Finite segments of the harmonic series

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From the immediately preceding two sets of inequalities we get that 1 1 − m + k − 1 m ′ < ln(1 + k′ − 1 m ′ k − 1 )/(1 + m ) + σ − σ′ , where σ and σ ′ are shorthand for σ(m, k) and σ(m ′ , k ′ ), respectively. Recall that σ − σ ′ ≤ 0. Since m + k ≤ m ′ , we have that Therefore 0 < L := The lemma follows. 1 1 − m + k − 1 m ′ < ln(1 + k′ − 1 m ′ )/(1 + 1 < e L < (1 + k′ − 1 m ′ )/(1 + 1 + k′ − 1 m ′ k − 1 ) , whence m k − 1 > 1 + m . Theorem 11. Let m < k. Then σ(m, k) = σ(m ′ , k ′ ). k − 1 ) . m Proof. Lemma 10 gives us that 1 ≤ (k−1)/m < (k ′ −1)/m ′ . So m ′ < k ′ −1 whence σ(m, k) = σ(m ′ , k ′ ) by Theorem 9. Corollary 12. Let σ > 1/m + ln 2. Then σ ′ = σ. Proof. In our proof of Lemma 10 we saw that σ < 1/m+ln(1+(k −1)/m). So 1/m + ln 2 < σ < 1/m + ln(1 + (k − 1)/m) by hypothesis. It follows that ln 2 < ln(1 + (k − 1)/m). Thus m < k − 1. So σ ′ = σ by Theorem 11. Given 〈m, k, m ′ 〉 ∈ N 3 , recall that k ′ is determined, and that if σ(m ′ , j) = σ(m, k) then j = k ′ . By Theorem 11, σ(m ′ , k ′ ) = σ(m, k) when the interval [m, m + k) is “long”; i.e., when m < k. Theorem 8 implies that for no m is there a 〈m ′ , k ′ 〉 such that σ(m ′ , k ′ ) = σ(m, 1). We have not established the injectivity of σ for 2 ≤ k ≤ m. But we now show that, for each m, there are at most finitely many m ′ > m for which the possibility σ(m ′ , k ′ ) = σ(m, k) will not yet have been eliminated. Theorem 13. Let k ′ ≥ 2(m + k). Then σ(m ′ , k ′ ) = σ(m, k). 6
Proof. By Lemma 3, 2 v ∈ S(m, k) for some v ∈ N. But 2 v ∈ S(m ′ , k ′ ) since 2 v+1 |µ(m ′ , k ′ ). So 2 v+1 |δ(m ′ , k ′ ) by Lemma 2, while 2 v δ(m, k). Since the fractions ν(m, k)/δ(m, k) and ν(m ′ , k ′ )/δ(m ′ , k ′ ) are in lowest terms, we have that σ(m ′ , k ′ ) = σ(m, k). Given m and k, clearly the integers m ′ and k ′ determine each other. For 2 ≤ k < m this implies that k ′ < 2(m+k) ⇒ m ′ < 4m 2 . In summary, σ(m ′ , k ′ ) = σ(m, k) remains unprecluded by Theorems 8, 11, and 13 only when all three of the following conditions obtain: m ≥ 3 and 2 ≤ k < m and k ′ < 2(m + k). Note that m ′ /m ≈ k ′ /k, with k ′ ≥ m ′ k/m > k ′ − 1. Thus, to prove σ injective, we need depose for each m ≥ 3 fewer than m 2 + m possible counterexamples. We have shown that S(X) = S(Y ) ⇒ σ(X) = σ(Y ), But regretably ¬ [S([x, y)) = S([w, z)) ⇒ [x, y) = [w, z)]. For example, S({5, 6, 7}) = {2, 3, 5, 7} = S({14, 15}). But S({14, 15, 16}) = {2 4 , 3, 5, 7} = S({5, 6, 7}). Indeed, S({5, 6, 7}) = S([14, 14 + k)) for every k ≥ 3. Echoing Sylvester, upon its verification the following narrowly germane guess would immediately establish that σ : N 2 → Q + is injective. Conjecture. If m + 2 ≤ m + k ≤ m ′ then S(m ′ , k ′ ) = S(m, k). Definition. We call a family D of finite subsets of N distinguished iff when X = Y are elements in D then S(X \ Y ) = S(Y \ X). From our work thus far, the following two theorems are easy exercises. Theorem 14. If D is distinguished then σ : D → Q + is injective. We do not allege the converse of Theorem 14. When B ⊂ N and n ∈ N, let B n := {j n : j ∈ B}. And, for D a family of finite nonempty subsets of N, let D (n) := {B n : B ∈ D}. Theorem 15. A family D of finite nonempty subsets of N is distinguished if and only if D (n) is distinguished for every n ∈ N. Hence, if D is distinguished then σ|`D (n) is injective for every n. 7
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Finite segments of the harmonic series

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