Bourbaki

Exercise 6, § 2

Functions of a Real Variable, VI, p. 24

a) The rational numbers bnb_n being defined by formula (20) of VI, p. 7, denote by dnd_n the denominator >0> 0 of bnb_n written in the form of an irreducible fraction. Now prove that no prime factor of dnd_n can be >n+1> n + 1 (use the recurrence formula (23) of VI, p. 8). b) Now prove that one has, for every integer p>0p > 0 and every integer n>0n > 0

Sn(p)=bnp+(n1)p2bn+1p+⋯+(nr)prr+1bn−rp+⋯+pn+1n+1. S_n(p) = b_n p + \binom{n}{1} \frac{p}{2} b_{n+1} p + \cdots + \binom{n}{r} \frac{p^r}{r+1} b_{n-r} p + \cdots + \frac{p^{n+1}}{n+1}.

c) Deduce from b) by induction on nn that, for every prime number pp the denominator of Sn(p)−bnpS_n(p) - b_n p written in the form of an irreducible fraction, is not divisible by pp (observe that pr+1p^{r+1} cannot divide r+1r + 1). d) Deduce from c) that the number

bn−∑pSn(p)p b_n - \sum_p \frac{S_n(p)}{p}

where pp runs through the set of prime numbers p≤n+1p \leq n + 1 and nn is even, is an integer. Conclude that

b2n+∑p1p b_{2n} + \sum_p \frac{1}{p}

where pp runs through the set of prime numbers such that p−1p - 1 divides 2n2n, is an integer (Clausen-von Staudt theorem; use Exercise 5 e).

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.