Here is a proof that for all but finitely many . You can extract an explicit bound for from the argument and check the smaller values by hand.
If without the floor, then , so it is positive and tends to 0. When you replace by its floor, changes by at most 2, hence the only chance for failure is to have when the fractional parts of and are very small and the fractional part of is very close to 1 (the difference is less than ).
Let denote the nearest integers to , and . Then because it is an integer very close to . Denote . Then and . Observe that
(the bounds are just the bounds for the derivative of on . Therefore
or, equivalently,
If is a multiple of 3, this inequality implies that , then is an integer and not slightly smaller than an integer as it should be. If is not divisible by 3, then
where is a fraction with denominator 9 and . From Taylor expansion
for , we have
where .
This cannot be close to an integer because it is close (from above) to a fraction with denominator 27.
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