THE BERNOULLI NUMBERS, POWER SUMS, and ZETA VALUES the Bernoulli Numbers Arise Naturally in the Context of Computing the Power S

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 ∈ 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, ∞ 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, ∞ 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,

∞ ∞ ∞  k  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  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 ∞ 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!) ∞ ∞ 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, ∞ 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−1|k

For k = 0, the assertion is that B0 ∈ Z (no positive multiple of any p − 1 is 0), and for k = 1 it is that B1 + 1/2 ∈ 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 | k) Sk(p) ≡ (mod p). 0 if p − 1 - k

−1 −1 Thus p (Sk(p) + 1) is integral if p − 1 | 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 | 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 | 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−1|k 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 ∞ 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 formula 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. Nonetheless, taking the expansion as granted, it is X z2e X (4) πz cot πz = 1 − 2 = 1 − 2 ζ(k)zk. n2e n,e≥1 k≥2 even Because the two power series expansions (3) and (4) of πz cot πz must match,

(2πi)k ζ(k) = − B , k ≥ 2 even. 2 k! k In particular ζ(2) = π2/6, ζ(4) = π4/90, and ζ(6) = π6/945. Alternatively to the approach taken here, one can obtain the boxed formula by complex contour integration, and then the second expansion of the cotangent follows. Now we go from ζ(k) to ζ(1 − k) for k ≥ 2 even. The Legendre duplication formula for the gamma function is

1 1 2 1−2s Γ(s)Γ(s + 2 ) = π 2 Γ(2s). Also, at any nonpositive integer −n ≤ 0 the gamma function has a simple pole of residue (−1)n/n!, giving for k ≥ 2 even, 1 2(−1)k/2 Γ(s) ∼ and Γ(s/2) ∼ as s → −k, k!(s + k) (k/2)!(s + k) and so we encode a limit as a finite quotient of two infinite values, Γ(−k/2) 2(−1)k/2k! = , k ≥ 2 even. Γ(−k) (k/2)! The work to follow will “multiply and divide by infinity,” but doing so is only convenient abbreviation of a tacit and legitimate limit process for the sake of clarity. The functional equation for the completed zeta function specializes to − 1−k 1 − k − k k π 2 Γ ζ(1 − k) = π 2 Γ ζ(k), k ≥ 2 even. 2 2 6 THE BERNOULLI NUMBERS, POWER SUMS, AND ZETA VALUES

Multiply the specialized functional equation by Γ(−k/2) and then divide it by k k + 1 1+k π 2 Γ(−k/2)Γ((1 − k)/2), which is π 2 2 2 Γ(−k) by the Legendre duplication formula, to get for any even integer k ≥ 2, Γ(−k/2)Γ(k/2) ζ(1 − k) = ζ(k). πk21+kΓ(−k) By our ﬁnite quotient of two inﬁnite values, and because Γ(k/2) = (k/2 − 1)!, this is 2(−1)k/2k!(k/2 − 1)! (−1)k/2k! ζ(1 − k) = ζ(k) = ζ(k). πk21+k(k/2)! πk2k(k/2) Substitute the boxed value of ζ(k) above to get (−1)k/2k!(2πi)k ζ(1 − k) = − B , πk2kk k! k and almost everything cancels, B ζ(1 − k) = − k , k ≥ 2 even. k This is tidier than the value of ζ(k), with no power of π and no factorial. For example, ζ(−1) = −1/12, ζ(−3) = 1/120, ζ(−5) = −1/252, etc. For elaborate computations with the zeta function and its variants that have similar functional equations, it is an indispensable gain of ease—and of likely-correct results—to move to the tidy divergent region of the functional equation, work there, and then take the answer back to the region of convergence.