Is the sum

¥ å n=1

{n x}/n

, where {x} = x - floor(x) divergent for all non-integer x? I've worked on this problem for about 2 weeks to little success. I've found a couple methods that come very close, but nothing that finishes the proof.

It's obviously divergent for rational x.

Nice question. More generally it is reasonable to conjecture:

If a_n is decreasing and

å

an

diverges then

å

{n x}an

diverges.

I think I see how to do this, though in your case (a_n = 1/n) we can probably simplify matters since n a_n is bounded away from zero. It suffices to prove (the very intuitive):

Lemma: If S_n is the number of {x}, ..., {n x} which are at least 1/2 then S_n/n ®1/2 as n®¥

because then by choosing N₁ sufficiently high you can guarentee that at least N₁/3 of {x}, ..., {N₁ x} are at least 1/2, then by choosing N₂ sufficiently high you can guarantee that at least N₂/3 of {(N₁+1)x}, ..., {N₂ x} are at least 1/2 and so on. A simple estimate shows that each of these groups contributes at least 1/6 to the series, so it diverges.

So it only remains to prove the lemma - to be honest I don't see how to do it immediately, but it looks too obvious to be false, and I'm sure it's not hard to prove.

Now for the more general case. sin x £ x for any positive x so:

a_n sin² 2pn x = a_n sin² 2p{n x} £ a_n (2p{n x})² £ 4p² a_n{n x}

So it suffices to prove that:

å

an sin2 2pn x

is divergent, provided x is irrational and a_n is decreasing and

å

an

diverges. We can assume a_n ® 0 otherwise the result is trivial.

Well a_n sin² 2pn x = a_n (e^{4pi n x} - 2 + e^{-4pi n x})/-4. It is clear that when we sum this, the middle term is going to diverge, so it is enough to show the other two terms converge. This boils down to showing that:

S a_n e^{n y i}

converges if y isn't an integer multiple of 2p, and a_n is decreasing and tending to zero.

I still don't see an easy proof of the lemma. Of course it's not that important since the proof of the general case doesn't involve the lemma, so we don't really need the lemma.

However the lemma does follow immediately from the result that if f is a periodic Riemann integrable function with period 1 then for any irrational x:

ò₀¹ f(t)dt = lim_n®¥(f(x) + f(2x) + ... + f(n x))/n (*)

because you then can just set f(x) = 1 if {x} > 1/2 and f(x) = 0 otherwise, and apply the result. This shows even more - that given a sub-interval I in [0,1] of length L, a proportion of exactly L of {n x} land in I in the long run.

Actually, the lemma you mention is proven in Hardy and Wright; page 390 - 392. They prove a good deal more, namely that {n x} is uniformly distributed in [0,1] for all irrational x, which is what you've written in your last message. I read the proof a while ago, but if I remember correctly, it uses only Dirichlet's theorem (that given an e, there is an n such that {n x} < e for any irrational x). In fact, I'm pretty sure the proof only uses calculus in a fairly superficial way (that is, you need analysis to justify some of the stuff with limits, but everything can be justified on intuitional grounds)

You can prove the convergence theorem you mention fairly straightforwardly using Abel's convergence condition. Is this the way you did it?

As for my attempt, I tried to use Fourier series to find an asymptotic formula for

x å n=1

{n x}/n

.

The basic idea was to use

{x} = 1/2 -

¥ å n=1

sin(2pn x)/ (pn)

, and with some manipulation

x å n=1

{n x}/n = 1/2Hx -

¥ å k=1

dx(k)sin(2pk x)/ kp

,

where

dx(k) =

å m n = k; n £ x

1

. I think the second term is bounded, and thus I could find a pretty good estimate; but proving this alluded me.

I few more conjectures, all related:

(1) Does

¥ å n=1

(1/2 - {n x})/n

always converge?

(2) Is it true that there are no integer sequences p_n such that

¥ å n=1

|x - pn/n|

converges?

(3) Is it true that for any integer sequence r_n,

¥ å n=1

{rn x}/rn

(which must be nonzero) diverges iff

å

1/rn

diverges? (the only if part is obvious)

Your Fourier series approach is nice, BTW. It's interesting how both of our approaches involved trig...

As for your conjectures. (1) is very interesting - I think it's true, and I suspect you can generalise it:

If a_n is decreasing and tending to zero then

å

(1/2 - {n x})an

converges.

I can't see how to do this yet.

Here's a link explaining Abel's Convergence Condition (they call it ''Test''). Just see that

å

ei y n

is bounded, and you have the proof.

Aha! That was going to be my next conjecture, but I didn't have a clue how to prove it especially since I took so long over the special case

å

an en y i

- and my proof of this can't be made to work in general :-(.

Unfortunately this probably means the trig was unnecessary . I thought at first the sum

å

an en y i

was special, but Abel's Convergence Condition shows that the only special thing about it is that

å

en y i

is bounded.

Actually here's a much simpler proof of the extension of your original conjecture, which doesn't involve Abel and also proves your more recent conjecture (2).

More generally, we prove that if a_n is decreasing and

å

ai

diverges then given any subset T of the natural numbers such that T has non-zero density (in other words there exists e > 0 such that T has at least Ne elements £ N for sufficiently high N) the restricted sum

å

ai

over i in T diverges.

This is straightforward. You can replace ''for sufficiently high N'' with ''for all N'' without loss of generality. Also WLOG a_n ® 0. Then take natural k with k > 1/e. The condition tells us that T has at least 1 element £ k, at least 2 elements £ 2k and so on. As a_n is decreasing a_k + a_2k + ... converges if the restricted sum over T converges. This means that:

a_k + a_2k + a_3k + ...

a_k+1 + a_2k+1 + a_3k+1 + ...

a_k+2 + a_2k+2 + a_3k+2 + ...

...

a_2k-1 + a_3k-1 + a_4k-1 + ...

all converge as well. Adding all these together we see that:

a_k + a_k+1 + a_k+2 + ...

converges, so

å

ai

converges, contradiction.

Now to verify that

å

{n x} an

diverges if a_n is decreasing and

å

an

diverges just let T be the set of all n such that {n x} > 1/2. We know this has a density of 1/2 which is non-zero, so the restricted sum diverges, and we're done.

To verify (2), prove more generally that

å

an|n x - pn|

diverges if a_n is decreasing and

å

an

diverges. Simply let T be the set of n such that 1/4 < {n x} < 3/4. This again has non-zero density, and |x - p_n/n| > 1/4 for any n in T so just apply the previous result.

Interesting generalization; is the lemma you've used (that a non-zero density subset of a divergent sum is divergent) well known? I've never seen it before, but it looks rather useful.

Also, as for my conjecture (1), I think it only holds for irrationals.

I tried proving 1/2 - {n x} is bounded (in fact, that was one of the first attempts I gave at proving the original theorem of this discussion) but I didn't have much luck. I also didn't give it much time though, so I might have another go.

If we could prove that

¥ å k=1

d(k)/k sin(2pk x)

converges, we would have a proof of (1). I'm not sure how to do this.

Brad, Sorry for halting you in the middle of your discussion, but how do you prove that

{x} = 1/2 -

¥ å n=1

sin(2pn x)/ (pn)

Thanks,

Yatir, it can be proven either directly with Fourier series, or it can be proven by considering the imaginary part of both the function and the power series of the function ln(1 - e^{i 2px}) [this neglects convergence concerns]. I know this is vague; I seem to recall seeing a webpage that had this on it, and I'll try to find it. If I haven't found it by tomorrow, I'll just go ahead and post the latter proof.

Also, here is yet another proof of the original theorem that I can't believe I missed. We know that

(

N å n=1

{n x})/N = 1/2 + o(1)

,

so

N å n=1

{n x} = N/2 + o(N)

. Using summation by parts, as the link I gave above calls it, we have

N å n=1

(1/n).{n x} = (N/2 + o(N))/N +

N å n=1

(n/2 + o(n)) (1/n - 1/n+1)

=1/2

å

n=1N 1/n +

N-1 å n=1

o(1/n+1) + o(1)

= 1/2

N å n=1

1/n + o(

N å n=1

1/n)

,

which proves the sum diverges. (Here we have simplified the sums in the second line.) A similar method of argument will show that the convergence of

å

(1/2 - {n x})/n

follows from

(

N å n=1

{n x})/N = 1/2 + O(N-d)

for some positive d. I have no idea how to prove this, however.

No luck here, either. Actually I've run some computer trials and it looks rather like

å

{n x}-1/2

is unbounded after all, meaning the generalisation of your conjecture would be false.

If however this is true, then we can ask more. Suppose you have a subinterval I of [-1/2,1/2] which has midpoint M. Consider the terms of {n x} which fall into I. Is it true that

å n £ N

{n x} = M N ' + o(N)

, where the sum only includes the values of n such that {n x} is in I and N ' is the number of such terms? If not, can you replace {n x} with *any* sequence in [-1/2,1/2] which makes this true? Must any such sequence be uniformly distributed?

Also, do you have an elementary proof of your first statement (

N å n=1

{n x})/N = 1/2 + o(1)

. Of course it follows immediately from the analysis theorem I stated in my second message, but I don't immediately see how to prove it from first principles. (Maybe it follows from the Fourier series for {x}? ...)

By the way: mistake in my message of 8:55 on Wednesday - in the penultimate paragraph, |x - p_n/n| > 1/4 should read |n x - p_n| > 1/4.

Yours,

Here's an outline of the proof:

(I) Prove that f(x) = Im( ln(1 - e^{i 2px}) ) = (x-1/2)p for 0 < x < 1, by differentiating f(x), and solving for imaginary parts as usual, then integrating and using the fact that f(1/2) = 0 to solve for the constant.

(II) Use (I) to prove that f(x) = ({x}-1/2)p for all non-integer x, by noting that f has a period of 1.

(III) Expand ln(1 - e^{i 2px}) using a power series, and take the imaginary part of this series using Eulers formula to find f(x).

(IV) Combine the two values of f(x) you've derived, and solve for {x}.

This neglects convergence; but you can prove that the series converges using Abel, if you so desire.

In (I), in case you're wondering, we have to have the 0 < x < 1 for the first part because otherwise, we would be integrating across a singularity.

Write back if anything is unclear,

Michael, what computer trials have you done that make is seem as though

å

{n x}-1/2

is unbounded? In all the ones I've done, the sum appears to be roughly periodic.

And, I used the analysis theorem to provide rigor to the result about the average of {n x}, but it's pretty intuitional without the rigor...

To answer your question, I don't have a first principles ''proof'', though.

I think I might have a proof of (1). It involves the identity I proved in my second message about d_x(k). Define f_N(x) as

fN(x) =

N å n=1

({n x}-1/2)/n

,

and F_N(x) as

FN(x) = -

¥ å k=1

dN(k) sin(2pk x)/k

,

Note that F_N(x) = f_N(x) for irrational x. Now, if ò₀¹ F_N²(x)dx is finite, we can conclude that F_N(x) was finite for all x in (0,1), thereby proving that f_N(x) is finite for all irrational x. Likewise, if the integral is bounded as N®¥, we may conclude that f_N(x) itself is bounded. We know, however, by Parseval's theorem, that

ò01 FN2(x) dx = 1/2

¥ å k=1

(dN(k)/n)2 £ 1/2

¥ å k=1

d2(k)/k2 £ 1/2

¥ å k=1

(log2(k))2/k2

,

which converges. Therefore, F_N(x) is bounded, and hence f_N(x) is bounded, hence the sum converges.

It's rather late over here in America; is this rigorous? It seems to be to me...

å

({n x} - 1/2)/n

diverges for some irrational x; just pick a sequence of rationals that converges sufficiently quickly. A friend of mine worked this out while we were both in Warsaw last week. I will post the details later.

Hmm, it may not be rigorous, after all. What it does show, though, is that

¥ å n=1

({n x}-1/2)/n

diverges only for a set of zero density. I'm pretty sure it doesn't diverge at all, still, but the proof above (unless someone has a clever alteration) doesn't seem to show it.

Oops, hadn't yet seen your post, David.

Theorem 1:

å n

( {n x} - 1/2)

diverges to -¥ whenever x is rational.

Proof: (I think you've already proved this, but I'm not sure.) Suppose x = p/q, [p,q]=1. Then {n x} repeats with period q, and takes on each of the values r/q for r=1..q exactly once.

So consider dividing these terms into pairs (t/q, (q-t)/q) = ({r₁ x}, {r₂ x} ). The corresponding terms {r₁ x}-1/2 and {r₂ x}-1/2 clearly sum to zero.

The terms left over are:

1) If q is even, a term (q/2)/q = 1/2, for which {r x}-1/2 = 0 anyway so it doesn't matter.

2) A term corresponding to r=q, for which {r x} -1/2 = -1/2.

So if we sum {n x}-1/2 from 1 to q, we get -1/2. As the sequence is periodic, the infinite sum clearly diverges.

Theorem 2:

å n

( {n x} - 1/2)/n

diverges for all rational x.

Proof: The same argument works again. If {r₁ x} + {r₂ x} = 1, we have ({r₁ x}-1/2)/r₁ + ({r₂ x}-1/2)/r₂

= ({r₁ x}-1/2)(r₂-r₁)/(r₁ r₂)

which has modulus at most q/(2r₁ r₂), so summing these pairs of terms gives a convergent series. The leftover terms for n º 0 mod q diverge harmonically, so the series is divergent at any rational x.

Lemma 3: Let (f_i)_i=1^¥ be a family of functions which are all everywhere right-continuous and such that

¥ å i=1

fi(x)

diverges to +¥ at every rational x. Then there exists an irrational x such that

¥ å i=1

diverges.

Confession: This isn't my proof, this is by a bloke called Barnaby who is also a mathematician at my college, and who was also on this year's Cambridge team for the International Maths Competition for University Students.

Proof: Let x₀ be an arbitrary rational number.

Then there is some N₀ such that

N0 å i=1

fi(x0) > 1

.

Furthermore, there is some interval I = [x₀, x₀ + e₀] such that for all y in I,

N0 å i=1

fi(y) > 1/2

(by right continuity). We may also assume e₀ < 1/q² where q is the denominator of x₀.

Now pick x₁ to be rational and in (x₀, x₀ + e₀) - note the open interval.

Repeat the procedure, replacing the sum from 1 to N₀ with a sum from N₀+1 to N₁, and imposing the additional constraint that the intervals I_n be nested, which we can always achieve by shortening them on the right. Then we have an increasing sequence of rationals x_n, bounded above by x₀ + e₀. So they tend to a limit y.

It's fairly obvious that the conditions of the lemma are satisfied by f_n(x) = (1/2-{n x})/n, so the above theorem follows.

I suspect that it is still true that the series converges except on a set of measure zero, but I wouldn't even like to contemplate proving this, certainly not unless someone can exhibit an explicit x such that it converges.

Nice proof! I'm guessing that you meant in the last line |x_n-y| < 1/q_n². (and also, more obviously, there should be a function you are summing in last part of the lemma)

What's wrong with my proof of zero-density using Fourier series? Supposing that f_N tended to infinity with density more than zero, the integral would be infinite. But it isn't...

It may still be true, in the sense that the set S of x where the series diverges might still be of zero measure. We can modify the proof above to show that S is everywhere dense and has uncountably many points in every interval, but it could still be of zero measure, I think. I've never seen a construction of an everywhere uncountably dense Lebesgue-null set, but maybe such a thing exists?

I have to admit that I don't know enough of the formal theory of Hilbert spaces and the like to know to what extent your integral proof is valid, since you seem to be interchanging the order of limits several times.

I know nothing of Hilbert spaces (I know the definition...), so I'm not sure of the connection. I'm also unsure what you mean by ''interchanging the order of limits.''

Sorry if this message seems overly terse; It isn't intended, it's just late.

Another approach to proving Brad's lemma (that ò₀¹ f_N(x)²dx is bounded) is to note that:

ò₀¹({m x}-1/2)({n x}-1/2) dx = HCF(m,n)²/(12m n)

Actually I can't see how to prove this in an elementary way, but it's easy by plugging in Fourier series, and I've checked it numerically and it seems to work.

Then we have:

ò01 fN(x)2 dx =

å

HCF(m,n)2/(12m2 n2)

where the sum is over naturals m, n £ N.

This is then easily seen to converge as N®¥. First sum it over those m, n which are coprime, show that is a finite number S then show that the sum over m, n with (m,n) = 2 is S/2² and the sum over m, n with (m,n) = 3 is S/3² etc, so the infinite sum is Sp²/6 i.e. finite.

OK, here's a first principles proof of the fact I stated in the 2nd line, but it's a bit painful. Without loss of generality (m,n) = 1, once we've got this then a simple substitution in the integral gives the general result.

So it suffices to show that if (m,n) = 1 then:

ò₀¹ {m x}{n x} dx = 1/(12m n) + 1/4

since all the other terms in the integral are easily evaluated.

Consider the integral ò{m x}{n x} dx between r/(m n) and (r+1)/(m n) where r is an integer. Write:

r = a + c m

r = b + d n

where a, b, c, d are integers and 0 £ a < m and 0 £ b < n. Then for x in the range (r/(m n),(r+1)/(m n)) we have r/n < m x < (r+1)/n and r/m < n x < (n+1)/m so d + b/n < m x < d + (b+1)/n and c + a/m < n x < c + (a+1)/m so:

[m x] = d = (r-b)/n

[n x] = c = (r-a)/m

So we have:

{m x} = m x - (r-b)/n

{n x} = n x - (r-a)/m

Substituting this in, a tedious but trivial calculation gives:

òr/(m n) (r+1)/(m n){m x}{n x}dx = (2 + 3a + 3b + 6a b)/(6m²n²)

(miraculously there is no explicit r dependence in the RHS)

We want to sum this over all r in {0,1,...,m n-1}. But since (m,n) = 1, when r runs through {0,1,...,m n-1}, a, b independently run through the sets {0,1,...,m-1} and {0,1,...,n-1} respectively (by the Chinese remainder theorem). So:

ò01{m x}{n x}dx =

m-1 å a=0

n-1 å b=0

(2 + 3a + 3b + 6a b)/(6m2n2) = (2m n + 3m n(n-1)/2 + 3n m(m-1)/2 + 3m(m-1)/2*n(n-1)/2)/ (6m2n2) = 1/(12m n) + 1/4

so we're done.

Oops; I had automatically assumed that if the sum does not diverge to an infinity, then it would converge, which seems intuitional, but certainly not obvious, probably not even true. That explains my usage of ''converge''.

Also, though I still don't see the error in my proof, I like your proof better, Michael; if nothing else it takes little foreknowledge. It also proves a bit more than I ''did'' (quotation marks because I don't seem to have convinced anyone but myself that the proof works); I proved that ò₀¹ F_N(x)² dx is bounded; it is a corollary of this that only a zero-density set of f_N(x) diverge off to +/-infinity, I think.

A pretty exercise: evaluate

n å m

HCF(m,n)r/(ms ns)

for all r, s for which it is defined (and characterise those r, s).

And that holds when s > 1 and 2s-r > 1. So in fact as n®¥ we have:

ò₀¹ f_n(x)²dx ® p²/12.

(Got cut off apparently, continuing). You're right that my proof is purer in that it is largely combinatorial rather than analytic, but I still rather like your Fourier series approach, because I've always found the application of analysis to problems of a largely combinatorial flavour interesting. The evaluation of ò₀¹({m x}-1/2)({n x}-1/2) dx is simpler using Fourier series and to be honest I used my new toy, Maple, to crunch the nasty algebra that appears in the first principles proof, and was rather relieved to see all the r's cancel out (paving the way for the chinese remainder theorem).

One other reason I wanted to check your result independently was that I'm still interested in the generalisation, that is for which x is

gN(x) =

å n £ N

{n x}-1/2

bounded. However, it turns out that

ò01 gN(x)2 =

å

HCF(m,n)/(12m n)

(over m, n £ N) and this tends to infinity as N ®¥ so this approach tells us very little about the general case.

On the subject of convergence, there are a number of things that

å

({n x}-1/2)/n

might do. First it could tend to infinity. We know that because your integral is bounded, this can't happen on a set with positive measure. We don't have proof that this actually happens for any x and I don't really believe it does.

Second it could oscillate infinitely. This means that it is not bounded but it doesn't tend to ±infinity either. It might swing between very high values and very low values, or it might be bounded in one direction only (but return often to low values). We know that the former does happen for an uncountable dense set. It could happen for all irrational x as far as I know. The integral approach doesn't help here, because for a specific high value of n it could be that only a very small portion of the x's have partial sums which are in their large phase - all the rest could have small partial sums, so the integral doesn't have to get very big.

The other two possibilities are that the partial sum oscillates finitely (that is, its bounded) or that it converges. We know this cannot happen for all irrational x, and in fact we don't know of a single x for which this happens.

It should be clear the right hand sum is

z(2s-r)

å

1/(ms ns)

where the sum is over (m,n) = 1. So it suffices to show that

å

hcf(m n)=11/(ms ns) = z(s)2/z(2s)

. But it is also obvious that

z(2s)

å

hcf(m n) = 11/(ms ns) =

å

1/ms ns

and the RHS is the Cartesian product

å

1/ms

å

1/ns = z(s)2

as required.

I know this doesn't matter, but isn't

ò01 gN(x)2 dx =

å m,n £ N

HCF(m,n)2/12m n

, not HCF(m,n)/12m n?

This leads one to the conclusion that if we use ({n x}-1/2)/n^t we must have t > 1/2 for the integral to converge, as we get

å m,n £ N

HCF(m,n)2/(m n)1+t

so we must have 2(1+t) - 2 > 1 for convergence.

It's a very nice conjecture though - that if a_n is square-summable and decreasing then

å m,n

am an HCF(m,n)2/m n

is convergent. Looks like a tough one!

Nevermind the last two posts; I've realized I was misreading the theorem (I apparently saw an n where there was a m). I'd still like to see the proof of the theorem you mention in your last post, though, Michael. (For some reason, the concept of hcf in series expansions throws me off in manipulations)

Silly me - it does work after all. The key is to observe:

Lemma: If a_i is decreasing and

å hcf(m n) = 1

am an

converges then so does

å

am an

. Further, there exists e > 0 such that for every decreasing sequence a_n we have

e

å

am an £

å hcf(m n) = 1

am an

whenever the right hand sum converges.

Proof: WLOG a_n > 0 for all n. Let h(N) be the number of ordered pairs (m,n) with 1 £ m,n £ N and HCF(m,n) = 1. We have h(N) > 0 for all N (since HCF(1,1) = 1) and h(N) ~ (6/p²)N² so there exists e > 0 such that h(N) > 3eN² for all N.

Now estimate

å

am an

summed over all (m,n) with max(m,n) = N and HCF(m,n) = 1. Well it has h(N) - h(N-1) terms in it, and they are at least a_N², so this sum is ³ (h(N) - h(N-1))a_N².

Plugging this into the finite sum:

å hcf(m n) = 1, 1 £ m n £ N

am an

³ h(1)a₁² + (h(2) - h(1))a₂² + ... (h(N) - h(N-1)) a_N²

= h(1)(a₁² - a₂²) + h(2)(a₂² - a₃²) + ... + h(N-1) (a_N-1² - a_N²) + h(N)a_N²

³ 3e(1²(a₁² - a₂²) + 2²(a₂² - a₃²) + ... + (N-1)²(a_N-1² - a_N²) + N² a_N²)

= 3e

N å n=1

(n2-(n-1)2)an2

³ e

N å n=1

(2n+1)an2

(as n² - (n-1)² ³ (2n+1)/3)

³ e

å 1 £ m n £ N+1

am an - eaN+12

³ e

å 1 £ m n £ N

am an

Let N®¥ and the lemma is proved.

So now what we want to show is that if a_n is decreasing then

å

am an HCF(m,n)2/(m n)

converges iff

¥ å r=1

(

¥ å m=1

ar m/m)2

converges.

The left hand side is:

¥ å r=1

å hcf(m n) = r

am an HCF(m,n)2/(m n)

=

¥ å r=1

å hcf(m n) = r

am an r2/(m n)

=

¥ å r=1

å hcf(m¢n¢) = 1

ar m¢ar n¢ r2/(r m¢r n¢)

=

¥ å r=1

å hcf(m n) = 1

ar m ar n/(m n)

But

e

å

ar mar n/(m n) £

å hcf(m n) = 1

ar mar n/(m n) £

å

ar mar n/(m n)

by the lemma. So the left hand side converges iff

¥ å r=1

å m n

ar m ar n/(m n)

converges, i.e. iff

¥ å r=1

(

¥ å m=1

ar m/m)2

converges.

It's been about a month since this discussion was still alive, but just rethinking through this problem the other day, I came up with the following:

Consider the sum

rN(x) =

N å n=1

({bn x} - 1/2)/bn

, where b_n is given, always an integer and b_n < b_n+1. We have from fourier series, for irrational x

rN(x) =

N å n=1

¥ å k=1

sin(2pbn k x)/(bn kp) =

¥ å m=1

jN(m) sin(2pm x)/(pm)

,

where

jN(m) =

å bn k = m,n £ N

1

(again, b_n is given). From Parseval's identity,

p2 ò01 rN2(x) dx =

¥ å m=1

jN2(m)/m2 £

¥ å m=1

dN2(m)/m2 £

¥ å m=1

d2(m)/m2 = O(1)

,

where d(n) and d_N(n) are still as indicated above. This provides, with a little thinking, the result that:

the set of x in [0,1] such that

N å n=1

({bn x} - 1/2)/bn = O(1)

is of density 1, for any set of integers b_n satisfying the above relations.

Nothing major, but nonetheless interesting.

The definition of j_N(m) doesn't show up well with this formatting; it should be the number of solutions to b_n k = m, for n £ N.