Show that $$\forall x \in ]-1,1[ \quad s(x)e^{x} = \frac{1}{1-x}$$ Deduce that $R = 1$.

Starting from the relation $(1-x)s(x) = e^{-x}$ for $x \in ]-1,1[$, express $\frac{d_n}{n!}$ for natural number $n$, in the form of a sum.

For natural number $n$, we set $r_n = \sum_{k=n+1}^{+\infty} \frac{1}{k!}$. Prove the bound $$r_n \leq \frac{1}{(n+1)!} \sum_{k=0}^{+\infty} \frac{1}{(n+2)^k}$$ Deduce a simple equivalent of $r_n$ as $n$ tends to $+\infty$.

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. We introduce the set $I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$ and assume that $n$ and $k$ vary such that $k \in I _ { n }$.
Show that we have $$k ! ( n - k ) ! = 2 \pi \mathrm { e } ^ { - n } k ^ { k + 1 / 2 } ( n - k ) ^ { n - k + 1 / 2 } \left( 1 + O \left( \frac { 1 } { n } \right) \right)$$ as $n$ tends to infinity. One may use Stirling's formula: $n ! = \left( \frac { n } { \mathrm { e } } \right) ^ { n } \sqrt { 2 \pi n } \left( 1 + O \left( \frac { 1 } { n } \right) \right)$.

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19. We assume $k \in I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$.
Deduce that, as $n$ tends to $+ \infty$, we have $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \frac { 1 + O \left( \frac { 1 } { n } \right) } { \left( \frac { 2 k } { n } \right) ^ { k + 1 / 2 } \left( 2 - \frac { 2 k } { n } \right) ^ { n - k + 1 / 2 } }$$

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19. We assume $k \in I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$.
Deduce that $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \frac { 1 + O \left( \frac { 1 } { n } \right) } { \left( 1 - \frac { x _ { n , k } ^ { 2 } } { n } \right) ^ { \frac { n + 1 } { 2 } } \left( 1 + \frac { x _ { n , k } } { \sqrt { n } } \right) ^ { \frac { x _ { n , k } } { 2 } \sqrt { n } } \left( 1 - \frac { x _ { n , k } } { \sqrt { n } } \right) ^ { - \frac { x _ { n , k } } { 2 } \sqrt { n } } }$$ then that $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \exp \left( - \frac { x _ { n , k } ^ { 2 } } { 2 } \right) \left( 1 + O \left( \frac { 1 } { \sqrt { n } } \right) \right)$$

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19, and $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$. We fix $\varepsilon > 0$ and $\ell \in \mathbb{R}^+$ such that $\varphi(\ell) \leqslant \frac{\varepsilon}{2}$.
Show that there exists a natural number $n _ { 1 }$ such that, for all integers $n \geqslant n _ { 1 }$, $$\sup _ { x \in [ 0 , \ell ] } \left| B _ { n } ( x ) - \varphi ( x ) \right| \leqslant \frac { \varepsilon } { 2 }$$

The function $B _ { n }$ is defined as in Q19, and $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$.
For all $\ell > 0$, show that there exists a natural number $n _ { 2 }$, such that, for all $n \geqslant n _ { 2 }$, $$B _ { n } ( \ell ) \leqslant 2 \varphi ( \ell )$$

The function $B _ { n }$ is defined as in Q19, $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$, and $\Delta _ { n } = \sup _ { x \in \mathbb { R } } \left| B _ { n } ( x ) - \varphi ( x ) \right|$.
Conclude that the sequence $\left( \Delta _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ converges to 0.

Let $I$ be an interval of $\mathbb { R }$ and $\left( f _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ a sequence of functions piecewise continuous on $I$ that converges uniformly on $I$ to a function $f$ also piecewise continuous on $I$.
If $\left( u _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ (respectively $\left( v _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$) is a sequence of real numbers belonging to $I$ that converges to $u \in I$ (respectively $v \in I$), show that $$\lim _ { n \rightarrow + \infty } \left( \int _ { u _ { n } } ^ { v _ { n } } f _ { n } ( x ) \mathrm { d } x \right) = \int _ { u } ^ { v } f ( x ) \mathrm { d } x$$

Exercise IV

Let $\left( u _ { n } \right) _ { n \in \mathbb { N } }$ be a sequence such that $u _ { n } \neq 0$ for every natural number $n$. For every natural number $n$, the sequence $\left( v _ { n } \right) _ { n \in \mathbb { N } }$ is defined by $v _ { n } = - \frac { 2 } { u _ { n } }$. IV-A- If $\left( u _ { n } \right) _ { n \in \mathbb { N } }$ is bounded below by $2$, then $\left( v _ { n } \right) _ { n \in \mathbb { N } }$ is bounded below by $-1$. IV-B- If $\left( u _ { n } \right) _ { n \in \mathbb { N } }$ is increasing, then $\left( v _ { n } \right) _ { n \in \mathbb { N } }$ is decreasing. IV-C- If $\left( u _ { n } \right) _ { n \in \mathbb { N } }$ converges, then $\left( v _ { n } \right) _ { n \in \mathbb { N } }$ converges.
For each statement, indicate whether it is TRUE or FALSE.

Mathematics Specialty - EXERCISE I (20 points)

First Part

Consider the sequence $\left( u _ { n } \right) _ { n \in \mathbb { N } }$ defined by $u _ { 0 } = 2$ and for every natural number $n$, $u _ { n + 1 } = f \left( u _ { n } \right)$ where $f$ is the function defined for every positive real $x$ by $f ( x ) = \frac { 3 x + 2 } { x + 4 }$. We admit that, for every natural number $\boldsymbol { n }$, $\boldsymbol { u } _ { \boldsymbol { n } }$ is greater than or equal to $1$.
I-1-a- Calculate the exact values of $u _ { 1 }$ and $u _ { 2 }$. Give the result as an irreducible fraction.
I-1-b- The graph below gives the representative curve in an orthonormal coordinate system of the function $f$. From this graph, what can be conjectured about the variations and convergence of the sequence $\left( u _ { n } \right) _ { n \in \mathbb { N } }$? Specify the possible limit.

Second Part - Method 1

I-2-a- Show that, for every natural number $n$, $u _ { n + 1 } - u _ { n } = \frac { \left( 1 - u _ { n } \right) \left( u _ { n } + 2 \right) } { u _ { n } + 4 }$. I-2-b- Deduce the direction of variation of the sequence $\left( u _ { n } \right) _ { n \in \mathbb { N } }$. Justify your answer. I-3- Prove that the sequence $\left( u _ { n } \right) _ { n \in \mathbb { N } }$ is convergent. Let $l$ denote its limit. I-4- Determine the value of $l$. Justify your answer.

Third Part - Method 2

Consider the sequence $\left( v _ { n } \right) _ { n \in \mathbb { N } }$ defined for every natural number $n$ by: $v _ { n } = \frac { u _ { n } - 1 } { u _ { n } + 2 }$. I-5- Calculate $v _ { 0 }$. I-6-a- Determine the constant $k$ in $] 0 ; 1 [$ such that $v _ { n + 1 } = k \times v _ { n }$ for every natural number $n$. Justify your answer. What can be deduced about the nature of the sequence $\left( v _ { n } \right) _ { n \in \mathbb { N } }$? For questions $\mathbf { I - 6 - b }$ and $\mathbf { I - 6 - c }$, answers may be expressed as a function of $k$ or its value. I-6-b- Deduce the expression of $v _ { n }$ as a function of $n$. I-6-c- Deduce the convergence of the sequence $\left( v _ { n } \right) _ { n \in \mathbb { N } }$ and its limit. Justify your answer. I-7-a- Express $u _ { n }$ as a function of $v _ { n }$ for every natural number $n$. I-7-b- Deduce the convergence of the sequence $\left( u _ { n } \right) _ { n \in \mathbb { N } }$ and its limit. Justify your answer.

Let $\left( u _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { C } ^ { \mathbb { N } }$ and $\ell \in \mathbb { C }$. Prove that $$\left( \lim _ { n \rightarrow + \infty } u _ { n } = \ell \right) \Rightarrow \left( \lim _ { n \rightarrow + \infty } \sigma _ { n } = \ell \right)$$ where $\sigma_n = \frac{1}{n+1}\sum_{k=0}^n u_k$.
If $\left( u _ { n } \right) _ { n \in \mathbb { N } }$ takes real values, prove that the result holds if $\ell = + \infty$ or $\ell = - \infty$.

Using (Cesàro), calculate the limit of the sequence $\left( v _ { n } \right) _ { n \geqslant 1 }$ defined by $v _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k n }$. Then, using a series-integral comparison, give an equivalent of $\left( v _ { n } \right) _ { n \geqslant 1 }$.

Let $\left( u _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { R } ^ { \mathbb { N } }$ and $\alpha \in \mathbb { R } ^ { * }$. Suppose that $\lim _ { n \rightarrow + \infty } e _ { n } = \alpha$, where $e_n = u_{n+1} - u_n$. Using (Cesàro), give an equivalent of $\left( u _ { n } \right) _ { n \in \mathbb { N } }$. Recover this result using a comparison theorem for series with positive terms.

Let $\left( u _ { n } \right) _ { n \in \mathbb { N } } \in \left] 0 , + \infty \right[ ^ { \mathbb { N } }$ and $\ell \in \left] 0 , + \infty \right[$. Suppose that $\lim _ { n \rightarrow + \infty } \frac { u _ { n + 1 } } { u _ { n } } = \ell$. Prove that $\lim _ { n \rightarrow + \infty } \sqrt [ n ] { u _ { n } } = \ell$. Prove that the result holds if $\ell = 0$ or $\ell = + \infty$. Deduce $\lim _ { n \rightarrow + \infty } \sqrt [ n ] { n ! }$ and $\lim _ { n \rightarrow + \infty } \sqrt [ n ] { \frac { n ^ { n } } { n ! } }$.

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { C } ^ { \mathbb { N } } , \left( b _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { C } ^ { \mathbb { N } } , a \in \mathbb { C }$ and $b \in \mathbb { C }$. Suppose that $\lim _ { n \rightarrow + \infty } a _ { n } = a$ and $\lim _ { n \rightarrow + \infty } b _ { n } = b$. Prove that $$\lim _ { n \rightarrow + \infty } \left( \frac { 1 } { n } \sum _ { k = 0 } ^ { n } a _ { k } b _ { n - k } \right) = a b$$

Let $\sum _ { n \geqslant 0 } a _ { n }$ and $\sum _ { n \geqslant 0 } b _ { n }$ be two series of complex numbers, convergent with respective sums $A$ and $B$. We denote $\left( c _ { n } \right) _ { n \in \mathbb { N } }$ the sequence with general term $c _ { n } = \sum _ { k = 0 } ^ { n } a _ { k } b _ { n - k }$ and $\left( C _ { n } \right) _ { n \in \mathbb { N } }$ the sequence of partial sums associated defined by $C _ { n } = \sum _ { k = 0 } ^ { n } c _ { k }$. Prove that $$\lim _ { n \rightarrow + \infty } \left( \frac { 1 } { n } \sum _ { k = 0 } ^ { n } C _ { k } \right) = A B \qquad \text{(Cauchy)}$$

Verify that the converse of (Cesàro) is not always true by exhibiting a sequence $\left( u _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { R } ^ { \mathbb{N} }$ that does not converge and such that $\left( \sigma _ { n } \right) _ { n \in \mathbb { N } }$ converges in $\mathbb { R }$.

Let $\left( u _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { R } ^ { \mathbb { N } }$ and $\ell \in \mathbb { R }$. Prove that $$\left( \lim _ { n \rightarrow + \infty } \sigma _ { n } = \ell \text { and } \left( u _ { n } \right) _ { n \in \mathbb { N } } \text { monotone } \right) \Rightarrow \left( \lim _ { n \rightarrow + \infty } u _ { n } = \ell \right) .$$ Prove that the result holds for $\ell = + \infty$ or $\ell = - \infty$.

Let $\left( u _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { C } ^ { \mathbb { N } }$ and $\ell \in \mathbb { C }$. Prove that $$\left( \lim _ { n \rightarrow + \infty } \sigma _ { n } = \ell \text { and } e _ { n } = o \left( \frac { 1 } { n } \right) \right) \Rightarrow \left( \lim _ { n \rightarrow + \infty } u _ { n } = \ell \right) \qquad \text{(Weak Hardy)}$$ Hint: one may prove that for all $n \geqslant 1$, $$\sum _ { k = 0 } ^ { n } k e _ { k } = n u _ { n + 1 } - \sum _ { k = 1 } ^ { n } u _ { k }$$

Let $\left( u _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { C } ^ { \mathbb { N } }$ and $\ell \in \mathbb { C }$. The purpose of this question is to prove that $$\left( \lim _ { n \rightarrow + \infty } \sigma _ { n } = \ell \text { and } e _ { n } = O \left( \frac { 1 } { n } \right) \right) \Rightarrow \left( \lim _ { n \rightarrow + \infty } u _ { n } = \ell \right) \qquad \text{(Strong Hardy)}$$
We suppose that $\lim _ { n \rightarrow + \infty } \sigma _ { n } = \ell$ and $e _ { n } = O \left( \frac { 1 } { n } \right)$.
(a) Let $0 \leqslant n < m$. Prove that $$\sum _ { k = n + 1 } ^ { m } u _ { k } - ( m - n ) u _ { n } = \sum _ { j = n } ^ { m - 1 } ( m - j ) e _ { j }$$
(b) Deduce that there exists a constant $C > 0$ such that for all $2 \leqslant n < m$, we have $$\left| \frac { ( m + 1 ) \sigma _ { m } - ( n + 1 ) \sigma _ { n } } { m - n } - u _ { n } \right| \leqslant C \ln \left( \frac { m - 1 } { n - 1 } \right)$$ and $$\left| u _ { n } - \ell \right| \leqslant C \ln \left( \frac { m - 1 } { n - 1 } \right) + \frac { m + 1 } { m - n } \left( \left| \sigma _ { m } - \ell \right| + \left| \sigma _ { n } - \ell \right| \right) .$$
(c) Deduce (Strong Hardy). Hint: one may take $m = 1 + \lfloor \alpha n \rfloor$ with a parameter $\alpha > 1$ to be chosen, where $\lfloor x \rfloor$ denotes the integer part of $x \in \mathbb { R }$.

By setting for every integer $k \geqslant 1$, $a_k = \frac{1}{k} - \frac{1}{2^{k+1}}$ and $b_k = \frac{1}{k} + \frac{1}{2^{k+1}}$, show that we can choose an integer $k_0 \geqslant 1$ such that: $$\forall k \geqslant k_0, \quad b_{k+1} < a_k.$$ Deduce that the function $f : ]0,1[ \longrightarrow \mathbf{R}$ defined by: $$f : t \longmapsto \begin{cases} k^2 \cdot 2^{k+1} \cdot (t - a_k), & \text{if there exists an integer } k \geqslant k_0 \text{ such that } t \in \left[a_k, a_k + \frac{1}{2^{k+1}}\right] \\ k^2 \cdot 2^{k+1} \cdot (b_k - t), & \text{if there exists an integer } k \geqslant k_0 \text{ such that } t \in \left[a_k + \frac{1}{2^{k+1}}, b_k\right] \\ 0, & \text{otherwise} \end{cases}$$ is a well-defined and continuous function on $]0,1[$, integrable on $]0,1[$ and that this function $f$ does not belong to the set $\mathscr{D}_{0,1}$.

Let $\sum _ { n \geqslant 0 } a _ { n } z ^ { n }$ be a power series with radius of convergence $R \geqslant 1$ and sum $f$. We denote $$\Delta _ { \theta _ { 0 } } = \left\{ z \in \mathbb { C } ; | z | < 1 \text { and } \exists \rho > 0 , \exists \theta \in \left[ - \theta _ { 0 } , \theta _ { 0 } \right] , z = 1 - \rho e ^ { i \theta } \right\}$$ for $\theta _ { 0 } \in [ 0 , \pi / 2 [$.
The purpose of this question is to prove that $$\left( \sum _ { n \geqslant 0 } a _ { n } \text { converges } \right) \Rightarrow \left( \lim _ { \substack { z \rightarrow 1 \\ z \in \Delta _ { \theta _ { 0 } } } } f ( z ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } \right) \qquad \text{(Abel)}$$
(a) Prove (Abel) for $R > 1$.
From now on, we assume that $R = 1$ and that $\sum _ { n \geqslant 0 } a _ { n }$ converges, and we are given a $\theta _ { 0 } \in [ 0 , \pi / 2 [$.
(b) Prove that for all $N \in \mathbb { N } ^ { * }$ and $z \in \mathbb { C } , | z | < 1$, we have $$\sum _ { n = 0 } ^ { N } a _ { n } z ^ { n } - S _ { N } = ( z - 1 ) \sum _ { n = 0 } ^ { N - 1 } R _ { n } z ^ { n } - R _ { N } \left( z ^ { N } - 1 \right)$$
(c) Deduce that for all $z \in \mathbb { C } , | z | < 1$, we have $$f ( z ) - S = ( z - 1 ) \sum _ { n = 0 } ^ { + \infty } R _ { n } z ^ { n }$$
(d) Let $\varepsilon > 0$. Prove that there exists $N _ { 0 } \in \mathbb { N }$ such that for all $z \in \mathbb { C } , | z | < 1$ $$| f ( z ) - S | \leqslant | z - 1 | \sum _ { n = 0 } ^ { N _ { 0 } } \left| R _ { n } \right| + \varepsilon \frac { | z - 1 | } { 1 - | z | }$$
(e) Prove that there exists $\rho \left( \theta _ { 0 } \right) > 0$ such that for all $z \in \Delta _ { \theta _ { 0 } }$ of the form $z = 1 - \rho e ^ { i \theta }$ with $0 < \rho \leqslant \rho \left( \theta _ { 0 } \right)$, we have $$\frac { | z - 1 | } { 1 - | z | } \leqslant \frac { 2 } { \cos \left( \theta _ { 0 } \right) }$$ Deduce (Abel).

Prove that $$\sum _ { n = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { n } } { 2 n + 1 } = \frac { \pi } { 4 }$$