We consider the sequence $x = (x_n)_{n \geqslant 0}$ defined by $$x_0 = 1, \quad x_1 = 1, \quad x_2 = 1, \quad x_3 = 0, \quad \text{and} \quad \forall n \in \mathbb{N}, x_{n+4} = x_{n+3} - 2x_{n+1}$$
Specify the rank of $H_n(x)$ for any integer $n$ in $\mathbb{N}^*$ and indicate the minimal order of the sequence $x$.

We consider the sequence $x = (x_n)_{n \geqslant 0}$ defined by $$x_0 = 1, \quad x_1 = 1, \quad x_2 = 1, \quad x_3 = 0, \quad \text{and} \quad \forall n \in \mathbb{N}, x_{n+4} = x_{n+3} - 2x_{n+1}$$
Determine the minimal recurrence relation of the sequence $x$.

We consider the sequence $x = (x_n)_{n \geqslant 0}$ defined by $$x_0 = 1, \quad x_1 = 1, \quad x_2 = 1, \quad x_3 = 0, \quad \text{and} \quad \forall n \in \mathbb{N}, x_{n+4} = x_{n+3} - 2x_{n+1}$$
Give a formula allowing for any $n \geqslant 1$ to directly compute $x_n$.

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 $x \rightarrow 1, x < 1$.
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.

Give the power series expansion of the function $t \mapsto \frac { 1 } { ( 1 - t ) ^ { 2 } }$ as well as its radius of convergence. Specify whether the series converges at the endpoints of the interval of convergence.

We consider $\psi : x \mapsto \frac { 1 } { ( 1 + x ) ^ { 2 } ( 1 - x ) }$ with power series expansion $\psi ( x ) = \sum _ { n = 0 } ^ { + \infty } v _ { n } x ^ { n }$ for $x \in ] - 1,1 [$. We denote $A_n = \sum_{k=0}^n a_k$ and $\widetilde{a}_n = \frac{A_n}{n+1}$.
Calculate $\widetilde { v } _ { n }$ (arithmetic mean of the numbers $v _ { 0 } , \ldots , v _ { n }$).

We consider $\psi : x \mapsto \frac { 1 } { ( 1 + x ) ^ { 2 } ( 1 - x ) }$ with power series expansion $\psi ( x ) = \sum _ { n = 0 } ^ { + \infty } v _ { n } x ^ { n }$ for $x \in ] - 1,1 [$. We denote $\widetilde{a}_n = \frac{1}{n+1}\sum_{k=0}^n a_k$.
Construct using $\psi$ 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 satisfying property II.3 ($\lim_{n\to\infty} \widetilde{a}_n = 1$).

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. We denote $A _ { n } = \sum _ { k = 0 } ^ { n } a _ { k }$.
For all $x \in \left[ 0,1 \left[ \right. \right.$ and all $n \in \mathbb { N }$, show that $f ( x ) \geqslant A _ { n } x ^ { n }$.

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$.
Show the existence of an integer $N > 0$ such that $$\forall n \geqslant N , \quad f \left( \mathrm { e } ^ { - 1 / n } \right) \leqslant \frac { 2 } { 1 - \mathrm { e } ^ { - 1 / n } }$$

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. We denote $A _ { n } = \sum _ { k = 0 } ^ { n } a _ { k }$ and $\widetilde { a } _ { n } = \frac { A _ { n } } { n + 1 }$.
Deduce that the sequence $\left( \widetilde { a } _ { n } \right) _ { n \geqslant 0 }$ is bounded above.

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. We denote $A _ { n } = \sum _ { k = 0 } ^ { n } a _ { k }$, $\widetilde { a } _ { n } = \frac { A _ { n } } { n + 1 }$, and $\mu > 0$ is an upper bound of the sequence $\left( \widetilde { a } _ { n } \right) _ { n \geqslant 0 }$: $\forall n \in \mathbb { N } , \widetilde { a } _ { n } \leqslant \mu$.
a) For all $x \in ] - 1,1 [$, show that $( 1 - x ) \sum _ { k = 0 } ^ { + \infty } A _ { k } x ^ { k } = f ( x )$. b) Deduce that for all $x \in \left[ 0,1 \left[ \right. \right.$ and all $N \in \mathbb { N } ^ { * }$ $$\frac { f ( x ) } { 1 - x } \leqslant A _ { N - 1 } \frac { 1 - x ^ { N } } { 1 - x } + \mu \sum _ { k = N } ^ { + \infty } ( k + 1 ) x ^ { k }$$ c) Deduce that for all $x \in \left[ 0,1 \left[ \right. \right.$ and all $N \in \mathbb { N } ^ { * }$ $$f ( x ) \leqslant A _ { N - 1 } + \mu \left( ( N + 1 ) x ^ { N } + \frac { x ^ { N + 1 } } { 1 - x } \right) .$$

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. We denote $A _ { n } = \sum _ { k = 0 } ^ { n } a _ { k }$, $\widetilde { a } _ { n } = \frac { A _ { n } } { n + 1 }$, and $\mu > 0$ is an upper bound of $\left( \widetilde { a } _ { n } \right)$.
Let $\lambda$ be a strictly positive real number. a) Show that there exists an integer $N _ { 0 } > 0$ such that for all $N \geqslant N _ { 0 }$, $$f \left( \mathrm { e } ^ { - \lambda / N } \right) \geqslant \frac { 1 } { 2 \left( 1 - \mathrm { e } ^ { - \lambda / N } \right) } \geqslant \frac { N } { 2 \lambda } .$$ b) Show that for all $N \geqslant N _ { 0 }$ $$\tilde { a } _ { N - 1 } \geqslant \frac { 1 } { 2 \lambda } - \mu \mathrm { e } ^ { - \lambda } \left( 1 + \frac { 1 } { N } + \mathrm { e } ^ { - \lambda / N } \frac { 1 } { N \left( 1 - \mathrm { e } ^ { - \lambda / N } \right) } \right)$$ c) Determine as a function of $\lambda$ the limit, when $N$ tends to infinity, of the right-hand side in the previous inequality. d) Show that there exists a real $\lambda > 0$ such that this limit is strictly positive.

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. We denote $\widetilde { a } _ { n } = \frac { A _ { n } } { n + 1 }$ where $A_n = \sum_{k=0}^n a_k$.
Conclude that there exists a real $\nu > 0$ such that from a certain rank onwards we have $\widetilde { a } _ { n } \geqslant \nu$.

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$.
Let $P$ be a polynomial with real coefficients. Show that $$( 1 - x ) \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } P \left( x ^ { n } \right) \underset { \substack { x \rightarrow 1 \\ x < 1 } } { \longrightarrow } \int _ { 0 } ^ { 1 } P ( t ) \mathrm { d } t$$ We will first consider the special case $P ( x ) = x ^ { k }$, where $k \in \mathbb { N }$.

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. Let $g$, $g^+$, $g^-$, $P$, $Q$ be as defined in II.E.1--II.E.3. For every integer $N > 0$ we set $x _ { N } = \mathrm { e } ^ { - 1 / N }$.
Establish the existence of an integer $N _ { 1 } > 0$ such that for every integer $N \geqslant N _ { 1 }$, $$\left( 1 - x _ { N } \right) \sum _ { n = 0 } ^ { + \infty } a _ { n } x _ { N } ^ { n } P \left( x _ { N } ^ { n } \right) \geqslant \int _ { 0 } ^ { 1 } P ( t ) \mathrm { d } t - \varepsilon$$ and $$\left( 1 - x _ { N } \right) \sum _ { n = 0 } ^ { + \infty } a _ { n } x _ { N } ^ { n } Q \left( x _ { N } ^ { n } \right) \leqslant \int _ { 0 } ^ { 1 } Q ( t ) \mathrm { d } t + \varepsilon$$

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. We denote $A_N = \sum_{k=0}^N a_k$. For every integer $N > 0$ we set $x _ { N } = \mathrm { e } ^ { - 1 / N }$. Let $N_1$ be as in II.E.4.
Deduce from the three previous questions that for every integer $N \geqslant N _ { 1 }$ $$1 - 5 \varepsilon \leqslant \left( 1 - x _ { N } \right) A _ { N } \leqslant 1 + 5 \varepsilon$$

Justify the equality
$$\forall t \in \mathbb { R } \quad G _ { x } ( t ) = e ^ { i x \sin t } = \sum _ { n = - \infty } ^ { + \infty } \varphi _ { n } ( x ) e ^ { i n t }$$
What can be said about the convergence of the Fourier series of $G _ { x }$ ?

Show that for all $k$ in $\mathbb { N } ^ { * } , \left| \varphi _ { n } ( x ) \right| = o \left( \frac { 1 } { n ^ { k } } \right)$ as $n$ tends to $+ \infty$.
Use Fourier series of successive derivatives of $G _ { x }$.

By expressing $G _ { x } ( - t )$ in terms of $G _ { x } ( t )$, show that for $n$ in $\mathbb { Z } , \varphi _ { n } ( x ) \in \mathbb { R }$.

Express $G _ { x } ( t + \pi )$ and deduce the following equalities for $n$ in $\mathbb { Z }$ :
$$\varphi _ { n } ( - x ) = ( - 1 ) ^ { n } \varphi _ { n } ( x ) = \varphi _ { - n } ( x )$$
What can be said about the parity of $\varphi _ { n }$ for $n \in \mathbb { Z }$ ?

Calculate $\sum _ { n = - \infty } ^ { + \infty } \left| \varphi _ { n } ( x ) \right| ^ { 2 }$.

Justify that for real $x$, $\left| \varphi _ { n } ( x ) \right| \leqslant 1$.

Show that for real $x$,
$$\varphi _ { n } ( x ) = \sum _ { k = 0 } ^ { + \infty } \frac { x ^ { k } } { k ! } I _ { n , k } \quad \text { with } \quad I _ { n , k } = \frac { 1 } { 2 \pi } \int _ { - \pi } ^ { \pi } i ^ { k } e ^ { - i n t } ( \sin t ) ^ { k } \mathrm {~d} t$$

Using Euler's formula, justify that for $( n , k )$ in $\mathbb { N } \times \mathbb { N }$,
$$I _ { n , k } = \sum _ { m = 0 } ^ { k } \frac { A _ { m , k } } { 2 \pi } \int _ { - \pi } ^ { \pi } e ^ { i t ( 2 m - k - n ) } \mathrm { d } t$$
with $A _ { m , k }$ constants to be determined.