Let $z \in \mathbb{C}$. Consider the function of the real variable $x$ $$G_z : x \mapsto \sum_{p=0}^{+\infty} \left(x^p(2z - x)^p\right)$$ Deduce (from II.C.5) that $G_z$ admits a Taylor expansion to any order at 0. We denote it $$G_z(x) = \sum_{k=0}^{n} a_k x^k + o\left(x^n\right) \quad x \to 0$$ Determine the coefficients $a_k$ for $k \in \mathbb{N}$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
Show that the sequence $(u_n)_{n\in\mathbb{N}}$ is increasing, then that it is convergent. We denote its limit by $l$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
Show that the equation $f(x)=x$ admits a smallest solution. In what follows, we denote it by $x_f$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We denote by $x_f$ the smallest solution of $f(x)=x$ and by $l$ the limit of $(u_n)$.
Show that $l=x_f$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We denote by $x_f$ the smallest solution of $f(x)=x$.
We assume $m>1$. Show that $x_f\in[0,1[$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We denote by $x_f$ the smallest solution of $f(x)=x$.
We now assume $m\leqslant 1$. Show that $x_f=1$ and that for all $n\in\mathbb{N}$, $u_n\neq 1$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Show that $\lim_{n\rightarrow+\infty}\left(\frac{1}{\varepsilon_{n+1}}-\frac{1}{\varepsilon_n}\right)=\frac{f''(1)}{2}$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Deduce that, as $n$ tends to infinity, $1-u_n\sim\frac{2}{f''(1)n}$.
One may use Cesaro's lemma: if $(a_n)_{n\in\mathbb{N}}$ is a sequence of real numbers converging to $l$ and if we set, for $n\in\mathbb{N}^*$, $b_n=\frac{1}{n}(a_1+\cdots+a_n)$, then the sequence $(b_n)_{n\geqslant 1}$ converges to $l$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
We now assume $m<1$ and we set again, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$.
Show that the series with general term $\varepsilon_n$ is absolutely convergent and deduce the convergence of the series with general term $\ln\left(\frac{m^{-(n+1)}\varepsilon_{n+1}}{m^{-n}\varepsilon_n}\right)$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
We now assume $m<1$ and we set again, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$.
Deduce that there exists $c>0$ such that, as $n$ tends to infinity, $1-u_n\sim cm^n$.

For $\lambda \neq 1$, we denote $C_{n} = \sum_{k=0}^{n} \frac{(n\lambda)^{k}}{k!}$ and $D_{n} = \sum_{k=n+1}^{+\infty} \frac{(n\lambda)^{k}}{k!}$.
Determine $\lim_{n \rightarrow +\infty} \mathrm{e}^{-n\lambda} C_{n}$ if $\lambda < 1$ and $\lim_{n \rightarrow +\infty} \mathrm{e}^{-n\lambda} D_{n}$ if $\lambda > 1$.

We assume $\lambda < 1$. Determine $\lim_{n \rightarrow +\infty} \left((n\lambda)^{-n} \int_{0}^{n\lambda} (n\lambda - t)^{n} \mathrm{e}^{t} \mathrm{~d}t\right)$.

We assume $\lambda < 1$, and $D_{n} = \sum_{k=n+1}^{+\infty} \frac{(n\lambda)^{k}}{k!}$. Using Taylor's formula with integral remainder, deduce an equivalent of $D_{n}$ when $n \rightarrow +\infty$.

If $\lambda > 1$, and $C_{n} = \sum_{k=0}^{n} \frac{(n\lambda)^{k}}{k!}$, determine an equivalent of $C_{n}$ when $n \rightarrow +\infty$.
Consider the integral $\frac{1}{n!} \int_{-\infty}^{0} (r - t)^{n} \mathrm{e}^{t} \mathrm{~d}t$ and choose the real number $r$ appropriately.

We study the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ defined by $$\forall n \in \mathbb{N}^{*}, \quad u_{n} = \int_{0}^{\infty} \frac{1 - (\cos t)^{n}}{t^{2}} \mathrm{~d}t$$ Justify the existence of the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ and specify the monotonicity of the subsequence $\left(u_{2n}\right)_{n \in \mathbb{N}^{*}}$.

We study the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ defined by $$\forall n \in \mathbb{N}^{*}, \quad u_{n} = \int_{0}^{\infty} \frac{1 - (\cos t)^{n}}{t^{2}} \mathrm{~d}t$$ We admit the relation $\int_{0}^{\infty} \frac{\mathrm{e}^{-u}}{\sqrt{u}} \mathrm{~d}u = \sqrt{\pi}$.
Conclude that $u_{n} \sim \sqrt{\frac{n\pi}{2}}$.

We consider a sequence $\left(X_{n}\right)_{n \in \mathbb{N}^{*}}$ of mutually independent random variables, taking values in $\{1, -1\}$ and such that, for all $k \in \mathbb{N}^{*}$, $$P\left(X_{k} = 1\right) = P\left(X_{k} = -1\right) = \frac{1}{2}$$ For all $n \in \mathbb{N}^{*}$, we set $S_{n} = X_{1} + \cdots + X_{n}$, and $u_{n} = \int_{0}^{\infty} \frac{1 - (\cos t)^{n}}{t^{2}} \mathrm{~d}t$.
Deduce from the previous question that, for all $n \in \mathbb{N}$, $u_{2n+1} = u_{2n+2}$.

Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum.
Let $m$, $q$ and $r$ be elements of $\mathbb{N}$. We set $n = mq + r$. Compare the two real numbers $u_{n}$ and $qu_{m} + u_{r}$ and show that $u_{n} - ns \geqslant q\left(u_{m} - ms\right) + u_{r} - rs$.

Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum.
We fix $m$ in $\mathbb{N}^{*}$ and $\varepsilon$ in $\mathbb{R}^{+*}$. Using the Euclidean division of $n$ by $m$, show that there exists an integer $N$ such that for all $n > N$, $$\frac{u_{n}}{n} \geqslant \frac{u_{m}}{m} - \varepsilon$$

Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum.
Show $\lim_{n \rightarrow \infty} \frac{u_{n}}{n} = s$.

Let $\left(u_{n}\right)_{n \in \mathbb{N}}$ be an element of $E$ whose convergence rate is of order $r$, where $r$ is a real strictly greater than 1. Show that the convergence of the sequence $\left(u_{n}\right)_{n \in \mathbb{N}}$ is fast.

Verify that a periodic sequence is bounded.

What can be said about 1-periodic sequences?

Verify that, if $\left( z _ { k } \right)$ is $p$-periodic, then $\forall n \in \mathbb { N } , \forall k \in \mathbb { N } , z _ { n + k p } = z _ { n }$.

What can be said about sequences that are both periodic and convergent?