Show $$\forall n \in \mathbb{N}^{\star}, \forall s \in ]1, +\infty[, \quad \sum_{k=n+1}^{+\infty} \frac{1}{(2k-1)^s} \leqslant \frac{1}{2(s-1)} \frac{1}{(2n-1)^{s-1}}.$$

For every natural integer $n$ and every real $x$ in $J = [0, 1/2[$, set $$S_n(x) = \sum_{p=1}^{+\infty} \left(\sum_{k=n+1}^{+\infty} \frac{2^{2p+1} x^{2p-1}}{(2k-1)^{2p}}\right).$$ Justify that, for every natural integer $n$, the function $S_n$ is defined on $J$.

In the general model of a Pólya urn ($b = c = 0$, $a = d$), the function $G$ is defined on $U$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ and admits the expansion $G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$ on $D_{\rho}$.
Justify that $G$ 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 $G$.

For every natural integer $n$ and every real $x$ in $J = [0, 1/2[$, set $$S_n(x) = \sum_{p=1}^{+\infty} \left(\sum_{k=n+1}^{+\infty} \frac{2^{2p+1} x^{2p-1}}{(2k-1)^{2p}}\right).$$ Show that the sequence $(S_n)$ converges pointwise on $J$ to the zero function.

In the general model of a Pólya urn ($b = c = 0$, $a = d$), the function $G$ is defined on $U$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ and admits the expansion $G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$ on $D_{\rho}$.
Prove that $G$ admits a first-order partial derivative with respect to $u$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $u$ of the expression for $G$.

For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$. By differentiating $x \mapsto \ln(\cos(\pi x))$, show $$\forall x \in J, \quad \pi \tan(\pi x) = -\frac{2I_{4n}^{\prime}(2x)}{I_{4n}(2x)} + \frac{I_{2n}^{\prime}(x)}{I_{2n}(x)} + \sum_{k=1}^{n} \frac{8x}{(2k-1)^2} \frac{1}{1 - \frac{4x^2}{(2k-1)^2}}.$$

In the general model of a Pólya urn ($b = c = 0$, $a = d$), the function $G$ is defined on $U$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ and admits the expansion $G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$ on $D_{\rho}$. The function $H(x,u,v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$ was defined in part III.
Deduce that, for all integers $n$, $P_{n} = Q_{n}$, and then that $H$ and $G$ coincide on $D_{\rho}$.

In Friedman's urn model ($a_{0} = 1, b_{0} = 0, a = d = 0, b = c = 1$), we have $\operatorname{card}(\Omega_{n}) = n!$. For $0 < u < v$ and $|x|$ sufficiently small, $$H(x,u,v) = \sum_{n=0}^{+\infty} \frac{x^{n}}{n!} \left( \sum_{p=1}^{+\infty} p^{n} (v-u)^{n+1} \left(\frac{u}{v}\right)^{p} \right)$$
Using question 6, justify that, for all integers $n$ and all $u$ and $v$ such that $0 < u < v$, the sum $$\sum_{p=1}^{+\infty} p^{n} (v-u)^{n+1} \left(\frac{u}{v}\right)^{p}$$ is a polynomial function of $u$ and $v$.

Using the power series expansion $\tan(x) = \sum_{n=0}^{+\infty} \frac{\alpha_{2n+1}}{(2n+1)!} x^{2n+1}$ and the formula $\pi \tan(\pi x) = \sum_{p=1}^{+\infty} 2(2^{2p}-1)\zeta(2p) x^{2p-1}$, show $$\forall n \in \mathbb{N}, \quad \alpha_{2n+1} = \frac{2\left(2^{2n+2} - 1\right)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2).$$

In Friedman's urn model ($a_{0} = 1, b_{0} = 0, a = d = 0, b = c = 1$), we have for all integers $n$ and all $t \in ]0,1[$, $$g_{n}(t) = \frac{1}{n!} \sum_{p=1}^{+\infty} p^{n} t^{p} (1-t)^{n+1}.$$ Fix an integer $n \geqslant 2$.
Show that $\sum_{p=n+1}^{+\infty} p^{n} t^{p} (1-t)^{n+1} \underset{t \rightarrow 0^{+}}{=} O(t^{n+1})$.

Using the result $\alpha_{2n+1} = \frac{2(2^{2n+2}-1)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2)$ and the fact that $\lim_{s \to +\infty} \zeta(s) = 1$, deduce an equivalent of $\alpha_{2n+1}$ as $n$ tends to infinity.

In Friedman's urn model ($a_{0} = 1, b_{0} = 0, a = d = 0, b = c = 1$), we have for all integers $n$ and all $t \in ]0,1[$, $$g_{n}(t) = \frac{1}{n!} \sum_{p=1}^{+\infty} p^{n} t^{p} (1-t)^{n+1}.$$ Fix an integer $n \geqslant 2$.
Using the result of question 30 and by expanding $(1-t)^{n+1}$, determine the Taylor expansion of $g_{n}$ to order $n$ at 0.

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
Calculate $y_{n,1}$.

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
We propose to show that, if $2 \leqslant j \leqslant n$, then $y_{n,j} = y_{n,j-1} + y_{n-j,\min(j,n-j)}$.
Prove that this equality is true for $j = n$.

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
We propose to show that, if $2 \leqslant j \leqslant n$, then $y_{n,j} = y_{n,j-1} + y_{n-j,\min(j,n-j)}$.
For $j < n$, verify that $y_{n,j} = y_{n,j-1} + y_{n-j,j}$. Conclude.

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
Calculate the $y_{n,j}$ for $1 \leqslant j \leqslant n \leqslant 5$ by presenting the results in the form of a table.

For $j \in \mathbb{N}$, we denote by $Y_{n,j}$ the set of partitions whose first term $\alpha_1$ is less than or equal to $j$ and by $y_{n,j}$ the cardinality of $Y_{n,j}$; we set $y_{0,0} = 1$.
Write a Python function that takes as argument an integer $n \geqslant 1$ and returns $y_{n,n}$.

Compare the result of question 48 (the value of $y_{n,n}$, the number of partitions of $n$) to that of question 40 (the maximum cardinality of a set of pairwise non-similar nilpotent matrices of size $n$).

1. We denote $E_n = \operatorname{Card} \operatorname{MD}(n)$ and $\mathscr{I}_n$ the set of odd numbers in $\Delta_n$. a. Prove that for $n \geq 1$: $$E_{n+1} = \sum_{i \in \mathscr{I}_{n+1}} \binom{n}{i-1} E_{i-1} E_{n+1-i}$$ b. Deduce that for $n \geq 1$: $$2E_{n+1} = \sum_{i=0}^{n} \binom{n}{i} E_i E_{n-i}$$

Recall Stirling's formula, then determine a real number $c > 0$ such that $$\binom { 2 n } { n } \underset { n \rightarrow + \infty } { \sim } c \frac { 4 ^ { n } } { \sqrt { n } }$$

If $\alpha$ is an element of $]0,1[$, show, for example by using a series-integral comparison, that $$\sum _ { k = 1 } ^ { n } \frac { 1 } { k ^ { \alpha } } \underset { n \rightarrow + \infty } { \sim } \frac { n ^ { 1 - \alpha } } { 1 - \alpha }$$ If $\alpha$ is an element of $]1 , + \infty[$, show similarly that $$\sum _ { k = n + 1 } ^ { + \infty } \frac { 1 } { k ^ { \alpha } } \underset { n \rightarrow + \infty } { \sim } \frac { 1 } { ( \alpha - 1 ) n ^ { \alpha - 1 } }$$

Show that the series of functions $\sum u_k$ where for all $k \in \mathbb{N}^*$, the function $u_k$ is defined on $[0, +\infty[$ by $u_k : x \mapsto (1 + kx)^{-k}(1/2)^k$ is normally convergent on $[0, +\infty[$.

Let $P ( X )$ be a monic polynomial of degree $d \geqslant 1$ with complex coefficients which we write in the form: $$P ( X ) = a _ { 0 } + a _ { 1 } X + a _ { 2 } X ^ { 2 } + \cdots + a _ { d - 1 } X ^ { d - 1 } + X ^ { d }$$ We assume that $a _ { 0 } \neq 0$. We denote by $\lambda _ { 1 } , \ldots , \lambda _ { d } \in \mathbb { C }$ the roots of $P ( X )$ (with multiplicity). For all integers $n \geqslant 1$, we define: $$N _ { n } = \lambda _ { 1 } ^ { n } + \lambda _ { 2 } ^ { n } + \cdots + \lambda _ { d } ^ { n }$$
We define the function $f : \mathbb { R } \backslash \left( \mathbb { R } \cap \left\{ \frac { 1 } { \lambda _ { 1 } } , \ldots , \frac { 1 } { \lambda _ { d } } \right\} \right) \rightarrow \mathbb { C }$ by $f ( x ) = \frac { Q ^ { \prime } ( x ) } { Q ( x ) }$.
Show that there exists $r > 0$ such that $f$ is expandable as a power series on $] - r , r [$, and that the power series expansion of $f$ at 0 is written: $$\forall x \in ] - r , r \left[ , \quad f ( x ) = - \sum _ { n = 0 } ^ { \infty } N _ { n + 1 } x ^ { n } \right.$$

Let $P ( X )$ be a monic polynomial of degree $d \geqslant 1$ with complex coefficients which we write in the form: $$P ( X ) = a _ { 0 } + a _ { 1 } X + a _ { 2 } X ^ { 2 } + \cdots + a _ { d - 1 } X ^ { d - 1 } + X ^ { d }$$ We assume that $a _ { 0 } \neq 0$. We denote by $\lambda _ { 1 } , \ldots , \lambda _ { d } \in \mathbb { C }$ the roots of $P ( X )$ (with multiplicity). For all integers $n \geqslant 1$, we define: $$N _ { n } = \lambda _ { 1 } ^ { n } + \lambda _ { 2 } ^ { n } + \cdots + \lambda _ { d } ^ { n }$$ Let $Q(X)$ be the reciprocal polynomial of $P(X)$ defined by $Q(X) = X^d P\left(\frac{1}{X}\right)$.
We define the function $f : \mathbb { R } \backslash \left( \mathbb { R } \cap \left\{ \frac { 1 } { \lambda _ { 1 } } , \ldots , \frac { 1 } { \lambda _ { d } } \right\} \right) \rightarrow \mathbb { C }$ by $f ( x ) = \frac { Q ^ { \prime } ( x ) } { Q ( x ) }$.
Show that there exists $r > 0$ such that $f$ is expandable as a power series on $] - r , r [$, and that the power series expansion of $f$ at 0 is written as: $$\forall x \in ] - r , r \left[ , \quad f ( x ) = - \sum _ { n = 0 } ^ { \infty } N _ { n + 1 } x ^ { n } \right.$$

Deduce that for all $n \in \mathbb{N}^*, \sum_{k \geqslant 1} \left(1 + \frac{k}{n}\right)^{-k} \left(\frac{1}{2}\right)^k$ converges and that $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{\infty} \left(1 + \frac{k}{n}\right)^{-k} \left(\frac{1}{2}\right)^k = 1.$$