For every integer $d\geq 2$, we denote by $\mathcal{P}(d)$ the set of prime numbers dividing $d$.
Show the inequality $$d \geq |\mathcal{P}(d)|!$$ Deduce that $$|\mathcal{P}(d)| = \underset{d\rightarrow+\infty}{o}(\log(d)).$$

The objective of this question is to prove that if $n$ is a non-zero natural integer, then $\prod_{\substack{p \leqslant n \\ p \text{ prime}}} p \leqslant 4^{n}$.
Handle the cases $n \in \{1, 2, 3\}$.

The objective of this question is to prove that if $n$ is a non-zero natural integer, then $\prod_{\substack{p \leqslant n \\ p \text{ prime}}} p \leqslant 4^{n}$.
We now assume $n \geqslant 4$ and the result is known at rank $k$ for any integer $k$ between 1 and $n-1$. Establish the result at rank $n$ if $n$ is even.

The objective of this question is to prove that if $n$ is a non-zero natural integer, then $\prod_{\substack{p \leqslant n \\ p \text{ prime}}} p \leqslant 4^{n}$.
We now assume $n \geqslant 4$ and the result is known at rank $k$ for any integer $k$ between 1 and $n-1$. Let $n = 2m+1$ with $m \in \mathbb{N}$. Justify that $\prod_{\substack{m+1 < p \leqslant 2m+1 \\ p \text{ prime}}} p$ divides $\binom{2m+1}{m}$ and show that $\binom{2m+1}{m} \leqslant 4^{m}$.

The objective of this question is to prove that if $n$ is a non-zero natural integer, then $\prod_{\substack{p \leqslant n \\ p \text{ prime}}} p \leqslant 4^{n}$.
Conclude.

Justify that $n! = \prod_{\substack{p \leqslant n \\ p \text{ prime}}} p^{\nu_{p}(n!)}$ and deduce that $$n \sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p} - n\ln(4) < \ln(n!) \leqslant n \sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p} + n \sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p(p-1)}.$$

Conclude that $\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p} \underset{n \rightarrow +\infty}{=} \ln(n) + O(1)$.

For all real $t \geqslant 2$, we define $$R(t) = \sum_{\substack{p \leqslant t \\ p \text{ prime}}} \frac{\ln(p)}{p} - \ln(t)$$ Show, using the result from question 16, that $$\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{1}{p} = 1 + \ln_{2}(n) - \ln_{2}(2) + \frac{R(n)}{\ln(n)} + \int_{2}^{n} \frac{R(t)}{t(\ln(t))^{2}} \mathrm{~d}t$$

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Show that $$\left(\sum_{\substack{p_{1}, p_{2} \leqslant x \\ p_{1} \neq p_{2} \text{ prime}}} \operatorname{Card}\left\{n \in \mathbb{N}^{*} : n \leqslant x, p_{1} \mid n \text{ and } p_{2} \mid n\right\}\right) - x\ln_{2}(x)^{2} \underset{x \rightarrow +\infty}{=} O\left(x\ln_{2}(x)\right)$$ One may estimate the cardinality of the set of pairs of prime numbers $(p_{1}, p_{2})$ such that $p_{1} p_{2} \leqslant x$ as $x$ tends to $+\infty$.

The objective of this question is to prove that if $n$ is a non-zero natural integer, then $\prod_{\substack{p \leqslant n \\ p \text{ prime}}} p \leqslant 4^n$.
Handle the cases $n \in \{1, 2, 3\}$.

The objective of this question is to prove that if $n$ is a non-zero natural integer, then $\prod_{\substack{p \leqslant n \\ p \text{ prime}}} p \leqslant 4^n$.
We assume $n \geqslant 4$ and the result is known at rank $k$ for any integer $k$ between 1 and $n-1$. Establish the result at rank $n$ if $n$ is even.

The objective of this question is to prove that if $n$ is a non-zero natural integer, then $\prod_{\substack{p \leqslant n \\ p \text{ prime}}} p \leqslant 4^n$.
We assume $n \geqslant 4$ and the result is known at rank $k$ for any integer $k$ between 1 and $n-1$. Let $n = 2m+1$ with $m \in \mathbb{N}$. Justify that $\prod_{\substack{m+1 < p < 2m+1 \\ p \text{ prime}}} p$ divides $\binom{2m+1}{m}$ and show that $\binom{2m+1}{m} \leqslant 4^m$.

The objective of this question is to prove that if $n$ is a non-zero natural integer, then $\prod_{\substack{p \leqslant n \\ p \text{ prime}}} p \leqslant 4^n$.
Conclude.

Justify that $n! = \prod_{\substack{p \leqslant n \\ p \text{ prime}}} p^{\nu_p(n!)}$ and deduce that $$n \sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p} - n\ln(4) < \ln(n!) \leqslant n \sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p} + n \sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p(p-1)}.$$

Conclude that $\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{\ln(p)}{p} \underset{n \rightarrow +\infty}{=} \ln(n) + O(1)$.

We set, for all real $t \geqslant 2$, $$R(t) = \sum_{\substack{p \leqslant t \\ p \text{ prime}}} \frac{\ln(p)}{p} - \ln(t)$$ Show, using the result from question 16, that $$\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{1}{p} = 1 + \ln_2(n) - \ln_2(2) + \frac{R(n)}{\ln(n)} + \int_2^n \frac{R(t)}{t(\ln(t))^2} \, \mathrm{d}t$$

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Show that $$\left(\sum_{\substack{p_1, p_2 \leqslant x \\ p_1 \neq p_2 \text{ prime}}} \operatorname{Card}\left\{n \in \mathbb{N}^* : n \leqslant x, p_1 \mid n \text{ and } p_2 \mid n\right\}\right) - x\ln_2(x)^2 \underset{x \rightarrow +\infty}{=} O\left(x\ln_2(x)\right)$$ One may estimate the cardinality of the set of pairs of prime numbers $(p_1, p_2)$ such that $p_1 p_2 \leqslant x$ when $x$ tends to $+\infty$.

Let $n \in \mathbb { N } ^ { * }$. Show that
$$\prod _ { \substack { n + 2 \leqslant p \leqslant 2 n + 1 \\ p \text { prime } } } p \leqslant \binom { 2 n + 1 } { n } \leqslant 4 ^ { n } .$$

Show that, for all $n \in \mathbb { N } ^ { * }$,
$$\prod _ { \substack { p \leqslant n \\ p \text { prime } } } p < 4 ^ { n } .$$
One may proceed by induction and perform the inductive step by discussing according to the parity of $n$.

Deduce that, for all real $x \geqslant 1$,
$$\prod _ { \substack { p \leqslant x \\ p \text { prime } } } p < 4 ^ { x } .$$

Let $n \in \mathbb { N } ^ { * }$. Justify that
$$\prod _ { \substack { p \leqslant n \\ p \text { prime } } } p \geqslant \prod _ { \substack { \sqrt { n } < p \leqslant n \\ p \text { prime } } } p$$

Deduce that, for all $n \in \mathbb { N } ^ { * }$,
$$n ^ { ( \pi ( n ) - \pi ( \sqrt { n } ) ) / 2 } < 4 ^ { n }$$

Let $n \in \mathbb { N } , n \geqslant 2$. Justify that
$$\pi ( \sqrt { n } ) \leqslant \sqrt { n } < \frac { n } { \ln ( n ) }$$
then deduce that
$$\pi ( n ) \leqslant 4 \frac { \ln ( n ) } { n }$$
One may note that $2 > \ln ( 4 )$.

Let $x \geqslant 3$. Using the monotonicity of the function $t \mapsto \frac { t } { \ln ( t ) }$ on the interval $[ \mathrm { e } , + \infty [$, show that
$$\pi ( x ) \leqslant 4 \frac { x } { \ln ( x ) }$$

Let $n \in \mathbb { N } ^ { * }$. Show that
$$\binom { 2 n } { n } \leqslant ( 2 n ) ^ { \pi ( 2 n ) }$$