THE BERNOULLI NUMBERS, POWER SUMS, AND ZETA VALUES The Bernoulli numbers arise naturally in the context of computing the power sums 10 + 20 + ··· + n0 = n; 1 11 + 21 + ··· + n1 = (n2 + n); 2 1 12 + 22 + ··· + n2 = (2n3 + 3n2 + n); 6 etc. Also they manifest in Euler's evaluation of the zeta function ζ(k) at even integers k ≥ 2, and then of the divergent series ζ(1 − k) for the same k-values. 1. The Bernoulli Numbers and Power Sums Let n be a positive integer, and introduce notation for the kth power sum up to n − 1 for any nonnegative integer k, n−1 X k Sk(n) = m ; k 2 N: m=0 k Thus S0(n) = n while for k ≥ 1 the term 0 of Sk(n) is 0. Having the sum start at 0 and stop at n − 1 neatens the ensuing calculation. The power series having these sums as its coefficients is their generating function, 1 X tk (n; t) = S (n) : S k k! k=0 Rearrange the generating function by reversing the double sum and putting a finite geometric sum into closed form, and then divide and multiply by t to get ent − 1 t (n; t) = : S t et − 1 The second term is independent of n. It is a power series in t whose coefficients are by definition the Bernoulli numbers, constants that can be computed once and for all, as will be explained further below, 1 t X tk = B : et − 1 k k! k=0 Again rearrange the generating function, this time by using the general formula P ` P j P P k k ` a`t =`! j bjt =j! = k( `+j=k j a`bj)t =k!, or, equivalently, summing over diagonal segments, 1 1 1 0 k 1 X t` X tj X 1 X k + 1 tk (n; t) = n`+1 B = B nk+1−j : S (` + 1)! j j! @k + 1 j j A k! `=0 j=0 k=0 j=0 1 2 THE BERNOULLI NUMBERS, POWER SUMS, AND ZETA VALUES Thus, if we define the kth Bernoulli polynomial as k X k B (X) = B Xk−j; k j j j=0 which again can be computed once and for all, then comparing the first and third expressions for S(n; t) shows that the kth power sum is a polynomial of degree k +1 in n, 1 S (n) = (B (n) − B ): k k + 1 k+1 k+1 Because k is fixed and we imagine n to be large or indeterminate, this expression of Sk(n) as a sum of k + 1 terms is a notable simplication of its original expression k+1 as a sum of n terms. The polynomial Sk(X) has leading term X =(k + 1) and lowest term BkX, because k + 1 k + 1 B (X) = Xk+1 + B Xk + ··· + B X + B : k+1 1 1 k k k+1 The first few Bernoulli numbers are B0 = 1, B1 = −1=2, B2 = 1=6, and B3 = 0, and so the first few Bernoulli polynomials are B0(X) = 1; 1 B (X) = X − ; 1 2 1 B (X) = X2 − X + ; 2 6 3 1 B (X) = X3 − X2 + X: 3 2 2 For example, the boxed formula gives B (n + 1) − B 12 + 22 + ··· + n2 = S (n + 1) = 3 3 ; 2 3 and indeed the right side works out to (2n3 + 3n2 + n)=6. 2. Computing the Bernoulli Numbers Because the Bernoulli numbers are defined by the formal power series expansion 1 t X tk = B ; et − 1 k k! k=0 they are calculable in succession by matching coefficients in the power series identity P j P k P P n n (again from j ajt =j! k bkt =k! = n( j+k=n k ajbk)t =n!) 1 1 n−1 ! X tk X X n tn t = (et − 1) B = B ; k k! k k n! k=0 n=1 k=0 i.e., the nth parenthesized sum is 1 if n = 1 and 0 otherwise. Thus the Bernoulli numbers are rational. Further, because the expression t t t et + 1 t et=2 + e−t=2 + = · = · et − 1 2 2 et − 1 2 et=2 − e−t=2 THE BERNOULLI NUMBERS, POWER SUMS, AND ZETA VALUES 3 is even, it follows that B1 = −1=2 and Bk = 0 for all other odd k. For example, 1 1 = B ; so B = 1; 0 0 0 2 2 0 = B + B ; so B = −1=2 (again); 0 0 1 1 1 3 3 3 0 = B + B + B ; so B = 1=6; 0 0 1 1 2 2 2 and similarly B4 = −1=30, B6 = 1=42, and so on. In hindsight, there are advantages to a slightly variant definition of the Bernoulli numbers, 1 tet X tk = B : et − 1 k k! k=0 Because tet t − = t; et − 1 et − 1 this produces the same Bernoulli numbers other than changing B1 from −1=2 to 1=2, and the latter value works better in some contexts. 3. Denominators of the Bernoulli Numbers We prove a result first shown by von Staudt and Clausen, independently, in 1840: X 1 B + is an integer for k = 0; 1; 2; 4; 6; 8;:::: k p p:p−1jk For k = 0, the assertion is that B0 2 Z (no positive multiple of any p − 1 is 0), and for k = 1 it is that B1 + 1=2 2 Z, and both of these are true by observation. So we may take k ≥ 2, k even. The argument to be given is due to Witt, as presented early in the book Local Fields by Cassels. Let k ≥ 2 be even. For any integer d ≥ 2, each of 0; 1; : : : ; pd − 1 uniquely takes the form r + qpd−1 with 0 ≤ r < pd−1 and 0 ≤ q < p, and so, using the binomial theorem for the first congruence to follow, pd−1−1 p−1 d X X d−1 k Sk(p ) = (r + qp ) r=0 q=0 pd−1−1 p−1 X X ≡ (rk + krk−1qpd−1) (mod pd) r=0 q=0 pd−1−1 pd−1−1 p−1 X X X = p rk + kpd−1 rk−1 q r=0 r=0 q=0 pd−1−1 k X = pS (pd−1) + (p − 1)pd rk−1 k 2 r=0 d−1 d ≡ pSk(p ) (mod p ): 4 THE BERNOULLI NUMBERS, POWER SUMS, AND ZETA VALUES Consequently, −d d −(d−1) d−1 p Sk(p ) − p Sk(p ) is integral for each d ≥ 2: Telescope this result for d = 2; : : : ; e for any e ≥ 2 to get −e e −1 (1) p Sk(p ) − p Sk(p) is integral for each e ≥ 2: Pp−1 k i Further, Sk(p) modulo p is the geometric sum i=1 (g ) with g a generator, giving (−1 if p − 1 j k) Sk(p) ≡ (mod p): 0 if p − 1 - k −1 −1 Thus p (Sk(p) + 1) is integral if p − 1 j k and p Sk(p) is integral if p − 1 - k. This combines with (1) to give ( −e e −1) ( ) p Sk(p ) + p p − 1 j k (2) −e e is integral if : p Sk(p ) p − 1 - k 1 As explained in section 1, Sk(X) = k+1 (Bk+1(X) − Bk+1) is Xk+1 k S (X) = + B Xk + ··· + B X2 + B X; k k + 1 1 2 k−1 k so that pke k p−eS (pe) = + B p(k−1)e + ··· + B pe + B ; k k + 1 1 2 k−1 k from which −e e Bk − p Sk(p ) is p-integral in Q for large e: This combines with (2) for large e to give ( −1) ( ) Bk + p p − 1 j k is p-integral in Q if : Bk p − 1 - k Finally, if p and p0 are distinct primes then p−1 is p0-integral in Q. Consequently P −1 Bk + p:p−1jk p is p-integral in Q for all p, and so it is an integer. Note that all occurrences of is integral in this argument could be replaced the weaker is p-integral in Q; that is, the argument is essentially p-adic until the p-adic results for all p are gathered at the end. 4. The Bernoulli Numbers and Zeta Values Euler famously evaluated the infinite negative power sums 1 X ζ(k) = n−k; k ≥ 2 even; n=1 with k understood to be an integer, and then used his functional equation for ζ to evaluate the divergent series ζ(1 − k) for those same k, the latter zeta values sim- pler than the former. We skim the ideas here, necessarily invoking an expansion of the cotangent function, the functional equation for ζ, the Legendre duplication for- mula for the gamma function, and the behavior of the gamma function at negative integers. THE BERNOULLI NUMBERS, POWER SUMS, AND ZETA VALUES 5 To compute ζ(k) for even k ≥ 2, first note an identity that we have essentially seen already above, eπi + e−πi 2πiz πz cot πz = πiz = πiz + : eπi − e−πi e2πiz − 1 The right side fits into the definition of the Bernoulli numbers, including the lone nonzero odd Bernoulli number B1 = −1=2, giving X (2πi)k X (2πi)k (3) πz cot πz = πiz + B zk = B zk: k! k k! k k≥0 k≥0 even But also the cotangent has a second expansion, X 1 1 X z2 1 πz cot πz = 1 + z + = 1 − 2 · : z − n z + n n2 1 − z2=n2 n≥1 n≥1 Although this expansion plausibly reproduces the cotangent, the fact that it does so is not trivial.