Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. If $n \in \mathbb{N}^*$, we denote $\{n \mid X\}$ the event ``$n$ divides $X$'' and $\{n \nmid X\}$ the complementary event. Calculate $P(n \mid X)$ for $n \in \mathbb{N}^*$.
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. If $n \in \mathbb{N}^*$, we denote $\{n \mid X\}$ the event ``$n$ divides $X$'' and $\{n \nmid X\}$ the complementary event. Let $\left(\alpha_i\right)_{i \in \mathbb{N}^*}$ be a sequence of natural integers. Show that the events $$\left\{p_1^{\alpha_1} \mid X\right\}, \left\{p_2^{\alpha_2} \mid X\right\}, \ldots, \left\{p_k^{\alpha_k} \mid X\right\}, \ldots$$ are mutually independent.
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. If $n \in \mathbb{N}^*$, we denote $\{n \mid X\}$ the event ``$n$ divides $X$'' and $\{n \nmid X\}$ the complementary event. Let $r \geqslant 1$ be an integer. Show that $$P\left(\bigcap_{i=1}^{r}\left\{p_i \nmid X\right\}\right) = \prod_{i=1}^{r}\left(1 - p_i^{-s}\right).$$
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. Deduce that $$\zeta(s)^{-1} = \lim_{n \rightarrow +\infty} \prod_{k=1}^{n}\left(1 - p_k^{-s}\right).$$
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. Show that for all $k \in \mathbb{N}^*$, the random variable $\nu_{p_k}(X) + 1$ follows the geometric distribution with parameter $\left(1 - p_k^{-s}\right)$.
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. Show that, for $r \in \mathbb{N}^*$, $k_1 < \cdots < k_r$ in $\mathbb{N}^*$ and $\left(n_1, \ldots, n_r\right) \in \mathbb{N}^r$, we have $$\begin{aligned}
& P\left(\nu_{p_{k_1}}(X) = n_1, \ldots, \nu_{p_{k_r}}(X) = n_r\right) = \\
& \sum_{\ell=0}^{r}(-1)^{\ell} \sum_{\substack{\left(\varepsilon_1, \ldots, \varepsilon_r\right) \in \{0,1\}^r \\ \varepsilon_1 + \cdots + \varepsilon_r = \ell}} P\left(\nu_{p_{k_1}}(X) \geqslant n_1 + \varepsilon_1, \nu_{p_{k_2}}(X) \geqslant n_2 + \varepsilon_2, \ldots, \nu_{p_{k_r}}(X) \geqslant n_r + \varepsilon_r\right).
\end{aligned}$$
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. Deduce that the random variables $\nu_{p_1}(X), \ldots, \nu_{p_k}(X), \ldots$ are mutually independent.
If $n \in \mathbb{N}^*$, we denote, for $i \in \{0,1,2,3\}$, $$r_i(n) = \operatorname{Card}\{d \in \mathbb{N} : d \equiv i [4] \text{ and } d \mid n\}$$ We set $g(n) = r_1(n) - r_3(n)$. Show that if $m$ and $n$ are two nonzero natural integers that are coprime, we have $g(mn) = g(m)g(n)$.
If $n \in \mathbb{N}^*$, we denote, for $i \in \{0,1,2,3\}$, $$r_i(n) = \operatorname{Card}\{d \in \mathbb{N} : d \equiv i [4] \text{ and } d \mid n\}$$ We set $g(n) = r_1(n) - r_3(n)$. Show that, for all $n \in \mathbb{N}$, and all prime number $p$, we have $$g\left(p^n\right) = \begin{cases} 1 & \text{if } p = 2 \\ n+1 & \text{if } p \equiv 1 [4] \\ \frac{1}{2}\left(1+(-1)^n\right) & \text{if } p \equiv 3 [4] \end{cases}$$
Let $\left(f_n\right)_{n \geqslant 1}$ be a sequence of functions from $\mathbb{N}^*$ to $\mathbb{R}$ such that, for all $x \in \mathbb{N}^*$, the sequence $\left(f_n(x)\right)_{n \geqslant 1}$ converges to a real number $f(x)$ as $n$ tends to $+\infty$. We assume that there exists a function $h : \mathbb{N}^* \rightarrow [0, +\infty[$ such that $h(X)$ has finite expectation and such that $\left|f_n(m)\right| \leqslant h(m)$ for all $m$ and $n$ in $\mathbb{N}^*$. Justify that $E(f(X))$ has finite expectation and show that $$\lim_{n \rightarrow +\infty} E\left(f_n(X)\right) = E(f(X)).$$
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. We denote $r(n)$ the number of divisors $d \geqslant 1$ of $n$. Show that the series $\sum_{n=1}^{+\infty} r(n) n^{-s}$ converges and that its sum equals $\zeta(s)^2$.
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. If $n \in \mathbb{N}^*$, we set $g(n) = r_1(n) - r_3(n)$ where $r_i(n) = \operatorname{Card}\{d \in \mathbb{N} : d \equiv i [4] \text{ and } d \mid n\}$. Deduce that the series $\sum_{n=1}^{+\infty} g(n) n^{-s}$ converges.
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. Show that the sequence of functions $\left(x \mapsto \prod_{k=1}^{n} p_k^{\nu_{p_k}(x)}\right)_{n \geqslant 1}$ from $\mathbb{N}^*$ to $\mathbb{N}^*$ converges pointwise to the identity function.
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. If $n \in \mathbb{N}^*$, we set $g(n) = r_1(n) - r_3(n)$ where $r_i(n) = \operatorname{Card}\{d \in \mathbb{N} : d \equiv i [4] \text{ and } d \mid n\}$. Show that $E(g(X)) = \lim_{n \rightarrow +\infty} \prod_{k=1}^{n} E\left(g\left(p_k^{\nu_{p_k}(X)}\right)\right)$.
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. Deduce $$E(g(X)) = \lim_{n \rightarrow +\infty} \prod_{k=1}^{n} \frac{1}{1 - \chi_4\left(p_k\right) p_k^{-s}}.$$
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. We recall that $\chi_4(2n) = 0$ and $\chi_4(2n-1) = (-1)^{n-1}$ for $n \in \mathbb{N}^*$. Show that, if $p$ is a prime number, $$E\left(\chi_4\left(p^{\nu_p(X)}\right)\right) = \frac{1 - p^{-s}}{1 - \chi_4(p) p^{-s}}.$$
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. We recall that $\chi_4(2n) = 0$ and $\chi_4(2n-1) = (-1)^{n-1}$ for $n \in \mathbb{N}^*$. Show that $$E\left(\chi_4(X)\right) = \frac{1}{\zeta(s)} \lim_{n \rightarrow +\infty} \prod_{k=1}^{n} \frac{1}{1 - \chi_4\left(p_k\right) p_k^{-s}}.$$
Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$. We recall that $\chi_4(2n) = 0$ and $\chi_4(2n-1) = (-1)^{n-1}$ for $n \in \mathbb{N}^*$. We set $g(n) = r_1(n) - r_3(n)$ where $r_i(n) = \operatorname{Card}\{d \in \mathbb{N} : d \equiv i [4] \text{ and } d \mid n\}$. Deduce that the series $$\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^s}$$ is convergent and that its sum equals $E(g(X))$.
Let $n \in \mathbb{N}$. Explicitly give a polynomial $P_n \in \mathbb{R}[X]$ such that, for all $\theta \in \mathbb{R}$, $$\sin((2n+1)\theta) = \sin(\theta) P_n\left(\sin^2(\theta)\right).$$ Hint: you may expand $(\cos(\theta) + i\sin(\theta))^{2n+1}$.
Let $n \in \mathbb{N}$ and let $P_n \in \mathbb{R}[X]$ be such that, for all $\theta \in \mathbb{R}$, $\sin((2n+1)\theta) = \sin(\theta) P_n\left(\sin^2(\theta)\right)$. Determine the roots of $P_n$ and deduce that, for all $x \in \mathbb{R}$, $$P_n(x) = (2n+1) \prod_{k=1}^{n}\left(1 - \frac{x}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$
Let $n \in \mathbb{N}$ and let $P_n \in \mathbb{R}[X]$ be such that, for all $\theta \in \mathbb{R}$, $\sin((2n+1)\theta) = \sin(\theta) P_n\left(\sin^2(\theta)\right)$, and $$P_n(x) = (2n+1) \prod_{k=1}^{n}\left(1 - \frac{x}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$ Deduce that, for all $x \in \mathbb{R}$, $$\sin(\pi x) = (2n+1)\sin\left(\frac{\pi x}{2n+1}\right) \prod_{k=1}^{n}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$
Let $x \in \mathbb{R} \backslash \mathbb{Z}$. Let $m \in \mathbb{N}$ such that $m > |x|$. We set, for $n \in \mathbb{N}$ such that $n > m$: $$u_{m,n}(x) = (2n+1)\sin\left(\frac{\pi x}{2n+1}\right) \prod_{k=1}^{m}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right)$$ and $$v_{m,n}(x) = \prod_{k=m+1}^{n}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$ Show that the sequences, indexed by $n$, $\left(u_{m,n}(x)\right)_{n > m}$ and $\left(v_{m,n}(x)\right)_{n > m}$ are convergent in $\mathbb{R}^*$.
Let $x \in \mathbb{R} \backslash \mathbb{Z}$. Let $m \in \mathbb{N}$ such that $m > |x|$. We set, for $n \in \mathbb{N}$ such that $n > m$: $$v_{m,n}(x) = \prod_{k=m+1}^{n}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$ We denote $v_m(x)$ the limit of $\left(v_{m,n}(x)\right)_{n > m}$. Show that, for $n \in \mathbb{N}$ such that $n > m$, we have $$1 \geqslant v_{m,n}(x) \geqslant \prod_{k=m+1}^{n}\left(1 - \frac{\pi^2 x^2}{4k^2}\right)$$ and deduce that $\lim_{m \rightarrow +\infty} v_m(x) = 1$.
We recall that the sequence $\left(\left(\sum_{k=1}^{n} k^{-1}\right) - \ln(n)\right)_{n \geqslant 1}$ converges. We denote $\gamma$ its limit. Let $n \in \mathbb{N}^*$. For $x \in ]0, +\infty[$, we set $$\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}.$$ Show that the sequence of functions $\left(\Gamma_n\right)_{n \geqslant 1}$ converges pointwise on $]0, +\infty[$ to a function $\Gamma$ from $]0, +\infty[$ to $]0, +\infty[$.
Let $\Gamma$ be the pointwise limit on $]0, +\infty[$ of the sequence $\left(\Gamma_n\right)_{n \geqslant 1}$ where $$\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}.$$ Show that, for all $x \in ]0, +\infty[$, we have $\Gamma(x+1) = x\Gamma(x)$.
Let $\Gamma$ be the pointwise limit on $]0, +\infty[$ of the sequence $\left(\Gamma_n\right)_{n \geqslant 1}$ where $$\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}.$$ Show that the function $\Gamma$ is of class $\mathscr{C}^2$ and that, for all $x \in ]0, +\infty[$, $$(\ln(\Gamma))''(x) = \sum_{k=0}^{+\infty} \frac{1}{(x+k)^2}.$$
Let $\Gamma$ be the pointwise limit on $]0, +\infty[$ of the sequence $\left(\Gamma_n\right)_{n \geqslant 1}$ where $$\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}.$$ Show that $\lim_{x \rightarrow +\infty} (\ln(\Gamma))''(x) = 0$.
Let $\Gamma$ be the pointwise limit on $]0, +\infty[$ of the sequence $\left(\Gamma_n\right)_{n \geqslant 1}$ where $$\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}.$$ Let $f : ]0, +\infty[ \rightarrow ]0, +\infty[$ be a function of class $\mathscr{C}^2$ such that the function $\ln(f)$ is convex and satisfies $f(1) = 1$ and $f(x+1) = xf(x)$ for all $x > 0$. The function $S(x) = \ln\left(\frac{f(x)}{\Gamma(x)}\right)$ is 1-periodic and convex. Deduce that $f = \Gamma$.
Let $\Gamma$ be the Gamma function defined as the pointwise limit on $]0, +\infty[$ of $\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}$. Show that for all $a \in ]0, +\infty[$ and $x \in ]0, +\infty[$: $$\int_0^{+\infty} \frac{t^{x-1}}{(1+t)^{x+a}} dt = \frac{\Gamma(x)\Gamma(a)}{\Gamma(x+a)}.$$ Hint: you may set, for $x \in ]0, +\infty[$, $f(x) = \frac{\Gamma(x+a)}{\Gamma(a)} \int_0^{+\infty} \frac{t^{x-1}}{(1+t)^{x+a}} dt$.
Using the result that for all $x \in ]0,1[$: $$\frac{\pi}{\sin(\pi x)} = \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+x} + \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+1-x},$$ deduce that, for $x \in ]-\frac{1}{2}, \frac{1}{2}[$: $$\frac{\pi}{\cos(\pi x)} = \sum_{k=0}^{+\infty} \left(\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}}\right) 2^{2k+2} x^{2k}.$$
Using the result that for $x \in ]-\frac{1}{2}, \frac{1}{2}[$: $$\frac{\pi}{\cos(\pi x)} = \sum_{k=0}^{+\infty} \left(\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}}\right) 2^{2k+2} x^{2k},$$ deduce that the function $$v : \begin{array}{ccc} ]-\frac{\pi}{2}, \frac{\pi}{2}[ & \longrightarrow & \mathbb{R} \\ x & \longmapsto & \frac{1}{\cos(x)} \end{array}$$ is expandable as a power series and that, for all $k \in \mathbb{N}$, $$\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}} = \frac{\pi^{2k+1}}{2^{2k+2}(2k)!} E_{2k}$$ where, for all $k \in \mathbb{N}$, $E_{2k} = v^{(2k)}(0)$.
Let $v(x) = \frac{1}{\cos(x)}$ on $]-\frac{\pi}{2}, \frac{\pi}{2}[$ and $E_{2k} = v^{(2k)}(0)$ for $k \in \mathbb{N}$. Show that, for $n \in \mathbb{N}^*$, $$\sum_{k=0}^{n} (-1)^k \binom{2n}{2k} E_{2k} = 0$$ and deduce the values of $E_0$, $E_2$ and $E_4$.