__mp 42 centrale-maths1__pc 36 centrale-maths1__psi 38 centrale-maths2__mp 2 centrale-maths2__pc 35 centrale-maths2__psi 39 mines-ponts-maths1__mp_cpge 22 mines-ponts-maths2__mp_cpge 19 x-ens-maths-a__mp_cpge 10 x-ens-maths-b__mp_cpge 19 x-ens-maths-c__mp 10 x-ens-maths-d__mp 13 x-ens-maths1__mp 13 x-ens-maths2__mp 20 x-ens-maths__pc 6
2019
centrale-maths1__mp 37 centrale-maths1__pc 40 centrale-maths1__psi 38 centrale-maths2__mp 37 centrale-maths2__pc 39 centrale-maths2__psi 46 x-ens-maths1__mp 24 x-ens-maths__pc 18 x-ens-maths__psi 9
2018
centrale-maths1__mp 21 centrale-maths1__pc 31 centrale-maths1__psi 39 centrale-maths2__mp 23 centrale-maths2__pc 35 centrale-maths2__psi 30 x-ens-maths1__mp 18 x-ens-maths2__mp 13 x-ens-maths__pc 17 x-ens-maths__psi 20
2017
centrale-maths1__mp 45 centrale-maths1__pc 22 centrale-maths1__psi 17 centrale-maths2__mp 30 centrale-maths2__pc 28 centrale-maths2__psi 44 x-ens-maths1__mp 24 x-ens-maths2__mp 7 x-ens-maths__pc 17 x-ens-maths__psi 19
2016
centrale-maths1__mp 41 centrale-maths1__pc 31 centrale-maths1__psi 33 centrale-maths2__mp 25 centrale-maths2__pc 42 centrale-maths2__psi 17 x-ens-maths1__mp 10 x-ens-maths2__mp 32 x-ens-maths__pc 1 x-ens-maths__psi 20
2015
centrale-maths1__mp 18 centrale-maths1__pc 11 centrale-maths1__psi 42 centrale-maths2__mp 44 centrale-maths2__pc 1 centrale-maths2__psi 14 x-ens-maths1__mp 16 x-ens-maths2__mp 19 x-ens-maths__pc 30 x-ens-maths__psi 20
2014
centrale-maths1__mp 28 centrale-maths1__pc 26 centrale-maths1__psi 36 centrale-maths2__mp 24 centrale-maths2__pc 23 centrale-maths2__psi 29 x-ens-maths2__mp 13
2013
centrale-maths1__mp 3 centrale-maths1__pc 45 centrale-maths1__psi 20 centrale-maths2__mp 32 centrale-maths2__pc 50 centrale-maths2__psi 32 x-ens-maths1__mp 14 x-ens-maths2__mp 10 x-ens-maths__pc 22 x-ens-maths__psi 9
2012
centrale-maths1__pc 23 centrale-maths1__psi 20 centrale-maths2__mp 27 centrale-maths2__psi 20
2011
centrale-maths1__mp 27 centrale-maths1__pc 15 centrale-maths1__psi 21 centrale-maths2__mp 29 centrale-maths2__pc 8 centrale-maths2__psi 28
2010
centrale-maths1__mp 7 centrale-maths1__pc 23 centrale-maths1__psi 9 centrale-maths2__mp 10 centrale-maths2__pc 36 centrale-maths2__psi 27
2024 centrale-maths1__official

21 maths questions

Q1 Indefinite & Definite Integrals Integral Inequalities and Limit of Integral Sequences View
Let $f : [ a , b ] \rightarrow \mathbb { R }$ be a piecewise continuous function taking values in an interval $J$. Let $\varphi$ be a continuous and convex function on $J$. Prove that
$$\varphi \left( \frac { 1 } { b - a } \int _ { a } ^ { b } f ( t ) \mathrm { d } t \right) \leqslant \frac { 1 } { b - a } \int _ { a } ^ { b } \varphi \circ f ( t ) \mathrm { d } t .$$
You may use Riemann sums.
Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Determine the limit of $g ( x )$ as $x$ tends to 0.
Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Determine the limit of $g ( x )$ as $x$ tends to $+ \infty$. Denoting by $\mathbb { 1 } _ { [ 0 , x ] }$ the indicator function of $[ 0 , x ]$, you may note that $g ( x ) = \int _ { 0 } ^ { + \infty } \frac { 1 } { x } t f ( t ) \mathbb { 1 } _ { [ 0 , x ] } ( t ) \mathrm { d } t$.
Q4 Integration with Partial Fractions Derive a Reduction/Recurrence Formula via Integration by Parts View
Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Deduce that the integral $\int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x$ converges and that
$$\int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x = \int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x .$$
You may use integration by parts.
Q5 Integration with Partial Fractions Integral Inequalities and Limit of Integral Sequences View
Let $f$ be a piecewise continuous function, strictly positive, integrable on $\mathbb { R } _ { + }$. Prove that, for all $x > 0$,
$$\exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right) \leqslant \frac { \mathrm { e } } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t$$
You may note that $\ln ( f ( t ) ) = \ln ( t f ( t ) ) - \ln ( t )$.
Q6 Integration with Partial Fractions Integral Inequalities and Limit of Integral Sequences View
Let $f$ be a piecewise continuous function, strictly positive, integrable on $\mathbb { R } _ { + }$. Deduce that $x \mapsto \exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right)$ is integrable on $\mathbb { R } _ { + } ^ { * }$ and that
$$\int _ { 0 } ^ { + \infty } \exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right) \mathrm { d } x \leqslant \mathrm { e } \int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x$$
Q7 Indefinite & Definite Integrals Maximizing or Optimizing a Definite Integral View
We assume that $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence of strictly positive real numbers. We denote by $f$ the step function which, for all $k \in \mathbb { N } ^ { * }$, equals $a _ { k }$ on the interval $[ k - 1 , k [$.
Let $k$ be in $\mathbb { N } ^ { * }$. Prove that the function $v _ { k }$ defined on $[ k - 1 , k ]$ by
$$\left\{ \begin{array} { l } v _ { k } ( x ) = \frac { 1 } { x } \sum _ { i = 1 } ^ { k - 1 } \ln \left( a _ { i } \right) + \frac { 1 } { x } ( x - k + 1 ) \ln \left( a _ { k } \right) \quad \text { if } k \geqslant 2 \\ v _ { 1 } ( x ) = \ln \left( a _ { 1 } \right) \end{array} \right.$$
is minimal for $x = k$.
Q8 Integration with Partial Fractions Bound or Estimate a Parametric Integral View
We assume that $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence of strictly positive real numbers. We denote by $f$ the step function which, for all $k \in \mathbb { N } ^ { * }$, equals $a _ { k }$ on the interval $[ k - 1 , k [$.
Prove that, for all $k$ in $\mathbb { N } ^ { * }$,
$$\int _ { k - 1 } ^ { k } \exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right) \mathrm { d } x \geqslant \exp \left( \frac { 1 } { k } \sum _ { i = 1 } ^ { k } \ln \left( a _ { i } \right) \right)$$
You may use the previous question.
Q9 Integration with Partial Fractions Proof of Inequalities Involving Series or Sequence Terms View
We assume that $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence of strictly positive real numbers. We denote by $f$ the step function which, for all $k \in \mathbb { N } ^ { * }$, equals $a _ { k }$ on the interval $[ k - 1 , k [$.
Deduce Carleman's inequality in the case where $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence.
Q10 Integration with Partial Fractions Proof of Inequalities Involving Series or Sequence Terms View
Explain how one can remove the decreasing hypothesis in Carleman's inequality.
Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
We admit that $f$ and $g _ { s }$ are of class $\mathcal { C } ^ { 1 }$ on $U _ { n }$. Give the expression of their gradient at a point $x = \left( x _ { 1 } , \ldots , x _ { n } \right)$ of $U _ { n }$.
Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
Prove that the restriction of $f$ to $X _ { s }$ admits a maximum on $X _ { s }$ and that this maximum is in fact attained on $X _ { s } \cap U _ { n }$.
You may verify that $f$ is strictly positive at certain points of $X _ { s } \cap U _ { n }$.
Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
We denote by $a = \left( a _ { 1 } , \ldots , a _ { n } \right)$ an element of $X _ { s } \cap U _ { n }$ at which the restriction of $f$ to $X _ { s }$ attains its maximum.
Prove that there exists a real number $\lambda > 0$ such that, for all $k$ in $\llbracket 1 , n \rrbracket , a _ { k } = \frac { f ( a ) } { \lambda }$.
Q14 Integration with Partial Fractions Proof of Inequalities Involving Series or Sequence Terms View
Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
Prove that, for all $\left( x _ { 1 } , \ldots , x _ { n } \right) \in U _ { n } \cap X _ { s } , \left( \prod _ { i = 1 } ^ { n } x _ { i } \right) ^ { 1 / n } \leqslant \frac { 1 } { n } \sum _ { i = 1 } ^ { n } x _ { i }$ and deduce the arithmetic-geometric inequality
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \left( \mathbb { R } _ { + } \right) ^ { n } , \quad \left( \prod _ { i = 1 } ^ { n } x _ { i } \right) ^ { 1 / n } \leqslant \frac { 1 } { n } \sum _ { i = 1 } ^ { n } x _ { i }$$
Q15 Stationary points and optimisation Partial derivatives and multivariable differentiation View
Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $h _ { n }$ the map from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad h _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \cdots + x _ { n } - 1 .$$
We admit that $F _ { n }$ and $h _ { n }$ are both of class $\mathcal { C } ^ { 1 }$ on $U _ { n }$. We denote by $H _ { n }$ the set $H _ { n } = \left\{ \left( x _ { 1 } , \ldots , x _ { n } \right) \in \mathbb { R } ^ { n } \mid x _ { 1 } + \cdots + x _ { n } = 1 \right\}$.
Determine the gradient of $F _ { n }$ and the gradient of $h _ { n }$ at every point of $U _ { n }$.
Q16 Stationary points and optimisation Partial derivatives and multivariable differentiation View
Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $H _ { n }$ the set $H _ { n } = \left\{ \left( x _ { 1 } , \ldots , x _ { n } \right) \in \mathbb { R } ^ { n } \mid x _ { 1 } + \cdots + x _ { n } = 1 \right\}$.
Prove that the restriction of $F _ { n }$ to $\overline { U _ { n } } \cap H _ { n }$ admits a maximum.
Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $h _ { n }$ the map from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad h _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \cdots + x _ { n } - 1 .$$
We denote by $H _ { n }$ the set $H _ { n } = \left\{ \left( x _ { 1 } , \ldots , x _ { n } \right) \in \mathbb { R } ^ { n } \mid x _ { 1 } + \cdots + x _ { n } = 1 \right\}$. We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$.
Prove that there exists a real number $\lambda > 0$ such that
$$\left\{ \begin{aligned} \gamma _ { 1 } + \frac { \gamma _ { 2 } } { 2 } + \cdots + \frac { \gamma _ { n } } { n } & = \lambda a _ { 1 } \\ \frac { \gamma _ { 2 } } { 2 } + \cdots + \frac { \gamma _ { n } } { n } & = \lambda a _ { 2 } \\ & \vdots \\ \frac { \gamma _ { n } } { n } & = \lambda a _ { n } \\ a _ { 1 } + a _ { 2 } + \cdots + a _ { n } & = 1 \end{aligned} \right.$$
Let $\varphi$ be the function defined by
$$\forall t \in ] - 1,1 \left[ \backslash \{ 0 \} , \quad \varphi ( t ) = ( 1 - t ) ^ { 1 - 1 / t } \right.$$
We define the sequence $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ by
$$\left\{ \begin{array} { l } b _ { 0 } = - 1 \\ \forall n \in \mathbb { N } ^ { * } , \quad b _ { n } = - \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \frac { 1 } { k + 1 } b _ { n - k } \end{array} \right.$$
Conclude that
$$\forall t \in ] - 1,1 \left[ , \quad \varphi ( t ) = \mathrm { e } \left( 1 - \sum _ { k = 1 } ^ { + \infty } b _ { k } t ^ { k } \right) \right.$$
Let $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ and $\left( c _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ be two sequences of strictly positive real numbers. Prove that
$$\sum _ { n = 1 } ^ { + \infty } \left( \prod _ { k = 1 } ^ { n } a _ { k } \right) ^ { 1 / n } \leqslant \sum _ { k = 1 } ^ { + \infty } c _ { k } a _ { k } \sum _ { n = k } ^ { + \infty } \frac { 1 } { n } \left( \prod _ { i = 1 } ^ { n } c _ { i } \right) ^ { - 1 / n }$$
Q28 Sequences and Series Recurrence Relations and Sequence Properties View
Let $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ and $\left( c _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ be two sequences of strictly positive real numbers. We define the sequence $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ by
$$\left\{ \begin{array} { l } b _ { 0 } = - 1 \\ \forall n \in \mathbb { N } ^ { * } , \quad b _ { n } = - \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \frac { 1 } { k + 1 } b _ { n - k } \end{array} \right.$$
By considering $c _ { n } = \frac { ( n + 1 ) ^ { n } } { n ^ { n - 1 } }$, deduce the Carleman-Yang inequality:
$$\sum _ { n = 1 } ^ { + \infty } \left( \prod _ { k = 1 } ^ { n } a _ { k } \right) ^ { 1 / n } \leqslant \mathrm { e } \sum _ { n = 1 } ^ { + \infty } \left( 1 - \sum _ { k = 1 } ^ { + \infty } \frac { b _ { k } } { ( n + 1 ) ^ { k } } \right) a _ { n }$$
We define the sequence $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ by
$$\left\{ \begin{array} { l } b _ { 0 } = - 1 \\ \forall n \in \mathbb { N } ^ { * } , \quad b _ { n } = - \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \frac { 1 } { k + 1 } b _ { n - k } \end{array} \right.$$
The Carleman-Yang inequality states:
$$\sum _ { n = 1 } ^ { + \infty } \left( \prod _ { k = 1 } ^ { n } a _ { k } \right) ^ { 1 / n } \leqslant \mathrm { e } \sum _ { n = 1 } ^ { + \infty } \left( 1 - \sum _ { k = 1 } ^ { + \infty } \frac { b _ { k } } { ( n + 1 ) ^ { k } } \right) a _ { n }$$
Prove that, for all $n$ in $\mathbb { N } ^ { * } , b _ { n } \geqslant 0$. In what way is the previous inequality a refinement of Carleman's inequality?