Verify that
$$\begin{cases} I _ { n , k } = 0 & \text { if } n > k \text { or if } k - n \text { is odd } \\ I _ { n , k } = \frac { ( - 1 ) ^ { p } } { 2 ^ { n + 2 p } } \binom { n + 2 p } { n + p } & \text { if } k = n + 2 p \text { with } p \geqslant 0 \end{cases}$$

In this question, $f$ denotes the linear form defined by $\forall M \in \mathcal{M}_n(\mathbb{R}), f(M) = \sum_{j=1}^n \sum_{i=j}^n m_{i,j}$, and $A$ is the matrix such that $\forall M \in \mathcal{M}_n(\mathbb{R}), f(M) = \operatorname{Tr}(AM)$.
Show that $M_n = \sum_{k=1}^n \dfrac{1}{2\cos\dfrac{k\pi}{2n+1}}$.

Let $\theta \in \mathbb { R } \backslash 2 \pi \mathbb { Z }$.
Deduce that $$\sum _ { k = 1 } ^ { + \infty } \frac { \mathrm { e } ^ { \mathrm { i } k \theta } } { k } = \int _ { 0 } ^ { 1 } \frac { \mathrm { e } ^ { \mathrm { i } \theta } } { 1 - \mathrm { e } ^ { \mathrm { i } \theta } t } \mathrm { ~d} t$$
One may use the dominated convergence theorem.

Let $\theta \in \mathbb { R } \backslash 2 \pi \mathbb { Z }$.
Deduce that $$\sum _ { k = 1 } ^ { + \infty } \frac { \mathrm { e } ^ { \mathrm { i } k \theta } } { k } = - \frac { 1 } { 2 } \ln ( 2 - 2 \cos \theta ) + \mathrm { i } \arctan \left( \frac { \sin \theta } { 1 - \cos \theta } \right)$$

Let $\theta \in \mathbb { R } \backslash 2 \pi \mathbb { Z }$.
Show that, for all $\theta \in ] 0 , \pi [$, $$\sum _ { k = 1 } ^ { + \infty } \frac { \sin ( k \theta ) } { k } = \frac { \pi - \theta } { 2 }$$

Let $r : \mathbb { R } \rightarrow \mathbb { R }$, be a $2 \pi$-periodic, odd function, such that $\forall \theta \in ] 0 , \pi ] , r ( \theta ) = \frac { \pi - \theta } { 2 }$.
Deduce that $\sum _ { n = 0 } ^ { + \infty } \frac { 1 } { ( 2 n + 1 ) ^ { 2 } } = \frac { \pi ^ { 2 } } { 8 }$.

Show that $\prod _ { k = 1 } ^ { n - 1 } \sin \frac { k \pi } { 2 n } = \frac { \sqrt { n } } { 2 ^ { n - 1 } }$.

We denote $\mathcal{I} = \{(j, k) \in \mathbf{N}^{2} \mid j \in \mathbf{N} \text{ and } 0 \leq k < 2^{j}\}$; for $j \in \mathbf{N}$, $\mathcal{T}_{j} = \{k \in \mathbf{N} \mid 0 \leq k < 2^{j}\}$. For all $(j, k) \in \mathcal{I}$, $\theta_{j,k} : [0,1] \rightarrow [0,1]$ is defined by $$\theta_{j,k}(x) = \left\{ \begin{array}{l} 1 - |2^{j+1} x - 2k - 1| \quad \text{if } x \in [k 2^{-j}, (k+1) 2^{-j}] \\ 0 \text{ otherwise} \end{array} \right.$$ Calculate $\theta_{j,k}(\ell 2^{-j-1})$ for all $j \in \mathbf{N}$, $k \in \mathcal{T}_{j}$, $\ell \in \mathcal{T}_{j+1}$.

We denote $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ for $r \geqslant 2$, $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$, and $\varphi ( x ) = ( \psi ( 1 + x ) - \psi ( 1 ) ) ^ { 2 } + \left( \psi ^ { \prime } ( 1 ) - \psi ^ { \prime } ( 1 + x ) \right)$. We have $x B(x) = \varphi(x)$ for $x > 0$, and $S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \lim _ { x \rightarrow 0 ^ { + } } B ^ { ( r - 2 ) } ( x )$. We have also shown that $\psi ( 1 + x ) = \psi ( 1 ) + \sum _ { n = 1 } ^ { + \infty } ( - 1 ) ^ { n + 1 } \zeta ( n + 1 ) x ^ { n }$ for $x \in ] -1, 1[$.
Conclude that, for every integer $r \geqslant 3$, $$2 S _ { r } = r \zeta ( r + 1 ) - \sum _ { k = 1 } ^ { r - 2 } \zeta ( k + 1 ) \zeta ( r - k )$$

We set for every integer $n \geqslant 0$, $$B _ { n } = \sum _ { k = 0 } ^ { n } S ( n , k )$$ where $S(n,k)$ denotes the number of partitions of $\llbracket 1, n \rrbracket$ into $k$ parts. Let $R$ be the radius of convergence of the power series $\sum _ { n \geqslant 0 } \frac { B _ { n } } { n ! } z ^ { n }$, and for $x \in ] - R , R [$, set $f ( x ) = \sum _ { n = 0 } ^ { + \infty } \frac { B _ { n } } { n ! } x ^ { n }$. It has been shown that $f ^ { \prime } ( x ) = \mathrm { e } ^ { x } f ( x )$ for all $x \in ] - R , R [$.
Deduce an expression for the function $f$ on $] - R , R [$.

We fix $n \in \mathbb { N }$. We define the linear map: $$\begin{aligned} \Delta : \mathbb { R } [ X ] & \rightarrow \mathbb { R } [ X ] \\ P ( X ) & \mapsto P ( X + 1 ) - P ( X ) \end{aligned}$$ We define $H _ { 0 } ( X ) = 1$ and, for all $k \in \mathbb { N } ^ { * }$, $H _ { k } ( X ) = X ( X - 1 ) \cdots ( X - k + 1 )$, and $S(n,k)$ denotes the number of partitions of $\llbracket 1, n \rrbracket$ into $k$ parts.
Deduce that $U _ { n } ( p ) = \sum _ { k = 0 } ^ { n } \frac { S ( n , k ) } { k + 1 } H _ { k + 1 } ( p + 1 )$.

Let $f$ be the function defined by $$f(x) = \sum_{n=1}^{+\infty} \left(\frac{1}{n+x} - \frac{1}{n}\right)$$ Let $k \in \mathbb{N}^{*}$. Calculate $f(k)$.

Using the results of the previous questions, deduce an integral expression of $\zeta(k+1)$ for all $k \in \mathbb{N}^{*}$.

Show that $$\forall k \in \mathbb{N}^{*}, \quad \zeta(k+1) = \frac{1}{k!} \int_{0}^{+\infty} \frac{u^{k}}{\mathrm{e}^{u} - 1} \mathrm{~d}u$$

We denote by $n$ the integer part of $\frac{N}{2}$. For $j \in \mathbb{N}$, the polynomials $P_j$ are defined by $P_{j}(X) = \frac{1}{2^{j} j!} \frac{d^{j}}{dX^{j}}\left[(X^{2}-1)^{j}\right]$, and $g_j = \int_{-1}^{1} P_j(x)^2\,dx$.
Discuss, depending on the parity of $n$, the value of $a_{N}$. We will give its explicit value.

We denote by $n$ the integer part of $\frac{N}{2}$. For $j \in \mathbb{N}$, the polynomials $P_j$ are defined by $P_{j}(X) = \frac{1}{2^{j} j!} \frac{d^{j}}{dX^{j}}\left[(X^{2}-1)^{j}\right]$.
Give the explicit formula for $R_{N}$, in terms of the polynomials $P_{j}$.

The sequence $(a_n)$ is defined by $$\left\{ \begin{array} { l } a _ { 0 } = 1 \\ a _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { ( - 1 ) ^ { n - k } } { ( n - k ) ! } H _ { k } \left( \frac { n + k } { 2 } - 1 \right) \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Demonstrate that all moments of order $p$ of the sequence $(a _ { n })$ are zero.

We have $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$ and $d_{n+1}(x) = 2^{n+1}(\pi_{n+1}(x) - \pi_n(x))$ for all $(x,n) \in \mathbb{R} \times \mathbb{N}$.
Justify $$\forall x \in [0,1[, \forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{k} \frac{d_j(x)}{2^j}.$$

Let $n \in \mathbb{N}^{\star}$ and $x = \sum_{j=1}^{n} \frac{x_j}{2^j}$ with $(x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n$. Show $$\forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{\min(n,k)} \frac{x_j}{2^j}.$$

For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Determine $C(s)$ such that $$\forall s \in ]1, +\infty[, \quad \sum_{k=1}^{+\infty} \frac{1}{(2k-1)^s} = C(s) \zeta(s).$$

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
Calculate $y_{n,1}$.

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
Calculate the $y_{n,j}$ for $1 \leqslant j \leqslant n \leqslant 5$ by presenting the results in the form of a table.

Compare the result of question 48 (the value of $y_{n,n}$, the number of partitions of $n$) to that of question 40 (the maximum cardinality of a set of pairwise non-similar nilpotent matrices of size $n$).

For $x \in ]-1,1[$, give the value of the sum of the power series $\sum_{p=1}^{+\infty} x^{p}$ as well as that of its derivative.

Let $f$ and $g$ be two arithmetic functions with finite abscissas of convergence. Show that if, for all $s > \max\left(A_c(f), A_c(g)\right), L_f(s) = L_g(s)$, then $f = g$.