The Riemann zeta function and Bernoulli numbers

1 Bernoulli numbers and power sums

Last time we observed that the exponential generating function for the power sums was expressed in terms of Bernoulli numbers. More precisely, if k > 0, then

n k X k X k Sk(n) = i = i . i=1 i=0 For k = 0, note that the two sums actually differ (since the sum starting at 1 has n terms while the sum starting at 0 has n + 1 terms). Nevertheless, we will use the latter expression to establish that

X tk ent − 1 t S (n − 1) = . k k! t et − 1 k≥0

t We defined the Bernoulli numbers as the coefficients in the Maclaurin series expansion of 1−et . In other words: ∞ t X tk = B . et − 1 k k! k=0 You can work out the first few Bernoulli numbers by explicitly computing derivatives of the function 1 1 on the left hand side at t = 0. In particular, we have B0 = 1, B1 = − 2 , B2 = 6 , B3 = 0 etc. 1−ent If we evaluate the series for t , we may put these two facts together to obtain a “closed form” expression for the power sums. Indeed ∞ ent − 1 1 X (nt)j = ( − 1). t t j! j=0 Since the term with j = 0 is 1, once we cancel the ones, every term within the parentheses contains a power of t that is ≥ 1. In other words, we may divide by t to get ∞ t X tj−1 = nj . 1 − ent j! j=1 Multiplying out we get ∞ ∞ ent − 1 t X tj−1 X ti = ( nj )( B ). t et − 1 j! i i! j=1 i=0

1 2 1 Bernoulli numbers and power sums

Writing out the first few terms we get: t t2 t3 t2 t3 (n + n2 + n3 + n4 + ··· )(B + B t + B + B + ··· ) 2! 3! 4! 0 1 2 2! 3 3! Multiplying these out yields B B B B nB + (n2 0 + nB )t + (n3 0 + n2 1 + n 2 )t2 0 2! 1 3! 2! 2! B B B B + (n4 0 + n3 1 + n2 2 + n 3 )t3 4! 3! (2!)(2!) 3!

tn To relate these to power sums, we want to write coefficients in terms of n! . Thus, in the third term, we get

B B B t2 B t2 1 3 3 3 2!(n3 0 +n2 1 +n 2 ) = (n3 0 +n2B +nB ) = (n3 B +n2 B +n B ) 3! 2! 2! 2! 3 1 2 2! 3 0 0 1 1 2 2 and in the fourth term we get: B B B B t3 B 3! t3 3!(n4 0 + n3 1 + n2 2 + n 3 ) = (n4 0 + n3B + n2 B + nB ) 4! 3! (2!)(2!) 3! 3! 4 1 2!2! 2 3 3! 1 4 4 4 4 t3 = (n4 B + n3 B + n2 B + n B ) . 4 0 0 1 1 2 2 3 3 3! These computations generalize to the following statement, which is sometimes known as Faul- haber’s formula. Theorem 1.1. For any integer k ≥ 0, we have

k X tk ent − 1 t X 1 X k + 1 tk S (n − 1) = = ( nk+1−j B ) , k k! t et − 1 k + 1 j j k! k≥0 k≥0 j=0 i.e., equating coefficients:

k 1 X k + 1 S (n − 1) = nk+1−j B . k k + 1 j j j=0 Proof. Exercise, generalize the observations above!

Remark 1.2. The above statement refines our prior assertions that Sk(n − 1) was a polynomial of 1 degree k + 1 in n with leading coefficient k+1 and constant term 0, by giving a precise formula for all the coefficients of the relevant polynomial! In the recursive approach to evaluating power sums with which we began, it was completely unclear that there should be any structural regularity to the form of the coefficients in the polynomial of degree k + 1 that we knew represented a power sum! Thus, the problem has been modified a bit: give more information about the Bernoulli numbers Bk. For example, one may show that Bn vanishes for all odd values of n that are strictly greater than 1, so many terms in the above sums are actually zero. 3 2 The zeta function

2 The zeta function

Above, we investigated the sums n X k Sk(n) := i , i=1 but the expression on the right makes sense for any value of k, even values that are integers. In particular, it makes sense for negative values of k, in which case we are summing reciprocals of powers. For negative values of k, this finite sum actually converges to a limit as n → ∞ as well, unlike the situation for positive powers of k. Bearing this in mind, we define X 1 ζ(s) := . ns n≥1 This function is one of the most important functions in number theory. It turns out the difference between the finite sum and the infinite sum can also be controlled using Bernoulli numbers via the Euler-Maclaurin summation formula, so we will focus on analyzing ζ(s).

Remark 2.1. The zeta function may also be thought of as a kind of generating function: if an, n ≥ 1 is a sequence, then we may consider the function

X an ; ns n≥0 a series of this kind is called a Dirichlet series. For example, X 1 ζ(2) = . n2 n≥1 The problem of evaluating this sum was known as the Basel problem. It was eventually solved by Euler, and our next goal will be to evaluate this sum. Observe that on the right hand side, we may break the sum into two pieces corresponding to n even and n odd. Indeed X 1 X 1 1 X 1 X 1 ζ(2) = + = ( ) + . (2n)2 (2n + 1)2 4 n2 (2n + 1)2 n≥1 n≥0 n≥0 n≥0

4 P 1 In other words, ζ(2) = 3 n≥0 (2n+1)2 , and it will suffice for us to evaluate this sum. We will do this solely by means of techniques you learned in Calculus II.

The general binomial theorem Previously, we stated the binomial theorem for integer values of n, i.e., we gave a formula for (a + b)n. However, if |x < 1|, we may also compute the Maclaurin series for (1 + x)a for a a real number that is not necessarily an integer. Indeed, if you repeatedly differentiate (1 + x)a using the power and chain rule, and plug in x = 0, then one obtains dn (1 + x)a| = a(a − 1) ··· (a − (n − 1)) dxn x=0 4 2 The zeta function

Then, we get the following fact. We will write

a a(a − 1) ··· (a − (n − 1)) := ; n n! this notion coincides with the usual binomial coefficient if a is an integer.

Theorem 2.2. For |x| < 1,

a(a − 1) X a (1 + x)a = 1 + ax + x2 + ··· = xn 2! n n≥0

Evaluation of ζ(2) Using the general binomial theorem we can compute a Maclaurin series for arcsin(x). The deriva- 2 − 1 tive of the arcsine function is (1 − x ) 2 . Therefore, the fundamental theorem of calculus tells us that: Z x 2 − 1 arcsin(x) := (1 − x ) 2 , 0 when |x| < 1. We may write a series for the expression inside the integrand on the right using the binomial theorem: replace x by (−x)2 and take a = −1/2. If we integrate the resulting series term by term, one obtains the following result. Remark 2.3. The products of odd and even numbers up to a point have nice expressions in terms of factorials and powers of 2. Indeed:

2 · 4 ··· 2n − 2 · 2n = 2nn!.

Similarly, 1 · 2 ··· 2n − 1 · 2n (2n)! 1 · 3 ···· 2n − 1 = = . 2 · 4 ··· 2n − 2 · 2n 2n(n!) You can (and should!) prove both of these formulas by induction.

Theorem 2.4. If |x| ≤ 1, then

∞ X (2n)! x2n+1 1 x3 1 · 3 x5 arcsin(x) = = x + + + ··· . 22n(n!)2 2n + 1 2 3 2 · 4 5 n=0 Proof. First, let us write the terms of the series. Each term takes the form

Z x  −1   −1  x2n+1 2 (−t2)ndt = 2 (−1)n . 0 n n 2n + 1 Thus, we just have to write an explicit formula for

 −1  (− 1 )(− 3 ) ··· (− 1 − n + 1) 2 = 2 2 2 . n n! 5 2 The zeta function

 −1  Now, ( 1 − n + 1) = −1−2n+2 = −2n+1 = − 2n−1 . In other words, the numerator of 2 is a 2 2 2 2 n product of odd numbers from −1 through −(2n − 1) divided by 2n, i.e., (−1)n times the product of the odd numbers from 1 to 2n − 1 divided by 2n. Since 2n · n! is the product of the first n  −1  even numbers, and since the (−1)n coming from 2 multiplied by the (−1)n coming from the n integral of (−t2)n gives (−1)2n = 1, we see that the formula for the coefficient of the n-th term takes the stated form. You can (and should) prove this by induction! The convergence of the series for |x| < 1 follows from the convergence statement in the general binomial theorem. In that case, convergence follows, e.g., by appeal to the ratio test. When |x| = 1, the ratio test is inconclusive. Nevertheless, a slight extension of the ratio test called Raabe’s test can be used to guarantee convergence in this case. Explicitly, Raabe’s test states that if an are the terms of a series, an > 0 and n are terms that tend to 0, and the successive ratios a β  n+1 = 1 − + n an n n P∞ where β is independent of n, then n=1 an converges if β > 1. We may then explicitly compute the ratios of terms in the series above: 4n2 + 4n + 1 4n2 + 10n + 6 − 6n − 5 6n 5 = = 1 − − . 4n2 + 10n + 6 4n2 + 10n + 6 4n2 + 10n + 6 4n2 + 10n + 6

3 Now, as n tends to infinity, the leading term is 1 and the next term is asympototic to − 2n ; the final 1 3 term behaves like n2 . The convergence then follows immediately from Raabe’s test with β = 2 . Since | sin x| ≤ 1, it follows that:

∞ X (2n)! (sin x)2n+1 x = arcsin sin x = . 22n(n!)2 2n + 1 n=0

If we integrate the left hand side from 0 to π/2, we get

π 2 Z 2 π x = . 0 8 On the other hand, we may compute:

Z π/2 sin2n+1(x)dx. 0

This integral can be computed using induction and integration by parts. Indeed, take u = sin2n x and dv = sin xdx. Then du = 2n sin2n−1 x cos xdx and v = − cos x, so

π π Z 2 π Z 2 2n+1 2n 2 2n−1 2 sin (x)dx = − sin x cos x|0 + (2n) sin x cos xdx. 0 0 6 2 The zeta function

π 2 2 Since sin 0 = 0 and cos 2 = 0, the first term is zero. The identity cos x = 1 − sin x then gives

π π π Z 2 Z 2 Z 2 sin2n+1 xdx = 2n sin2n−1 xdx − 2n sin2n+1 xdx, 0 0 0 and we may solve to get the identity

π π Z 2 2n Z 2 sin2n+1 xdx = sin2n−1 xdx 0 2n + 1 0 By induction, one deduces the following result.

Theorem 2.5. The identity

π 2n 2 Z 2 2 (n!) 1 sin2n+1(x)dx = . 0 (2n)! 2n + 1 The following evaluation of ζ(2) is due to to Boo Rim Choe (Amer. Math. Monthly 94 (1987), no. 7, 662-663). It is rather different from Euler’s evaluation of ζ(2), which used more advanced techniques.

Corollary 2.6. The identity: ∞ π2 X 1 = 8 (2n + 1)2 n=0 holds. Thus, π2 ζ(2) = . 6 1 2 In fact, the factor of 6 = B2, thus ζ(2) = π B2. This formula generalizes to higher even integer values of the zeta function as follows: B (2π)2n ζ(2n) = (−1)n+1 2n , 2(2n)! but one requires different techniques to establish this statement.