Number theory: smallest fractional part of a power series?

Let (3/2)^n = b(n) + a(n) where the a(n) is as you defined so b(n) is the integer part. These definitions mean that

(3/2) * (b(n) + a(n)) = b(n+1) + a(n+1).

However, we can expand the left hand side to get

(3/2) * b(n) + (3/2) * a(n) = b(n+1) + a(n+1).

If b(n) is even and a(n) is small, then b(n+1) = (3/2) * b(n) and a(n+1) = (3/2) * a(n). When b(n) is odd and a(n) is small, then b(n+1) = (3/2) * b(n) - 1/2 and a(n+1) = (3/2) * a(n) + 1/2.

Either way, when a(n) is small, a(n+1) is bigger. So, a(n) can't converge to zero.

Interestingly, if you try to check the sequence with floating point arithmetic by calculating a(n) = frac(1.5^n), it will look as if a(n) is converging. This is because the significant digits that floating point arithmetic uses run out in the integer part, before it gets to the fractional part.

Edit: Thanks, String. You're right. I must have overlooked the minimum. I'll have another think.

Edit: Okay, Hardy and Littlewood to the rescue, er, sort of. They proved that {frac(x^n)} is equidistributed for almost all real numbers x > 1. If this is true for x = 3/2, then inf(a(n)) = 0 as suspected. It is, however, unknown whether this is true. Your problem is a little simpler than this, though.

@cpbm:

I am afraid your proof, although interesting, is not valid. The problem lies in the conclusion of your analysis:

"Either way, when a(n) is small, a(n+1) is bigger. So a(n) can't converge to zero."

But maybe the mistake lies in interpretation of the question. As I understand it, david asks whether the optimal lower bound for a(n) is zero. And we know by number theory that 3^n/2^n is never an integer so a(n)>0 for all n. We are not to prove that a(n) converges to zero - only that a subsequence of it does.

Just for clarification: by "a(n) is small", I read that 1.5·a(n)+0.5 < 1 <=> a(n) < 1/3.

Oh, Dondi, you're a devil with words. You relatively understand how you can make the women cry. I consider might be some men too, however the macho forms would never admit it. Wonderful. It's no wonder you're so loved.