Bourbaki

Exercise 2, § 2

Functions of a Real Variable, VI, p. 23

Prove the formula

(n−1)Bn(X)=n(X−1)Bn−1(X)−∑k=0n(nk)bkBn−k(X) (n-1)B_n(X) = n(X-1)B_{n-1}(X) - \sum_{k=0}^{n} \binom{n}{k} b_k B_{n-k}(X)

(differentiate the series SesX/(es−1)S e^{sX}/(e^s-1) with respect to SS). Deduce the formula

(2n+1)b2n=−∑k=1n−1(2n2k)b2kb2n−2k (2n+1)b_{2n} = - \sum_{k=1}^{n-1} \binom{2n}{2k} b_{2k} b_{2n-2k}

for the Bernoulli numbers.

Solution

No solution has been written yet.

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.