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I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts".

Edit: There is a classic reference: G. H. Hardy & J. E. Littlewood, "Some problems of Diophantine approximation: The lattice points of a right-angled triangle," (1st memoir), Proc. London Math. Soc. (2), v. 20, 1922, pp. 15-36. They consider the modified fractional part sum, with {x} set as x - [x] - 1/2, of the ${k\theta}$ up to n, where to be compatible with their notation $\theta$ would be the reciprocal of $\tau$, not that this matters at all. The bound they get is O(log n) (Hardy's Works vol. I p. 145), which depends only on the continued fraction having partial quotients. The particular case relevant to $\tau$ is worked out in detail over the next few pages. The result is sharp. Off the top of my head this looks enough to get the error term O(nlog n) for the sum as posted, by breaking into at most n sums of this type.

I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts".

Edit: There is a classic reference: G. H. Hardy & J. E. Littlewood, "Some problems of Diophantine approximation: The lattice points of a right-angled triangle," (1st memoir), Proc. London Math. Soc. (2), v. 20, 1922, pp. 15-36. They consider the modified fractional part sum, with {x} set as x - [x] - 1/2, of the ${k\theta}$ up to n, where to be compatible with their notation $\theta$ would be the reciprocal of $\tau$, not that this matters at all. The bound they get is O(log n) (Hardy's Works vol. I p. 145), which depends only on the continued fraction having bounded partial quotients. The particular case relevant to $\tau$ is worked out in detail over the next few pages. The result is sharp. Off the top of my head this looks enough to get the error term O(nlog n) for the sum as posted, by breaking into at most n sums of this type.

I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts". Then we know the fractional parts are uniformly distributed mod 1. On the face of it, though, this won't be enough, i.e. the average fractional part will therefore be 0.5. But the problem is set up so that this "main term" contribution drops out? Otherwise why should the error term be small?

So the next recourse should be to exploit the Fourier series of the sawtooth function (not just its constant term). The Fourier series should be truncated at around n terms, and the error estimated away. Then there are geometric progressions to sum. This is what brings in the diophantine approximation properties of $\tau$, in the denominators, which basically can be small at Fibonacci numbers. We do know how small. We do know how many Fibonacci numbers are involved for the terms under consideration.

Forgive me if I'm not writing down the many summations and estimates here. I don't find MO well adapted to talking about analytic number theory. We're basically talking here about the technology of Weyl's equidistribution theorem being applied to a particular case, to say more than simply that the fractional parts are spread uniformly on [0, 1], but to dig down a bit further using exponential sums.

Edit: So I want to rely on the ability to estimate sums like

$\sum_{j=1}^{M} exp(2\pi i\tau rj)/j$

as bounded by

$|cosec(\pi \tau r)|$

and the corresponding

$\sum_{j=M+1}^{infinity}$

by log M.

I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts".

Edit: There is a classic reference: G. H. Hardy & J. E. Littlewood, "Some problems of Diophantine approximation: The lattice points of a right-angled triangle," (1st memoir), Proc. London Math. Soc. (2), v. 20, 1922, pp. 15-36. They consider the modified fractional part sum, with {x} set as x - [x] - 1/2, of the ${k\theta}$ up to n, where to be compatible with their notation $\theta$ would be the reciprocal of $\tau$, not that this matters at all. The bound they get is O(log n) (Hardy's Works vol. I p. 145), which depends only on the continued fraction having partial quotients. The particular case relevant to $\tau$ is worked out in detail over the next few pages. The result is sharp. Off the top of my head this looks enough to get the error term O(nlog n) for the sum as posted, by breaking into at most n sums of this type.

I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts". Then we know the fractional parts are uniformly distributed mod 1. On the face of it, though, this won't be enough, i.e. the average fractional part will therefore be 0.5. But the problem is set up so that this "main term" contribution drops out? Otherwise why should the error term be small?

So the next recourse should be to exploit the Fourier series of the sawtooth function (not just its constant term). The Fourier series should be truncated at around n terms, and the error estimated away. Then there are geometric progressions to sum. This is what brings in the diophantine approximation properties of $\tau$, in the denominators, which basically can be small at Fibonacci numbers. We do know how small. We do know how many Fibonacci numbers are involved for the terms under consideration.

Forgive me if I'm not writing down the many summations and estimates here. I don't find MO well adapted to talking about analytic number theory. We're basically talking here about the technology of Weyl's equidistribution theorem being applied to a particular case, to say more than simply that the fractional parts are spread uniformly on [0, 1], but to dig down a bit further using exponential sums.

I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts". Then we know the fractional parts are uniformly distributed mod 1. On the face of it, though, this won't be enough, i.e. the average fractional part will therefore be 0.5. But the problem is set up so that this "main term" contribution drops out? Otherwise why should the error term be small?

So the next recourse should be to exploit the Fourier series of the sawtooth function (not just its constant term). The Fourier series should be truncated at around n terms, and the error estimated away. Then there are geometric progressions to sum. This is what brings in the diophantine approximation properties of $\tau$, in the denominators, which basically can be small at Fibonacci numbers. We do know how small. We do know how many Fibonacci numbers are involved for the terms under consideration.

Forgive me if I'm not writing down the many summations and estimates here. I don't find MO well adapted to talking about analytic number theory. We're basically talking here about the technology of Weyl's equidistribution theorem being applied to a particular case, to say more than simply that the fractional parts are spread uniformly on [0, 1], but to dig down a bit further using exponential sums.

Edit: So I want to rely on the ability to estimate sums like

$\sum_{j=1}^{M} exp(2\pi i\tau rj)/j$

as bounded by

$|cosec(\pi \tau r)|$

and the corresponding

$\sum_{j=M+1}^{infinity}$

by log M.