Bernoulli Numbers Johar M. Ashfaque The Bernoulli Numbers Bn are defined to be the coefficients in the series expansion ∞ X Bn xn x = . ex − 1 n=0 n! It is a result from complex analysis that this series converges for |x| < 2π but, other than this, we cannot gain much of an understanding from the implicit definition. Please note that although the left hand side would appear to become infinite at x = 0, it is not actually the case. Corollary 0.1. We can calculate the Bernoulli numbers by the recursion formula 0= k−1 X k j j=0 Bj , where B0 = 1. Proof. Let us first replace ex − 1 with it’s Taylor series to see that ! ! ∞ ∞ ∞ ∞ X X Bn xn X xj xj X Bn xn x= −1 = . j n! j! n! n=0 n=0 j=0 j=1 If we compare coefficients of powers of x we can clearly see that, except for x1 , xk : 0 = B0 B1 B2 Bk−2 Bk−1 + + + ... + + . k! (k − 1)! 2!(k − 2)! 2!(k − 2)! (k − 1)! Hence 0= k−1 X j=0 k−1 Bj 1 X k Bj . = (k − j)!j! k! j=0 j Note that the inverse k! term is irrelevant to the recursion formula. 1 The first few Bernoulli numbers are therefore B0 = 1, B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30, B5 = 0, B6 = 1/42, B7 = 0. Lemma 0.2. The values of the odd Bernoulli numbers, except B1 , are zero. Proof. As we know the values of B0 and B1 , we can remove the first two terms and rearrange to get ∞ X x x Bn xn + = 1 + , ex − 1 2 n! n=2 which then simplifies to give x 2 ex + 1 ex − 1 =1+ ∞ X Bn xn n=2 n! . We can now multiply both the numerator and denominator of the left hand side by e−x/2 to obtain x 2 ex/2 + e−x/2 ex/2 − e−x/2 =1+ ∞ X Bn xn n=2 n! . (0.1) By substituting x → −x into the left hand side of this equation we can see that it is an even function and hence invariant under this transformation. Hence, as the odd Bernoulli numbers multiply odd powers of x, the right hand side can only be invariant under the same transformation if the value of the odd coefficients are all zero. 2
