Finite segments of the harmonic series

Corollary 6. Let a/b ∈ H with (a, b) = 1. Then there is a pair 〈m, k〉
such that b|µ(m, k), and such that µ(m, k)/b = σ(m, k)µ(m, k)/a.
Proof. There exists 〈m, k〉 ∈ N 2 for which 〈a, b〉 = 〈ν(m, k), δ(m, k)〉.
But ν/δ is the lowest-terms form of the rational number σµ/µ. It follows
that µ/b = q = σµ/a for some q ∈ N.
Corollary 7. The integer µ/δ = σµ/ν is a product of powers p n of odd
primes p ≤ k.
The next result engenders our guess that σ : N 2 → Q + is injective.
Theorem 8. If k ≥ 2 then σ(m, k) ∈ 1/N.
Proof. Pretend the theorem fails for a given k ≥ 2 and m ≥ 1. That is,
we pretend that σ = 1/δ.
If kδ < m then 1/kδ > 1/m > 1/(m+1) > · · · > 1/(m+k−1), whence
1/δ = k · 1/kδ > 1/m + 1/(m + 1) + · · · + 1/(m + k − 1) =: σ. Similarly, if
kδ > m + k − 1 then 1/δ < σ. Hence m ≤ kδ ≤ m + k − 1 if σ = 1/δ.
From k ≥ 2 we get 1/m < σ = 1/δ. Thus 2 ≤ δ < m. So
kδ ≤ m + k − 1 implies k ≤ k(δ − 1) ≤ m − 1 < m, giving us m > k.
Sylvester’s theorem implies that there is a prime p > k and v ∈ N such
that p v ∈ S(m, k). Then p v xp for exactly one xp ∈ [m, m + k), and
(p, x) = 1 for every x ∈ [m, m + k) \ {xp}. Hence (µ/xp, p) = 1, while
p|µ/x for each x ∈ [m, m + k) \ {xp}. So (µσ, p) = 1 while p v µ.
Therefore p v δ, since µσ/µ = 1/δ. We must infer that xp = kδ.
Define σ ′ := σ − 1/kδ = c/d with (c, d) = 1. Then σ ′ = 1/δ − 1/kδ =
(k − 1)/kδ. Now p|kδ, but (k − 1, p) = 1 since p > k. So p|d. On
the other hand, σ ′ = (µσ − µ/kδ)/µ, and p v | (µσ − µ/kδ) since p v µ/x
for every x ∈ [m, m + k) \ {kδ}. Thus, recalling that p v µ we infer that
(p, d) = 1, reaching a contradiction: p|d ∧ (p, d) = 1. So σ = 1/δ.
Notice that Theorem 8 implies that Hn ⊃ Hn+1 for all n ∈ N.
Henceforth fix 〈m, k, m ′ 〉 ∈ N 3 with k ≥ 2 and m ′ ≥ m + k; define
k ′ = min{j : σ(m ′ , j) ≥ σ(m, k)}. Observe that 0 ≤ σ(m ′ , k ′ ) − σ(m, k) <
σ(m ′ , k ′ ) − σ(m ′ , k ′ − 1) = 1/(m ′ + k ′ − 1) ≤ 1/(m + 2k).
Theorem 9. Let m ′ ≤ k ′ . Then σ(m ′ , k ′ ) = σ(m, k).
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