Exhibit a power series $\sum _ { n \geqslant 0 } a _ { n } z ^ { n }$ with radius of convergence 1 and sum $f$, such that $f ( z )$ converges when $z \rightarrow 1 , | z | < 1$ and such that the series $\sum _ { n \geqslant 0 } a _ { n }$ does not converge.

Let $\sum _ { n \geqslant 0 } a _ { n } z ^ { n }$ be a power series with radius of convergence 1 and sum $f$. Let $S \in \mathbb { C }$. The purpose of this question is to prove that $$\left( \lim _ { \substack { x \rightarrow 1 ^ { - } \\ x \in \mathbb { R } } } f ( x ) = S \text { and } a _ { n } = o \left( \frac { 1 } { n } \right) \right) \Rightarrow \left( \sum _ { n \geqslant 0 } a _ { n } \text { converges and } \sum _ { n = 0 } ^ { + \infty } a _ { n } = S \right) . \quad \text{(Weak Tauberian)}$$
In the rest of this question we suppose that $\lim _ { \substack { x \rightarrow 1 ^ { - } \\ x \in \mathbb { R } } } f ( x ) = S$ and that $a _ { n } = o \left( \frac { 1 } { n } \right)$.
(a) Prove that for all $n \in \mathbb { N } ^ { * }$ and $x \in \left] 0,1 \right[$, we have $$\left| S _ { n } - f ( x ) \right| \leqslant ( 1 - x ) \sum _ { k = 1 } ^ { n } k \left| a _ { k } \right| + \frac { \sup _ { k > n } \left( k \left| a _ { k } \right| \right) } { n ( 1 - x ) }$$
(b) Deduce (Weak Tauberian) by specifying $x = x _ { n } = 1 - 1 / n$ for $n \in \mathbb { N } ^ { * }$.

Let $\sum _ { n \geqslant 0 } a _ { n } z ^ { n }$ be a power series with radius of convergence 1 and sum $f$. Let $S \in \mathbb { C }$. The purpose of this question is to prove that $$\left( \lim _ { \substack { x \rightarrow 1 ^ { - } \\ x \in \mathbb { R } } } f ( x ) = S \text { and } a _ { n } = O \left( \frac { 1 } { n } \right) \right) \Rightarrow \left( \sum _ { n \geqslant 0 } a _ { n } \text { converges and } \sum _ { n = 0 } ^ { + \infty } a _ { n } = S \right) . \quad \text{(Strong Tauberian)}$$
(a) Prove that, without loss of generality, we can assume that $S = 0$.
We now suppose that $\lim _ { \substack { x \rightarrow 1 ^ { - } \\ x \in \mathbb { R } } } f ( x ) = S$ and that $a _ { n } = O \left( \frac { 1 } { n } \right)$, with $S = 0$.
(b) We define $\Theta$ as follows $$\Theta = \left\{ \theta : [ 0,1 ] \rightarrow \mathbb { R } ; \forall x \in \left[ 0,1 \left[ , \sum _ { n \geqslant 0 } a _ { n } \theta \left( x ^ { n } \right) \text { converges and } \lim _ { x \rightarrow 1 ^ { - } } \sum _ { n = 0 } ^ { + \infty } a _ { n } \theta \left( x ^ { n } \right) = 0 \right\} . \right. \right.$$ Prove that $\Theta$ is a vector space over $\mathbb { R }$.
(c) Let $P \in \mathbb { R } [ X ]$ such that $P ( 0 ) = 0$. Prove that $P \in \Theta$.
(d) Prove that $$\forall P \in \mathbb { R } [ X ] , \quad \lim _ { \substack { x \rightarrow 1 ^ { - } \\ x \in \mathbb { R } } } ( 1 - x ) \cdot \sum _ { n = 0 } ^ { + \infty } x ^ { n } P \left( x ^ { n } \right) = \int _ { 0 } ^ { 1 } P ( t ) d t$$
We define the function $g : \mathbb { R } \rightarrow \mathbb { R }$ by $$g ( x ) = \begin{cases} 1 & \text { if } x \in [ 1 / 2,1 ] \\ 0 & \text { otherwise } \end{cases}$$
(e) Prove that to establish (Strong Tauberian), it suffices to prove that $g \in \Theta$.
(f) Let $$h ( x ) = \begin{cases} - 1 & \text { if } x = 0 \\ \frac { g ( x ) - x } { x ( 1 - x ) } & \text { if } x \in ] 0,1 [ \\ 1 & \text { if } x = 1 \end{cases}$$ Given $\varepsilon > 0$, prove that there exist $s _ { 1 } , s _ { 2 } \in \mathcal { C } ^ { 0 } ( [ 0,1 ] )$ satisfying $$s _ { 1 } \leqslant h \leqslant s _ { 2 } \text { and } \int _ { 0 } ^ { 1 } \left( s _ { 2 } ( x ) - s _ { 1 } ( x ) \right) d x \leqslant \varepsilon$$ Represent graphically $h$ and two such functions $s _ { 1 } , s _ { 2 }$.
From now on, $\varepsilon > 0 , s _ { 1 }$ and $s _ { 2 }$ are fixed.
(g) Prove that there exist $T _ { 1 } , T _ { 2 } \in \mathbb { R } [ X ]$ such that $$\sup _ { x \in [ 0,1 ] } \left| T _ { 1 } ( x ) - s _ { 1 } ( x ) \right| \leqslant \varepsilon \quad \text { and } \quad \sup _ { x \in [ 0,1 ] } \left| T _ { 2 } ( x ) - s _ { 2 } ( x ) \right| \leqslant \varepsilon$$
We set, for all $x \in [ 0,1 ]$, $$P _ { 1 } ( x ) = x + x ( 1 - x ) \left( T _ { 1 } ( x ) - \varepsilon \right) , \quad P _ { 2 } ( x ) = x + x ( 1 - x ) \left( T _ { 2 } ( x ) + \varepsilon \right) \quad \text{and} \quad Q ( x ) = \frac { P _ { 2 } ( x ) - P _ { 1 } ( x ) } { x ( 1 - x ) }$$
(h) Prove that $$P _ { 1 } ( 0 ) = P _ { 2 } ( 0 ) = 0 , \quad P _ { 1 } ( 1 ) = P _ { 2 } ( 1 ) = 1 , \quad P _ { 1 } \leqslant g \leqslant P _ { 2 } \quad \text{and} \quad 0 \leqslant \int _ { 0 } ^ { 1 } Q ( x ) d x \leqslant 5 \varepsilon$$
(i) Prove that there exists $M > 0$ such that for all $x \in ] 0,1 [$, $$\left| \sum _ { n = 0 } ^ { + \infty } a _ { n } g \left( x ^ { n } \right) - \sum _ { n = 0 } ^ { + \infty } a _ { n } P _ { 1 } \left( x ^ { n } \right) \right| \leqslant M ( 1 - x ) \sum _ { n = 1 } ^ { + \infty } x ^ { n } Q \left( x ^ { n } \right)$$
(j) Conclude.

Let $f \in \mathcal { C } ^ { 0 } ( [ 0 , + \infty [ )$ and $\ell \in \mathbb { R }$. Prove that $$\left( \lim _ { x \rightarrow + \infty } f ( x ) = \ell \right) \Rightarrow \left( \lim _ { x \rightarrow + \infty } \frac { 1 } { x } \int _ { 0 } ^ { x } f ( t ) d t = \ell \right)$$

Using a counterexample, prove that the converse of the result in question 3.1 is false, i.e. that $$\left( \lim _ { x \rightarrow + \infty } \frac { 1 } { x } \int _ { 0 } ^ { x } f ( t ) d t = \ell \right) \not\Rightarrow \left( \lim _ { x \rightarrow + \infty } f ( x ) = \ell \right)$$ for $f \in \mathcal{C}^0([0,+\infty[)$.

Consider a sequence $(\varepsilon_n)_{n \in \mathbb{N}}$ of real numbers strictly greater than $-1$, convergent with limit zero. Deduce that: $$\lim_{n \rightarrow +\infty} \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}} \cdot \left(\frac{(1+\varepsilon_i)(1+\varepsilon_{n-i})}{1+\varepsilon_n} - 1\right) = 0.$$

Let $(c_n)_{n \in \mathbf{N}^*}$ and $(d_n)_{n \in \mathbf{N}^*}$ be two sequences of strictly positive real numbers such that: $c_n \underset{n \to +\infty}{\sim} d_n$ and the series $\sum_n c_n$ diverges.
We admit without proof the following result:
Theorem 1. Let $(a_n)_{n \in \mathbf{N}^*}$ and $(b_n)_{n \in \mathbf{N}^*}$ be two sequences of nonzero real numbers such that $a_n = o(b_n)$ as $n \to +\infty$ and the series $\sum_n |b_n|$ is divergent. Then: $$\sum_{k=1}^n a_k = o\!\left(\sum_{k=1}^n |b_k|\right) \text{ as } n \to +\infty.$$
By using this result, show that the series $\sum_n d_n$ is divergent and that: $$\sum_{k=1}^n c_k \underset{n \rightarrow +\infty}{\sim} \sum_{k=1}^n d_k.$$

We consider a sequence of random variables $(X_n : \Omega \longrightarrow \{-1,1\})_{n \in \mathbf{N}}$ defined on the same probability space $(\Omega, \mathscr{A}, P)$, taking values in $\{-1,1\}$, mutually independent and centered. The random variable $N_n$ counts the number of equality indices of the path $(X_1(\omega), \cdots, X_{2n}(\omega))$, and it has been shown that: $$\mathbb{E}(N_n) = \sum_{i=1}^n \frac{\binom{2i}{i}}{4^i}.$$ Deduce the equivalent: $$\mathbb{E}(N_n) \underset{n \to +\infty}{\sim} \frac{2}{\sqrt{\pi}} \sqrt{n}.$$

In an urn containing $n$ white balls and $n$ black balls, we proceed to draw balls without replacement, until the urn is completely empty. The random variable $M_n$ counts the number of equality indices $k$ between $1$ and $2n$, and it has been shown that: $$\mathbb{E}(M_n) = \sum_{i=0}^{n-1} \frac{\binom{2i}{i} \cdot \binom{2n-2i}{n-i}}{\binom{2n}{n}}.$$ Deduce the equivalent: $$\mathbb{E}(M_n) \underset{n \to +\infty}{\sim} \sqrt{\pi n}.$$

Recall the hypothesis $b_1 e^{a_1} + \cdots + b_r e^{a_r} = 0$ of Proposition 1. Define rational numbers $s_1, \ldots, s_r$ by the formula $$(T - a_1) \cdots (T - a_r) = T^r - s_1 T^{r-1} - \cdots - s_{r-1} T - s_r.$$ Show the equality: for all $n \geq 0$, $$u_{n+r} = s_1 u_{n+r-1} + \cdots + s_r u_n.$$

Let $D$ be a common denominator of the rational numbers $a_1, \ldots, a_r$ and let $A = \max(1, |a_1|, \ldots, |a_r|)$. Show that $D^n u_n \in \mathbf{Z}$ for all $n \geq 0$.

Show that there exists a real number $c_1 > 0$ such that: for all $n \geq 0$, $$|v_n| \leq c_1 \frac{A^{n+1}}{n+1}.$$

For all integers $n, k \geq 0$, define the rational number $v_n(k)$ as the coefficient of degree $n$ in the power series $$\left(1 - s_1 x - \cdots - s_r x^r\right)^k v(x) = \sum_{n=0}^{\infty} v_n(k) x^n.$$ Observe the equality: for all $n \geq r$ and $k \geq 0$, $$v_n(k+1) = v_n(k) - s_1 v_{n-1}(k) - \cdots - s_r v_{n-r}(k).$$

Let $C = 1 + |s_1| + \cdots + |s_r|$. Show that $D^n v_n(k) \in \mathbf{Z}$ and that there exists a real number $c_2 > 0$ such that $$|v_n(k)| \leq c_2 A^n C^k \text{ for all } n \geq kr.$$

Show that for all $k \in \mathbf { N }$, the real numbers $b _ { k } = \sum _ { n = 1 } ^ { + \infty } \lambda _ { n } ^ { k } a _ { n }$ are well-defined.
The sequences satisfy: $\left| a _ { n } \right| \leq \frac { M } { 2 ^ { n } }$ for some $M \in \mathbf{R}_+^*$, and $\lambda_n$ is strictly increasing with $\lambda_0 = 0$, $\lim_{n\to+\infty} \lambda_n = +\infty$, and $\lambda_n \underset{n\to+\infty}{=} O(n)$.

Show that every Dirichlet series $\sum _ { n \geq 0 } f _ { n }$ converges uniformly on $\mathbf { R } _ { + }$. We then denote its sum by $f$. Justify that $f$ is continuous on $\mathbf { R } _ { + }$.
A Dirichlet series satisfies $f_n(x) = a_n e^{-\lambda_n x}$ with $\left| a _ { n } \right| \leq \frac { M } { 2 ^ { n } }$, $\lambda_0 = 0$, $\lim_{n\to+\infty}\lambda_n = +\infty$, and $\lambda_n = O(n)$.

Express $f ( 0 )$ and $\lim _ { x \rightarrow + \infty } f ( x )$ in terms of $a _ { 0 }$ and $b _ { 0 }$.
Here $f = \sum_{n\geq 0} f_n$ is the sum of a Dirichlet series with $f_n(x) = a_n e^{-\lambda_n x}$, $\lambda_0 = 0$, and $b_0 = \sum_{n=1}^{+\infty} a_n$.

Let $k \in \mathbf { N } ^ { * }$. Show that $f \in \mathcal { C } ^ { k } \left( \mathbf { R } _ { + } , \mathbf { R } \right)$ and give an expression for $x \mapsto f ^ { ( k ) } ( x )$. Then express $f ^ { ( k ) } ( 0 )$ in terms of $b _ { k }$.
Here $f = \sum_{n\geq 0} f_n$ is the sum of a Dirichlet series with $f_n(x) = a_n e^{-\lambda_n x}$, and $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.

7. For $n \geqslant 0$ we define the vector $U _ { n } \in \mathbb { C } ^ { d }$ by $U _ { n } = \left( u _ { n } , \ldots , u _ { n + d - 1 } \right)$ (recall that $U _ { n }$ is identified with a column vector). Show that the sequence $( U _ { n } )$ satisfies a recurrence relation of the form $U _ { n + 1 } = A U _ { n } + B$, with $A \in \mathrm { M } _ { d } ( \mathbb { C } )$ and $B \in \mathbb { C } ^ { d }$ are elements that we will specify.

Show that if $f ( x ) = 0$ for all $x \in \mathbf { R } _ { + }$ then $a _ { n } = 0$ for all $n \in \mathbf { N }$.
Here $f = \sum_{n\geq 0} a_n e^{-\lambda_n x}$ is the sum of a Dirichlet series.

Suppose that $y$ is the sum of a Dirichlet series: $$\forall x \in \mathbf { R } _ { + } \quad y ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } \mathrm { e } ^ { - \lambda _ { n } x },$$ where $y(0) = 0$ and $\lim_{x\to+\infty} y(x) = c$. Express $a _ { 0 }$ and $b _ { 0 }$ in terms of the constant $c$ introduced in Part I.

Problem 2, Part 2: Linear recurrence sequences with constant coefficients
We consider a sequence $\left( u _ { n } \right) _ { n \geqslant 0 }$ of complex numbers defined by the data of $u _ { 0 } , \ldots , u _ { d }$ and the linear recurrence relation $$u _ { n + d } = \sum _ { i = 0 } ^ { d - 1 } a _ { i } u _ { n + i } + b ,$$ where the $a _ { i }$ and $b$ are complex numbers. We define $P \in \mathbb { C } [ X ]$ by $P ( X ) = X ^ { d } - \sum _ { i = 0 } ^ { d - 1 } a _ { i } X ^ { i }$ and we assume that all complex roots of $P$ have modulus strictly less than 1.
Suppose in this question that $b = 0$. Show that $(u _ { n })$ tends to 0.

9. Suppose in this question that $b = 0$. Show that $( u _ { n } )$ tends to 0.

Using equation $(E)$ satisfied by $y$, calculate $b _ { 1 }$.
The equation $(E)$ is $y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }$, and $y$ is the sum of a Dirichlet series $y(x) = \sum_{n=0}^{+\infty} a_n e^{-\lambda_n x}$ with $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.

Problem 2, Part 2: Linear recurrence sequences with constant coefficients
We consider a sequence $\left( u _ { n } \right) _ { n \geqslant 0 }$ of complex numbers defined by the data of $u _ { 0 } , \ldots , u _ { d }$ and the linear recurrence relation $$u _ { n + d } = \sum _ { i = 0 } ^ { d - 1 } a _ { i } u _ { n + i } + b ,$$ where the $a _ { i }$ and $b$ are complex numbers. We define $P \in \mathbb { C } [ X ]$ by $P ( X ) = X ^ { d } - \sum _ { i = 0 } ^ { d - 1 } a _ { i } X ^ { i }$ and we assume that all complex roots of $P$ have modulus strictly less than 1.
In the general case, show that $(u _ { n })$ converges and specify its limit.