Exercise 20, § 1

\* Let $\pi(x)$ be the number of primes less than the real number $x > 0$. For each integer $n > 0$, show that

(notice that $\binom{2n}{n}$ is a multiple of the product of all the primes q such that $n < q \leq 2n$); on the other hand, use Ex. 19 to show that $\binom{2n}{n}$ divides the product $\prod p^{r(p)}$, where p runs through the set of prime numbers $\leq 2n$, and where $r(p)$ is the largest integer such that $p^{r(p)} \leq 2n$. Deduce from this result that there exist two real numbers a and b such that $0 < a < b$ and (notice that $2^{2n}(2n + 1)^1 \leq (2n)!/(n!)^{-2} \leq 2'''$).

1 It has been shown that $\pi(x) \sim \frac{x}{\log x}$ as x tends to $+ \infty$ (cf. for example A. E. Ingham, The distribution of prime numbers (Cambridge tracts, No. 30), Cambridge University Press, 1932).

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