Euler Product Formula Converges for Re(s) > 1

To move ahead on Kummer's class number formula for ideal numbers, it is necessary to review the proof for the convergence of Euler's Product Formula, that is:

∑ n^-s = ∏ (1 - p^-s)^-1

Ernst Kummer proposed an equivalent formula for ideal numbers which I will talk about in a future blog. If you are not familiar with the notation Re(s), start here.

Theorem 1: For s greater than 1, ∑ (n=1, ∞) n^-s converges absolutely.

Proof:

(1) Let σ = Re(s)

(2) abs(n^-s) = n^-σ [See Theorem, here]

(3) Let M be any integer ≥ 1.

(4) Let N be any integer greater than M

(5) Using the triangle inequality for complex numbers (See Lemma 3, here for more details) abs[∑ (n=M+1,N) n^-s)]≤ ∑ (n=M+1,N) abs(n^-s) = ∑ (n=M+1,N) n^-σ

(6) From a previous result (See Corollary 1.2, here for more details), we know that ∑ (n=M+1,N) n^-σ converges.

*(6) Since f(x)=1/x^s is positive, continuous, and decreasing for x ≥ 1, I can use a result from an earlier blog (See Corollary 1.2, here for more details) to establish that ∑ (n=M+1,N) n^-σ converges.

(7) Since all n^-s ≥ 0, we can see that it converges absolutely. [See here for definition and properties of absolute convergence]

QED

Theorem 2: For S greater than 1, ∏(1 - p^-s) converges absolutely

Proof:

(1) ∑ abs(p^-s) = ∑(p^-σ) where σ = Re(s). [See Theorem 2, here for proof]

(2) ∑(p^-σ) ≤ ∑(n^-σ) and since ∑ (n^-σ) is convergent (by Theorem 1 above), ∑ abs(p^-s) is also convergent.

(4) ∑ abs(p^-s) is convergent implies that ∏ (1 - p^-s) is convergent. [See Lemma 6, here]

(5) And this means that ∏ (1 - p^-s)^-1 is also convergent. [See Lemma 6, here]

(6) Since ∑ abs(p^-s) is convergent, it follows that ∏(1 - p^-s) is absolutely convergent and therefore ∏ (1 - p^-s)^-1 is absolutely convergent since:

(a) Assume ∑ abs(p^-s) is convergent

(b) It follows that ∏(1 - abs[p^-s]) is convergent [See Lemma 3, here]

(c) Which means that ∏(1 - p^-s) is absolutely convergent. [See Definition 1, here]

(d) Using the Reciprocal Property of limits (see Lemma 6, here), we can see that 1/∏(1 - abs[p^-s]) is also convergent.

(e) Using the definition of absolute convergence for infinite products (see step #6c above), we see that ∏ (1 - p^-s)^-1 absolutely convergent.

QED