Chapter 1 Euler’s Product Formula 1.1 The Product Formula The whole of analytic number theory rests on one marvellous formula due to Leonhard Euler (1707-1783): −1 X n−s = Y 1 − p−s . n∈N, n>0 primes p Informally, we can understand the formula as follows. By the Funda- mental Theorem of Arithmetic, each n ≥ 1 is uniquely expressible in the form n = 2e2 3e3 5e5 ··· , where e2, e3, e5,... are non-negative integers (all but a finite number being 0). Raising this to the power −s, n−s = 2−e2s3−e3s5−e5s ··· . Adding for n = 1, 2, 3,... , X n−s = 1 + 2−s + 2−2s + ··· 1 + 3−s + 3−2s + ··· 1 + 5−s + 5−2s + ··· ··· , each term on the left arising from just one product on the right. But for each prime p, −1 1 + p−s + p−2s + ··· = 1 − p−s , and the result follows. Euler’s Product Formula equates the Dirichlet series P n−s on the left with the infinite product on the right. 1–1 1.2. INFINITE PRODUCTS 1–2 To make the formula precise, we must develop the theory of infinite prod- ucts, which we do in the next Section. To understand the implications of the formula, we must develop the the- ory of Dirichlet series, which we do in the next Chapter. 1.2 Infinite products 1.2.1 Definition Infinite products are less familiar than infinite series, but are no more com- plicated. Both are examples of limits of sequences. Definition 1.1. The infinite product Y cn n∈N is said to converge to ` 6= 0 if the partial products Y Pn = cm → ` as n → ∞. 0≤m≤n We say that the infinite product diverges if either the partial products do not converge, or else they converge to 0 (as would be the case for example if any factor were 0). Q Proposition 1.1. If cn is convergent then cn → 1. Proof I We have Pn cn = . Pn−1 Since Pn → ` and Pn−1 → `, it follows that ` c → = 1. n ` J It is usually more convenient to write the factors in the form cn = 1 + an. In these terms, the last Proposition states that Y (1 + an) convergent =⇒ an → 0. 1.2. INFINITE PRODUCTS 1–3 1.2.2 The complex logarithm The theory of infinite products requires some knowledge of the complex log- arithmic function. Suppose z 6= 0. Let z = reiθ, where r > 0. We are interested in solutions w ∈ C of ew = z. If w = x + iy then ex = r, e−iy = e−iθ, ie x = log r, y = θ + 2nπ for some n ∈ Z. Just one of these solutions to ew = z satisfies −π < y = =(w) ≤ π. We call this value of w the principal logarithm of z, and denote it by Log z. Thus eLog z = z, −π < =(z) ≤ π. The general solution of ew = z is w = Log z + 2nπi (n ∈ Z). Now suppose w1 = Log z1, w2 = Log z2. Then w1+w2 Log(z1z2) e = z1z2 = e It follows that Log(z1z2) = Log z1 + Log z2 + 2nπi, where it is easy to see that n = 0, −1 or 1. If <(z) > 0 then z = reiθ with −π/2 < θ < π/2. It follows that <(z1), <(z2) > 0 =⇒ −π/2 < =(Log z1), =(Log z2) < π/2; and so −π < =(Log z1 + Log z2) < π. Thus <(z1), <(z2) > 0 =⇒ Log(z1z2) = Log z1 + Log z2. In particular, this holds if |z1|, |z2| < 1 (Fig 1.1). 1.2. INFINITE PRODUCTS 1–4 z r θ 1 Figure 1.1: |z − 1| < 1, Log z = log r + iθ 1.2.3 Convergence Proposition 1.2. Suppose an 6= −1 for n ∈ N. Then Y X (1 + an) converges ⇐⇒ Log(1 + an) converges. P Proof I Suppose Log(1 + an) converges to S. Let X Sn = Log(1 + am). m≤n Then Sn Y e = (1 + am). m≤n But Sn S Sn → S =⇒ e → e . Q s Thus (1 + an) converges to e . Q Conversely, suppose (1 + an) converges. Let Y Pn = (1 + an). m≤N Given > 0 there exists N such that P | n − 1| < Pm if m, n ≥ N. 1.2. INFINITE PRODUCTS 1–5 It follows that if m, n ≥ N then Log(Pn/PN ) = Log(Pm/PN ) + Log(Pn/Pm). In particular (taking m = n − 1), Log(Pn/PN ) = Log(Pn−1/PN ) + Log(1 + an). Hence X Log(Pn/PN ) = Log(1 + am). N<m≤m Since Pn → ` =⇒ Log(Pn/LN ) → Log(`/PN ), P we conclude that n>N (1 + an) converges to Log(`/PN ); and in particular, P n≥0 Log(1 + an) is convergent. J Proposition 1.3. Suppose an 6= −1 for n ∈ N. Then X Y |an| convergent =⇒ (1 + an) convergent. Proof I The function Log(1 + z) is holomorphic in |z| < 1, with Taylor expansion Log(1 + z) = z − z2/2 + z3/3 − · · · . Thus if |z| < 1/2 then |Log(1 + z)| ≤ |z| + |z|2 + |z|3 + ··· |z| = 1 − |z| ≤ 2|z|. P Now suppose |an| converges. Then an → 0; and so |an| ≤ 1/2 for n ≥ N. It follows that |Log(1 + an)| ≤ 2|an| for n ≥ N. Hence X Log(1 + an) converges. J 1.3. PROOF OF THE PRODUCT FORMULA 1–6 1.3 Proof of the product formula Proposition 1.4. For <(s) > 1, −1 X n−s = Y 1 − p−s , n∈N, n>0 primes p in the sense that each side converges to the same value. Proof I Let σ = <(s). Then |n−s| = n−σ. Thus N N | X n−s| ≤ X n−σ. M+1 M+1 Now Z n n−σ ≤ x−σdx; n−1 and so N Z N X n−σ ≤ x−σdx M+1 M 1 = M −σ − N −σ σ → 0 as M, N → ∞. Hence P n−s is convergent, by Cauchy’s criterion. On the other hand, Y(1 − p−s) is absolutely convergent, since X|p−s| = X p−σ ≤ X n−σ, which we just saw is convergent. Hence Q(1 − p−s) is convergent, by Propo- sition 1.3; and so therefore is −1 Y 1 − p−s . To see that the two sides are equal note that 0 −1 Y 1 − χ(p)p−s = X χ(n)n−s + X χ(n)n−s, p≤N n≤N where the second sum on the right extends over those n > N all of whose prime factors are ≤ N. As N → ∞, the right-hand side → P n−s, since this sum is absolutely convergent; while by definition, the left-hand side → Q(1 − p−s)−1. We conclude that the two sides converge to the same value. J 1.4. EULER’S THEOREM 1–7 1.4 Euler’s Theorem Proposition 1.5. (Euler’s Theorem) 1 X = ∞. primes p p P Proof I Suppose 1/p is convergent. Then 1! Y 1 − p is absolutely convergent, and so converges to ` say, by Proposition ?? It follows that 1! Y 1 − → `−1. p≤N p But N 1 1! X ≤ Y 1 − , 1 n p≤N p since each n on the left is expressible in the form e1 er n = p1 ··· pr with p1, . , pr ≤ N. Hence P 1/n is convergent. But 1 Z n+1 dx > . n n x Thus N Z N+1 dx X n−1 ≥ = log(N + 1). 1 1 x Since log N → ∞ as N → ∞ it follows that P 1/n is divergent. Our hypothesis is therefore untenable, and 1 X diverges. p J P 1 This is a remarkably accurate result; p only just diverges. For it follows from the Prime Number Theorem, x π(x) ∼ , log x 1.4. EULER’S THEOREM 1–8 th that if pn denotes the n prime (so that p2 = 3, p5 = 11, etc) then pn ∼ n log n. To see that, note that π(pn) = n (ie the number of primes ≤ pn is n). Thus setting x = pn in the Prime Number Theorem, p n ∼ n , log pn ie p n → 1. n log pn Taking logarithms, log pn − log n − log log pn → 0; hence log n → 1, log pn ie log pn ∼ log n. We conclude that pn ∼ n log pn ∼ n log n. Returning to Euler’s Theorem, we see that P 1/p behaves like P 1/n log n. The latter diverges, but only just, as we see by comparison with Z dx = log log x. x log x On the other hand, X 1 p p log p converges for any > 0, since X 1 1+ n n log n converges by comparison with Z dx 1 = − log− x. x log1+ x What is perhaps surprising is that it is so difficult to pass from Euler’s Theorem to the Prime Number Theorem. Chapter 2 Dirichlet series 2.1 Definition Definition 2.1. A Dirichlet series is a series of the form −s −s −s a11 + a22 + a33 + ··· , where ai ∈ C. Remarks. 1. For n ∈ N we set n−s = e−s log n, taking the usual real-valued logarithm. Thus n−s is uniquely defined for all s ∈ C. Moreover, m−sn−s = (mn)−s, n−sn−s0 = n−(s+s0); while 1−s = 1 for all s. 2. The use of −s rather than s is simply a matter of tradition. The series may of course equally well be written a a a + 2 + 3 + ··· . 1 2s 3s 3. The term ‘Dirichlet series’ is often applied to the more general series −s −s −s a0λ0 + a1λ1 + a2λ2 + ··· , 2–1 2.2.