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 $g : [ 0,1 ] \rightarrow \mathbb { R }$ be the function such that $g ( x ) = 1 / x$ if $x \geqslant \mathrm { e } ^ { - 1 }$ and $g ( x ) = 0$ otherwise. We fix a real $\varepsilon \in ] 0 , \mathrm { e } ^ { - 1 } [$. We define two continuous applications $g ^ { + } , g ^ { - } : [ 0,1 ] \rightarrow \mathbb { R }$ as follows:

• $g ^ { + }$ is affine on $\left[ \mathrm { e } ^ { - 1 } - \varepsilon , \mathrm { e } ^ { - 1 } \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } - \varepsilon \right] \cup \left[ \mathrm { e } ^ { - 1 } , 1 \right]$;
• $g ^ { - }$ is affine on $\left[ \mathrm { e } ^ { - 1 } , \mathrm { e } ^ { - 1 } + \varepsilon \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } \left[ \cup \left[ \mathrm { e } ^ { - 1 } + \varepsilon , 1 \right] \right. \right.$.

Calculate $\int _ { 0 } ^ { 1 } g ^ { + } ( t ) \mathrm { d } t$ and $\int _ { 0 } ^ { 1 } g ^ { - } ( t ) \mathrm { d } t$.

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.

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

Deduce the power series development, for $n \geqslant 0$ and $x \in \mathbb { R }$ :
$$\varphi _ { n } ( x ) = \sum _ { p = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { p } } { p ! ( n + p ) ! } \left( \frac { x } { 2 } \right) ^ { n + 2 p }$$
Specify the radius of convergence.

Show that $\varphi _ { n }$ is of class $\mathcal { C } ^ { \infty }$ on $\mathbb { R }$.

Let $n$ in $\mathbb { N } ^ { * }$, verify that for real $x$
$$\frac { \mathrm { d } } { \mathrm {~d} x } \left( x ^ { n } \varphi _ { n } ( x ) \right) = x ^ { n } \varphi _ { n - 1 } ( x )$$

We approximate $\varphi _ { n } ( x )$ using partial sums
$$S _ { m } = \sum _ { p = 0 } ^ { m } ( - 1 ) ^ { p } a _ { p } \quad \text { with } \quad m \in \mathbb { N } \quad \text { and } \quad a _ { p } = \frac { 1 } { p ! ( n + p ) ! } \left( \frac { x } { 2 } \right) ^ { n + 2 p }$$
From which value $p _ { 0 }$ of $p$ is the sequence $\left( a _ { p } \right) _ { p \in \mathbb { N } }$ decreasing?

We approximate $\varphi _ { n } ( x )$ using partial sums
$$S _ { m } = \sum _ { p = 0 } ^ { m } ( - 1 ) ^ { p } a _ { p } \quad \text { with } \quad m \in \mathbb { N } \quad \text { and } \quad a _ { p } = \frac { 1 } { p ! ( n + p ) ! } \left( \frac { x } { 2 } \right) ^ { n + 2 p }$$
Assume $N > p _ { 0 }$. Bound $\left| R _ { N } \right|$ as a function of $(N, n, x)$ with
$$R _ { N } = \sum _ { p = N + 1 } ^ { + \infty } ( - 1 ) ^ { p } a _ { p }$$
Deduce, for fixed $\varepsilon > 0$, a sufficient condition on $N$ for $\left| \varphi _ { n } ( x ) - S _ { N } \right| < \varepsilon$.

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 $\forall \theta \in ] 0 , \pi ] , r ( \theta ) = \frac { \pi - \theta } { 2 }$.
Justify the existence and uniqueness of $r$.

Let $r : \mathbb { R } \rightarrow \mathbb { R }$, be a $2 \pi$-periodic, odd function, such that $\forall \theta \in ] 0 , \pi ] , r ( \theta ) = \frac { \pi - \theta } { 2 }$.
Determine the Fourier series of $r$.

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 $\sum _ { n = 0 } ^ { + \infty } \frac { 1 } { ( 2 n + 1 ) ^ { 2 } } = \frac { \pi ^ { 2 } } { 8 }$.

Let $x \in ] - 1,1 [$.
Determine the Fourier series of the function $\widetilde { h } : \mathbb { R } \rightarrow \mathbb { R }$ defined by $\widetilde { h } ( \theta ) = \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right)$.
One may use the result from question II.A.2.

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a $2 \pi$-periodic function of class $\mathcal { C } ^ { 1 }$. We consider the Fourier series of $f$ in cosines and sines, denoted $$c _ { 0 } + \sum _ { n \geq 1 } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right)$$
Show that, for all $x \in ] - 1,1 [$ and all $t \in \mathbb { R }$, $$c _ { 0 } + \sum _ { n = 1 } ^ { + \infty } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right) x ^ { n } = \frac { 1 } { 2 \pi } \int _ { 0 } ^ { 2 \pi } \frac { \left( 1 - x ^ { 2 } \right) f ( u ) } { x ^ { 2 } - 2 x \cos ( t - u ) + 1 } \mathrm { ~d} u$$

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a $2 \pi$-periodic function of class $\mathcal { C } ^ { 1 }$. We consider the Fourier series of $f$ in cosines and sines, denoted $$c _ { 0 } + \sum _ { n \geq 1 } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right)$$
Deduce that, for all $t \in \mathbb { R }$, $$f ( t ) = \lim _ { x \rightarrow 1 ^ { - } } \frac { 1 } { 2 \pi } \int _ { 0 } ^ { 2 \pi } \frac { \left( 1 - x ^ { 2 } \right) f ( u ) } { x ^ { 2 } - 2 x \cos ( t - u ) + 1 } \mathrm { ~d} u$$

We recall that $\phi$ is defined on $] - 1 , + \infty [$ by $\phi ( s ) = s - \ln ( 1 + s )$.
Show that if $s \in ] - 1,1 [$, $$\phi ( s ) = s ^ { 2 } \sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } } { k + 2 } s ^ { k }$$
We admit the existence of two power series $\sum _ { k \geqslant 1 } b _ { k } q ^ { k }$ and $\sum _ { k \geqslant 1 } c _ { k } q ^ { k }$, in the variable $q$, and of strictly positive radius of convergence, where $b _ { 1 } > 0$ and $c _ { 1 } < 0$, and such that we have, for $q$ in a neighborhood of 0 in $[ 0 , + \infty [$, $$\phi \left( \sum _ { k = 1 } ^ { \infty } b _ { k } q ^ { k } \right) = \phi \left( \sum _ { k = 1 } ^ { \infty } c _ { k } q ^ { k } \right) = q ^ { 2 } .$$

We consider the function $F : ] 0 , + \infty [ \rightarrow \mathbb { R }$ defined by $F ( x ) = \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - 1 } d t$.
For $N \in \mathbb { N } ^ { * }$ and $x > 0$, we set $$\begin{aligned} r _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N + 1 } e ^ { - 1 / x } \\ S _ { N } ( x ) & = \sum _ { k = 1 } ^ { N } ( - 1 ) ^ { k - 1 } ( k - 1 ) ! x ^ { k } e ^ { - 1 / x } \\ R _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N } \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - ( N + 1 ) } d t \end{aligned}$$
Show that, for all $N \geqslant 1$ and all $x > 0 , F ( x ) = S _ { N } ( x ) + R _ { N } ( x )$.

We consider the function $F : ] 0 , + \infty [ \rightarrow \mathbb { R }$ defined by $F ( x ) = \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - 1 } d t$.
For $N \in \mathbb { N } ^ { * }$ and $x > 0$, we set $$R _ { N } ( x ) = ( - 1 ) ^ { N } N ! x ^ { N } \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - ( N + 1 ) } d t$$
Specify the domain of convergence of the series $\sum _ { k \geqslant 1 } ( - 1 ) ^ { k - 1 } ( k - 1 ) ! x ^ { k }$ and show that the sequence $\left( R _ { N } ( x ) \right) _ { N \geqslant 1 }$ is not bounded.

We consider the function $F : ] 0 , + \infty [ \rightarrow \mathbb { R }$ defined by $F ( x ) = \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - 1 } d t$.
For $N \in \mathbb { N } ^ { * }$ and $x > 0$, we set $$r _ { N } ( x ) = ( - 1 ) ^ { N } N ! x ^ { N + 1 } e ^ { - 1 / x }, \quad R _ { N } ( x ) = ( - 1 ) ^ { N } N ! x ^ { N } \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - ( N + 1 ) } d t$$
Show that, if $N \in \mathbb { N } ^ { * }$ and $x > 0$, $$\left| R _ { N } ( x ) \right| \leqslant \left| r _ { N } ( x ) \right|$$ Deduce that $R _ { N + 1 } ( x ) = o \left( r _ { N } ( x ) \right)$ when $x \rightarrow 0$.