Recall the definition of the Cauchy product of two power series and state the theorem relating to it.

Using the Cauchy product of power series, deduce that, for all integers $n$ and all real numbers $\alpha$ and $\beta$, $$L_{n}(\alpha + \beta) = \sum_{k=0}^{n} \binom{n}{k} L_{k}(\alpha) L_{n-k}(\beta).$$

For $x \in ]-1,1[$, give the value of the sum of the power series $\sum_{p=1}^{+\infty} x^{p}$ as well as that of its derivative.

Using the result of Q6, deduce the lower bound $R \geqslant \pi/2$ for the radius of convergence $R$ of the power series $\sum_{n \in \mathbb{N}} \frac{\alpha_n}{n!} x^n$.

We have $D_n = \left\{\sum_{j=1}^{n} \frac{x_j}{2^j}, (x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n\right\}$ and $D = \bigcup_{n \in \mathbb{N}^{\star}} D_n$.
Establish the monotonicity in the sense of inclusion of the sequence $(D_n)_{n \geqslant 1}$ and then verify $D \subset [0,1[$.

We denote $$D_n = \left\{ \sum_{j=1}^{n} \frac{x_j}{2^j},\, (x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n \right\} \quad \text{and} \quad D = \bigcup_{n \in \mathbb{N}^{\star}} D_n.$$
Establish the monotonicity in the sense of inclusion of the sequence $(D_n)_{n \geqslant 1}$ then verify $D \subset [0,1[$.

We have $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$ for all $(x,n) \in \mathbb{R} \times \mathbb{N}$.
Establish $$\forall (x,n) \in \mathbb{R} \times \mathbb{N}, \quad \pi_n(x) \leqslant x < \pi_n(x) + \frac{1}{2^n}.$$

For all $(x,n) \in \mathbb{R} \times \mathbb{N}$, we define $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$.
Establish $$\forall (x,n) \in \mathbb{R} \times \mathbb{N}, \quad \pi_n(x) \leqslant x < \pi_n(x) + \frac{1}{2^n}.$$

We have $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$ and $d_{n+1}(x) = 2^{n+1}(\pi_{n+1}(x) - \pi_n(x))$ for all $(x,n) \in \mathbb{R} \times \mathbb{N}$.
Justify $$\forall x \in [0,1[, \forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{k} \frac{d_j(x)}{2^j}.$$

For all $(x,n) \in \mathbb{R} \times \mathbb{N}$, we define $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$ and $d_{n+1}(x) = 2^{n+1}(\pi_{n+1}(x) - \pi_n(x))$.
Justify $$\forall x \in [0,1[,\, \forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{k} \frac{d_j(x)}{2^j}.$$

We have $d_{n+1}(x) = 2^{n+1}(\pi_{n+1}(x) - \pi_n(x))$ for all $(x,n) \in \mathbb{R} \times \mathbb{N}$.
Establish $$\forall (x,j) \in \mathbb{R} \times \mathbb{N}^{\star}, \quad d_j(x) \in \{0,1\}.$$

For all $(x,n) \in \mathbb{R} \times \mathbb{N}$, we define $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$ and $d_{n+1}(x) = 2^{n+1}(\pi_{n+1}(x) - \pi_n(x))$.
Establish $$\forall (x,j) \in \mathbb{R} \times \mathbb{N}^{\star}, \quad d_j(x) \in \{0,1\}.$$

We have $D_n = \left\{\sum_{j=1}^{n} \frac{x_j}{2^j}, (x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n\right\}$ and $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$.
Let $n \in \mathbb{N}^{\star}$. Justify $x \in D_n \Longleftrightarrow 2^n x \in \llbracket 0, 2^n - 1 \rrbracket$.

For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Show that $\zeta$ is continuous on $]1, +\infty[$.

We denote $D_n = \left\{ \sum_{j=1}^{n} \frac{x_j}{2^j},\, (x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n \right\}$ and $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$.
Let $n \in \mathbb{N}^{\star}$. Justify $x \in D_n \Longleftrightarrow 2^n x \in \llbracket 0, 2^n - 1 \rrbracket$.

Let $n \in \mathbb{N}^{\star}$. Show that the application $$\Psi_n : \begin{gathered} \{0,1\}^n \rightarrow D_n \\ (x_j)_{j \in \llbracket 1,n \rrbracket} \mapsto \sum_{j=1}^{n} \frac{x_j}{2^j} \end{gathered}$$ is bijective.

For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Bound $\sum_{n=2}^{+\infty} \frac{1}{n^s}$ by two integrals and deduce $\lim_{s \rightarrow +\infty} \zeta(s) = 1$.

We set $L_n = \lfloor n^{1/3} \rfloor$, where $\lfloor t \rfloor$ denotes the integer part of a real number $t$. Let $x, y \in \{0,1\}^n$ such that $x \neq y$. Set for $z \in \mathbb{C}$, $A_{x,y}(z) = \sum_{k=0}^{n-1}(x_k - y_k)z^k$.
a. Justify the existence of $\theta_0 \in \left[-\frac{\pi}{L_n}, \frac{\pi}{L_n}\right]$ such that $\left|A_{x,y}\left(e^{i\theta_0}\right)\right| \geq \frac{1}{n^{L_n - 1}}$.
b. Prove that $$\sum_{j=0}^{n-1} \left|\mathbb{E}\left[O_j(x)\right] - \mathbb{E}\left[O_j(y)\right]\right| \cdot \left|\frac{e^{i\theta_0} - (1-p)}{p}\right|^j \geq \frac{p}{n^{L_n - 1}}$$
c. Justify the existence of an integer $j_n(x,y)$ such that $0 \leq j_n(x,y) \leq n-1$ and $$\left|\mathbb{E}\left[O_{j_n(x,y)}(x)\right] - \mathbb{E}\left[O_{j_n(x,y)}(y)\right]\right| \geq \frac{p}{n^{L_n}} \exp\left(-\frac{1-p}{2p^2} \cdot \frac{\pi^2}{L_n^2} n\right).$$

We consider a general balanced urn with parameters $a_{0}, b_{0}, a, b, c, d \in \mathbb{N}$ satisfying $a + b = c + d = s$. For all real $u$ and $v$, we set $P_{0}(u,v) = u^{a_{0}} v^{b_{0}}$ and $P_{n}(u,v) = \sum_{\omega \in \Omega_{n}} u^{b(\omega)} v^{n(\omega)}$.
Prove that, for all integers $n$, $$P_{n+1}(u,v) = u^{a+1} v^{b} \frac{\partial P_{n}}{\partial u}(u,v) + u^{c} v^{d+1} \frac{\partial P_{n}}{\partial v}(u,v)$$

Let $n \in \mathbb{N}^{\star}$ and $x = \sum_{j=1}^{n} \frac{x_j}{2^j}$ with $(x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n$. Show $$\forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{\min(n,k)} \frac{x_j}{2^j}.$$

For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Determine $C(s)$ such that $$\forall s \in ]1, +\infty[, \quad \sum_{k=1}^{+\infty} \frac{1}{(2k-1)^s} = C(s) \zeta(s).$$

We consider a general balanced urn. For all real $x, u$ and $v$, we set, provided it exists, $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ Let $\rho > 0$. We set $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2} = \{(x,u,v) \in \mathbb{R}^{3} ; |x| < \rho, 0 < u < 2, 0 < v < 2\}$.
Justify that, for $\rho$ sufficiently small, the function $H$ is well defined on $D_{\rho}$.

We denote $\pi_k(x) = \frac{\lfloor 2^k x \rfloor}{2^k}$.
Let $n \in \mathbb{N}^{\star}$ and $x = \sum_{j=1}^{n} \frac{x_j}{2^j}$ with $(x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n$. Show $$\forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{\min(n,k)} \frac{x_j}{2^j}.$$

We consider a general balanced urn. For all real $x, u$ and $v$, we set $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ defined on $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$ for $\rho$ sufficiently small.
Justify that $H$ admits a first-order partial derivative with respect to $x$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $x$ of the expression for $H$.

Let $k$ be a non-negative integer. We define the function $f _ { k }$ from the interval $[ - 1,1 ]$ to itself by $f _ { k } ( x ) = \cos ( k \arccos x )$.
a) Expand the expression $f _ { k + 1 } ( x ) + f _ { k - 1 } ( x )$, and deduce the relation $$\forall x \in [ - 1,1 ] \quad f _ { k + 1 } ( x ) = 2 x f _ { k } ( x ) - f _ { k - 1 } ( x )$$
b) Deduce that $f _ { k }$ identifies on $[ - 1,1 ]$ with a polynomial $T _ { k }$, of degree $k$, with the same parity as $k$.