Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence. We assume that the associated power series $\sum a _ { n } x ^ { n }$ has radius of convergence $R _ { a } = 1$ and that the sum $f$ of this series satisfies $f ( x ) \sim \frac { 1 } { 1 - x }$ when $x \rightarrow 1, x < 1$.
Determine a real sequence $\left( b _ { n } \right) _ { n \geqslant 0 }$ such that $$\forall x \in ] - 1,1 [ , \quad \frac { 1 } { 1 - x ^ { 2 } } = \sum _ { n = 0 } ^ { + \infty } b _ { n } x ^ { n }$$

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence. We assume that the associated power series $\sum a _ { n } x ^ { n }$ has radius of convergence $R _ { a } = 1$ and that the sum $f$ of this series satisfies $f ( x ) \sim \frac { 1 } { 1 - x }$ when $x \rightarrow 1, x < 1$.
Deduce an example of a sequence $\left( a _ { n } \right) _ { n \geqslant 0 }$ satisfying hypothesis II.1 ($f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$) but not converging to 1.

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 $M_n = \sum_{k=1}^n \dfrac{1}{2\cos\dfrac{k\pi}{2n+1}}$.

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 $M_n = \sum_{k=1}^n \dfrac{1}{2\cos\dfrac{k\pi}{2n+1}}$.
Give an equivalent of $M_n$ as $n$ tends to $+\infty$.

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ be a real decreasing sequence converging to 0, and $\left( b _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ be a complex sequence such that the sequence $\left( B _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ defined for all $n \in \mathbb { N } ^ { * }$ by $B _ { n } = b _ { 1 } + \cdots + b _ { n }$ is bounded.
Show that, for all integer $n \geqslant 2$, $$\sum _ { k = 1 } ^ { n } a _ { k } b _ { k } = a _ { n } B _ { n } + \sum _ { k = 1 } ^ { n - 1 } \left( a _ { k } - a _ { k + 1 } \right) B _ { k }$$

Show that $E \in \mathrm{GL}(V)$.

Determine $T_0, T_1, T_2$ and $T_3$, where the Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $$\forall n \in \mathbb{N}, \quad \forall \theta \in \mathbb{R}, \quad T_n(\cos\theta) = \cos(n\theta)$$

The Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $T_n(\cos\theta) = \cos(n\theta)$.
By noting that for every real $\theta$, we have $e^{in\theta} = \left(e^{i\theta}\right)^n$, show that $$\forall n \in \mathbb{N}, \quad T_n = \sum_{0 \leqslant k \leqslant n/2} \binom{n}{2k} \left(X^2 - 1\right)^k X^{n-2k}$$
Write in Maple or Mathematica language a function $T$ taking as argument a natural integer $n$ and returning the expanded expression of the polynomial $T_n$.

The Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $T_n(\cos\theta) = \cos(n\theta)$.
Show that the sequence $(T_n)_{n \in \mathbb{N}}$ satisfies the recurrence relation $$\forall n \in \mathbb{N}, \quad T_{n+2} = 2X T_{n+1} - T_n$$
Deduce, for every natural integer $n$, the degree and the leading coefficient of $T_n$. Find this result again with the expression from question I.A.2.

The Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $T_n(\cos\theta) = \cos(n\theta)$.
Show that, for every natural integer $n$, the polynomial $T_n$ is split over $\mathbb{R}$, with simple roots belonging to $]-1,1[$. Determine the roots of $T_n$.

The Chebyshev polynomials of the second kind $(U_n)_{n \in \mathbb{N}}$ are defined by $$\forall n \in \mathbb{N}, \quad U_n = \frac{1}{n+1} T_{n+1}'$$ where $(T_n)_{n \in \mathbb{N}}$ are the Chebyshev polynomials of the first kind defined by $T_n(\cos\theta) = \cos(n\theta)$.
Show that $$\forall n \in \mathbb{N}, \quad \forall \theta \in \mathbb{R} \backslash \pi\mathbb{Z}, \quad U_n(\cos\theta) = \frac{\sin((n+1)\theta)}{\sin\theta}$$

The Chebyshev polynomials of the second kind $(U_n)_{n \in \mathbb{N}}$ are defined by $U_n = \frac{1}{n+1} T_{n+1}'$, and satisfy $U_n(\cos\theta) = \frac{\sin((n+1)\theta)}{\sin\theta}$ for $\theta \notin \pi\mathbb{Z}$.
Deduce the following properties:
a) The sequence $(U_n)_{n \in \mathbb{N}}$ satisfies the same recurrence relation $T_{n+2} = 2X T_{n+1} - T_n$ as the sequence $(T_n)_{n \in \mathbb{N}}$.
b) For every natural integer $n$, the polynomial $U_n$ is split over $\mathbb{R}$ with simple roots belonging to $]-1,1[$. Determine the roots of $U_n$.

The Chebyshev polynomials of the first and second kind are defined by $T_n(\cos\theta) = \cos(n\theta)$ and $U_n(\cos\theta) = \frac{\sin((n+1)\theta)}{\sin\theta}$ for $\theta \notin \pi\mathbb{Z}$.
Show that $$\begin{cases} T_m \cdot T_n = \frac{1}{2}\left(T_{n+m} + T_{n-m}\right) & \text{for all integers } 0 \leqslant m \leqslant n \\ T_m \cdot U_{n-1} = \frac{1}{2}\left(U_{n+m-1} + U_{n-m-1}\right) & \text{for all integers } 0 \leqslant m < n \end{cases}$$

The Chebyshev polynomials of the first kind satisfy $T_m \cdot T_n = \frac{1}{2}(T_{n+m} + T_{n-m})$ for $0 \leqslant m \leqslant n$, and $T_m \cdot U_{n-1} = \frac{1}{2}(U_{n+m-1} + U_{n-m-1})$ for $0 \leqslant m < n$.
For $m$ and $n$ natural integers such that $m \leqslant n$, we propose to determine the quotient $Q_{n,m}$ and the remainder $R_{n,m}$ of the Euclidean division of $T_n$ by $T_m$.
a) Suppose $m < n < 3m$. Show that $$Q_{n,m} = 2T_{n-m} \quad \text{and} \quad R_{n,m} = -T_{|n-2m|}$$
b) Determine $Q_{n,m}$ and $R_{n,m}$ when $n$ is of the form $(2p+1)m$ with $p \in \mathbb{N}^*$.
c) Suppose that $m > 0$ and that $n$ is not the product of $m$ by an odd integer. Show that there exists a unique integer $p \geqslant 1$ such that $|n - 2pm| < m$ and that $$Q_{n,m} = 2\left(T_{n-m} - T_{n-3m} + \cdots + (-1)^{p-1} T_{n-(2p-1)m}\right) \quad \text{and} \quad R_{n,m} = (-1)^p T_{|n-2pm|}$$

Throughout this subsection II.B, we fix two natural integers $m$ and $n$. Let $g > 0$ be the $\gcd$ of $m$ and $n$. We set $m_1 = m/g$ and $n_1 = n/g$.
a) Show that if $m_1$ and $n_1$ are odd, then $T_g$ is a $\gcd$ of $T_n$ and $T_m$.
b) Show that if one of the two integers $m_1$ or $n_1$ is even, then $T_n$ and $T_m$ are coprime.
c) What can be said about the gcds of $T_n$ and $T_m$ when $m$ and $n$ are odd? When $n$ and $m$ are two distinct powers of 2?

We equip $\mathbb{C}[X]$ with the internal composition law given by composition, denoted $\circ$. We seek families $(F_n)_{n \in \mathbb{N}}$ of complex polynomials satisfying $$\forall n \in \mathbb{N}, \quad \deg F_n = n \quad \text{and} \quad \forall (m,n) \in \mathbb{N}^2, \quad F_n \circ F_m = F_m \circ F_n \tag{III.1}$$
Show that the family $(T_n)_{n \in \mathbb{N}}$ satisfies property (III.1). One may compare $T_n \circ T_m$ and $T_{mn}$.

We introduce the Dickson polynomials of the first and second kind, $(D_n)_{n \in \mathbb{N}}$ and $(E_n)_{n \in \mathbb{N}}$, defined in the form of polynomial functions of two variables by $$D_0(x,a) = 2 \quad D_1(x,a) = x \quad E_0(x,a) = 1 \quad E_1(x,a) = x$$ then, for every integer $n \in \mathbb{N}$, $$D_{n+2}(x,a) = x D_{n+1}(x,a) - a D_n(x,a) \quad \text{and} \quad E_{n+2}(x,a) = x E_{n+1}(x,a) - a E_n(x,a)$$
Justify the following relation with Chebyshev polynomials $$\forall (x,a) \in \mathbb{C}^2, \quad D_n\left(2xa, a^2\right) = 2a^n T_n(x) \quad \text{and} \quad E_n\left(2xa, a^2\right) = a^n U_n(x)$$ as well as the following two relations, valid for every natural integer $n$ and every $(x,a) \in \mathbb{C}^* \times \mathbb{C}$ $$D_n\left(x + \frac{a}{x}, a\right) = x^n + \frac{a^n}{x^n} \quad \text{and} \quad \left(x - \frac{a}{x}\right) E_n\left(x + \frac{a}{x}, a\right) = \left(x^{n+1} - \frac{a^{n+1}}{x^{n+1}}\right)$$

Justify that the series with general term $a _ { n } = \frac { 1 } { n } - \int _ { n - 1 } ^ { n } \frac { \mathrm {~d} t } { t }$ converges.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k } = 1 + \frac { 1 } { 2 } + \cdots + \frac { 1 } { n }$.
Show that there exists a real constant $A$ such that $H _ { n } \underset { + \infty } { = } \ln n + A + o ( 1 )$. Deduce that $H _ { n } \sim \ln n$.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k } = 1 + \frac { 1 } { 2 } + \cdots + \frac { 1 } { n }$. We denote $\zeta$ the function defined for $x > 1$ by $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$.
Let $r$ be a natural integer. For which values of $r$ is the series $\sum _ { n \geqslant 1 } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ convergent?
In the rest of the problem we will denote $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ when the series converges.

Give without proof the power series expansions of the functions $t \mapsto \ln ( 1 - t )$ and $t \mapsto \frac { 1 } { 1 - t }$ as well as their radius of convergence.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k } = 1 + \frac { 1 } { 2 } + \cdots + \frac { 1 } { n }$.
Deduce that the function $$t \mapsto - \frac { \ln ( 1 - t ) } { 1 - t }$$ is expandable as a power series on $] - 1,1 [$ and specify its power series expansion using the real numbers $H _ { n }$.

For every pair of natural integers $( p , q )$ and for every $\varepsilon \in ] 0,1 [$, we denote $$I _ { p , q } = \int _ { 0 } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t \quad \text { and } \quad I _ { p , q } ^ { \varepsilon } = \int _ { \varepsilon } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t$$
Show that the integral $I _ { p , q }$ exists for every pair of natural integers $( p , q )$.

For every pair of natural integers $( p , q )$ and for every $\varepsilon \in ] 0,1 [$, we denote $$I _ { p , q } = \int _ { 0 } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t \quad \text { and } \quad I _ { p , q } ^ { \varepsilon } = \int _ { \varepsilon } ^ { 1 } t ^ { p } ( \ln t ) ^ { q } \mathrm {~d} t$$
Show that $\left. \forall p \in \mathbb { N } , \forall q \in \mathbb { N } ^ { * } , \forall \varepsilon \in \right] 0,1 \left[ , \quad I _ { p , q } ^ { \varepsilon } = - \frac { q } { p + 1 } I _ { p , q - 1 } ^ { \varepsilon } - \frac { \varepsilon ^ { p + 1 } ( \ln \varepsilon ) ^ { q } } { p + 1 } \right.$.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$, $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ when the series converges, and $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$.
Deduce that $S _ { 2 } = \frac { 1 } { 2 } \int _ { 0 } ^ { 1 } \frac { ( \ln t ) ^ { 2 } } { 1 - t } \mathrm {~d} t$ then find the value of $S _ { 2 }$ as a function of $\zeta ( 3 )$.