Recall that $x$ is a fixed element of $]0;1[$. Show that:
$$\int _ { 0 } ^ { 1 } \frac { t ^ { x - 1 } } { 1 + t } \mathrm {~d} t = \sum _ { k = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { k } } { k + x }$$

Recall that $x$ is a fixed element of $]0;1[$. Establish the identity
$$\int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t } \mathrm {~d} t = \sum _ { n = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { n } } { n + x } + \sum _ { n = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { n } } { n + 1 - x }$$

Recall that $x$ is a fixed element of $]0;1[$. Deduce that:
$$\frac { \pi } { \sin ( \pi x ) } = \frac { 1 } { x } - \sum _ { n = 1 } ^ { + \infty } \frac { 2 ( - 1 ) ^ { n } x } { n ^ { 2 } - x ^ { 2 } }$$

Let $n$ be a non-zero natural integer. For an integer $k$ at most $n$, we denote by $s(n,k)$ the number of permutations of $\mathfrak{S}_n$ such that $\omega(\sigma) = k$.
Establish that, for any real $x$, $$\prod_{i=0}^{n-1}(x+i) = \sum_{k=1}^{n} s(n,k) x^k.$$

Recall that $x$ is a fixed element of $]0;1[$. Finally deduce that:
$$\forall y \in ] 0 ; \pi \left[ , \quad \sum _ { n = 1 } ^ { + \infty } \frac { 2 ( - 1 ) ^ { n } y \sin ( y ) } { y ^ { 2 } - n ^ { 2 } \pi ^ { 2 } } = 1 - \frac { \sin ( y ) } { y } . \right.$$

Justify that the series $\sum_{k \geqslant 2} \frac{\ln(k)}{k(k-1)}$ converges.

Conclude that $\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p} \underset{n \rightarrow +\infty}{=} \ln(n) + O(1)$.

We set, for all real $t \geqslant 2$, $$R(t) = \sum_{\substack{p \leqslant t \\ p \text{ prime}}} \frac{\ln(p)}{p} - \ln(t)$$ Show, using the result from question 16, that $$\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{1}{p} = 1 + \ln_2(n) - \ln_2(2) + \frac{R(n)}{\ln(n)} + \int_2^n \frac{R(t)}{t(\ln(t))^2} \, \mathrm{d}t$$

We set, for all real $t \geqslant 2$, $$R(t) = \sum_{\substack{p \leqslant t \\ p \text{ prime}}} \frac{\ln(p)}{p} - \ln(t)$$ Justify that the function $t \mapsto \frac{R(t)}{t(\ln(t))^2}$ is integrable on $[2, +\infty[$.

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

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?

Let $\ell \geq 0$ be an integer. Show that there exists a polynomial $P_\ell \in \mathbf{Q}[x]$ of degree $< r(\ell+1)$ satisfying $$\sum_{n=0}^{\infty} n(n-1)\cdots(n-\ell+1)\, u_{n-\ell}\, x^n = \frac{P_\ell(x)}{\left(1 - s_1 x - \cdots - s_r x^r\right)^{\ell+1}}.$$

Define two sequences $(w_{n,k})_{n,k \geq 0}$ and $(w_n(k))_{n,k \geq 0}$ by the formulas $$w_{n,k} = n! \sum_{i=0}^{n-k} \frac{u_i}{i!} \quad \text{and} \quad \sum_{n=0}^{\infty} w_n(k) x^n = \left(1 - s_1 x - \cdots - s_r x^r\right)^k \sum_{n=0}^{\infty} w_{n,k}\, x^n.$$ Show the equality $w_n(k) = v_n(k)$ for all $n$ and $k$ such that $n \geq kr$.

Deduce that $k!$ divides $D^n v_n(k)$ for all $n$ and $k$ such that $n \geq kr$.

Show the equality $$v_n(k) = \sum_{i=1}^{r} b_i e^{a_i} \int_{a_i}^{\infty} e^{-t} t^{n-kr} (t - a_1)^k \cdots (t - a_r)^k\, dt.$$

Show that if $v_n(k)$ is non-zero and $n \geq kr$, then $$k! \leq |D^n v_n(k)| \leq c_2 (AD)^n C^k.$$

Deduce that there exists an integer $k_0$ such that $$v_n(k) = 0 \text{ for all } k \geq k_0 \text{ and } kr \leq n \leq 10kr.$$

Conclude that $v_n(k_0) = 0$ for all $n \geq k_0 r$.

A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$.
Show that if $f$ is a function $E$, then the numerical series $\sum_{n=0}^{\infty} \frac{b_n}{n!} \alpha^n$ converges for every real number $\alpha$. We denote its value by $f(\alpha)$.

A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$.
Let $f$ be a function $E$ that is not a polynomial. Show that there exists $R > 0$ such that the numerical series $\sum_{n=0}^{\infty} b_n \alpha^n$ diverges for every real number $\alpha$ with $|\alpha| > R$.

Show that a function $P : \mathbb{Z} \rightarrow \mathbb{C}$ is quasi-polynomial if and only if there exist an integer $m \in \mathbb{N}^*$ and $m$ polynomials $P_0, \ldots, P_{m-1}$ with complex coefficients such that for all $j \in \{0, \ldots, m-1\}$ and for all $n \in \mathbb{Z}$ congruent to $j$ modulo $m$, we have $P(n) = P_j(n)$.

Let $\omega$ be a root of unity and $p \in \mathbb{N}^*$. Let $\sum_{n=0}^{+\infty} R(n) x^n$ denote the power series expansion of $\frac{1}{(1 - \omega x)^p}$. Show that $R$ is a quasi-polynomial function then determine its degree and its leading coefficient.

Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\},$$ then we define the power series $F(x) = \sum_{n=0}^{\infty} P(n) x^n$.
Show that the radius of convergence of $F$ is greater than or equal to 1.