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.

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 $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
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
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \left( \mathbb { R } _ { + } \right) ^ { n } , \quad \left( \prod _ { i = 1 } ^ { n } x _ { i } \right) ^ { 1 / n } \leqslant \frac { 1 } { n } \sum _ { i = 1 } ^ { n } x _ { i }$$

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 $M _ { n }$ 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 $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$. We assume by contradiction that $\lambda > \mathrm { e }$, where $\lambda$ is the real number from Q17, and $\omega_k$ is as defined in Q18b.
Prove that $\omega _ { 1 } \leqslant \frac { 1 } { \mathrm { e } }$ and that, for all $k$ in $\llbracket 1 , n \rrbracket , \omega _ { k } \leqslant \frac { k } { k + 1 }$.
You may prove, for $k \in \llbracket 1 , n - 1 \rrbracket$, that $\omega _ { k + 1 } ^ { k + 1 } = \frac { 1 } { \lambda } \omega _ { k } ^ { k } \left( 1 - \frac { \omega _ { k } } { k } \right) ^ { - k }$.

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 $M _ { n }$ 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 $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$. We assume by contradiction that $\lambda > \mathrm { e }$, where $\lambda$ is the real number from Q17, and $\omega_k$ is as defined in Q18b.
Reach a contradiction on $\omega _ { n }$. Deduce that, for all $n$ in $\mathbb { N } ^ { * }$, for all $\left( x _ { 1 } , \ldots , x _ { n } \right) \in \left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$ such that $x _ { 1 } + \cdots + x _ { n } = 1$,
$$\sum _ { k = 1 } ^ { n } \left( x _ { 1 } x _ { 2 } \cdots x _ { k } \right) ^ { 1 / k } \leqslant \mathrm { e }$$

Deduce Carleman's inequality:
$$\sum _ { n = 1 } ^ { + \infty } \left( \prod _ { k = 1 } ^ { n } a _ { k } \right) ^ { 1 / n } \leqslant \mathrm { e } \sum _ { n = 1 } ^ { + \infty } a _ { n }$$
for any sequence $\left( a _ { k } \right) _ { k \in \mathbb { N } ^ { * } }$ of strictly positive real numbers such that $\sum a _ { n }$ converges.

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
$$\sum _ { n = 1 } ^ { + \infty } \left( \prod _ { k = 1 } ^ { n } a _ { k } \right) ^ { 1 / n } \leqslant \sum _ { k = 1 } ^ { + \infty } c _ { k } a _ { k } \sum _ { n = k } ^ { + \infty } \frac { 1 } { n } \left( \prod _ { i = 1 } ^ { n } c _ { i } \right) ^ { - 1 / n }$$

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?

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(A) $a _ { n } < b _ { n }$ for all $n > 1$
(B) $a _ { n } > b _ { n }$ for all $n > 1$
(C) $a _ { n } = b _ { n }$ for infinitely many $n$
(D) None of the above

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(a) $a _ { n } < b _ { n }$ for all $n > 1$
(b) $a _ { n } > b _ { n }$ for all $n > 1$
(c) $a _ { n } = b _ { n }$ for infinitely many $n$
(d) none of the above.

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(a) $a _ { n } < b _ { n }$ for all $n > 1$
(b) $a _ { n } > b _ { n }$ for all $n > 1$
(c) $a _ { n } = b _ { n }$ for infinitely many $n$
(d) none of the above.

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(A) $a _ { n } < b _ { n }$ for all $n > 1$
(B) $a _ { n } > b _ { n }$ for all $n > 1$
(C) $a _ { n } = b _ { n }$ for infinitely many $n$
(D) None of the above

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(A) $a _ { n } < b _ { n }$ for all $n > 1$
(B) $a _ { n } > b _ { n }$ for all $n > 1$
(C) $a _ { n } = b _ { n }$ for infinitely many $n$
(D) None of the above

Define $S _ { n } = \frac { 1 } { 2 } \cdot \frac { 3 } { 4 } \cdots \cdot \frac { 2 n - 1 } { 2 n }$ where $n$ is a positive integer. Then
(A) $S _ { n } < \frac { 1 } { \sqrt { 4 n + 2 } }$ for some $n > 2$.
(B) $S _ { n } < \frac { 1 } { \sqrt { 2 n + 1 } }$ for all $n \geq 2$.
(C) $S _ { n } < \frac { 1 } { \sqrt { 2 n + 5 } }$ for all $n \geq 2$.
(D) $S _ { n } > \frac { 1 } { \sqrt { 4 n + 2 } }$ for all $n \geq 2$.

Let $$S _ { n } = \sum _ { k = 1 } ^ { n } \frac { n } { n ^ { 2 } + k n + k ^ { 2 } } \quad \text { and } \quad T _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { n } { n ^ { 2 } + k n + k ^ { 2 } } ,$$ for $n = 1,2,3 , \cdots$. Then,
(A) $\quad S _ { n } < \frac { \pi } { 3 \sqrt { 3 } }$
(B) $\quad S _ { n } > \frac { \pi } { 3 \sqrt { 3 } }$
(C) $T _ { n } < \frac { \pi } { 3 \sqrt { 3 } }$
(D) $T _ { n } > \frac { \pi } { 3 \sqrt { 3 } }$