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I was playing around with sums the other day, and started fiddling with the function $$ f(x) = \sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n}\, . $$ Now, obviously this is a very jagged function. (I think the derivative doesn't exist anywhere.) However, it seems to go to a finite, positive limit as $x\rightarrow 0_+$. Furthermore, just looking at the first few decimal places of this limit, it looks like it may be $\pi/4$. It seems plausible that such a limit might go to a "nice" number like $\pi/4$, but I can't prove it.

A few known things about this problem:

1) $f(x)$ is odd, so $\lim_{x\rightarrow 0_{-}} f(x) = -\lim_{x\rightarrow 0_{+}} f(x)$.

2) $f(x)$ is $2\pi$-periodic (obviously).

3) One possible way that occurs to me to evaluate this limit (if it exists), is to replace it with the following: \begin{align} \lim_{x\rightarrow 0_{+}} f(x) &= \lim_{x\rightarrow 0_{+}} \frac{1}{x}\int_{0}^{x} dy\, f(y)\\ &= \lim_{x\rightarrow 0_{+}} \frac{2}{x}\sum_{n=1}^{\infty} \frac{\sin^2\left(n^2 x/2\right)}{n^3} \end{align} The latter series is smoother and converges more quickly than the original one, so it's better-suited to numerics. If I use $x = 0.0001$ in this series and sum the first $100,000$ terms in Mathematica, I get $0.785393$, whereas $\pi/4 = 0.785398...$ I don't know where to go from there. (I tried the Poisson summation formula to no avail.)

Can anyone here prove this conjecture? Or disprove it? Or show that the question is somehow ill-posed?

John Barber
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  • You can try, maybe (I dont tested) express $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ – Masacroso Jul 20 '15 at 17:18
  • My first though would be to see if the series is the fourier series of a known function, and from there use Parseval – user2520938 Jul 20 '15 at 17:18
  • Your integral is wrong... – David C. Ullrich Jul 20 '15 at 17:23
  • @DavidC.Ullrich How so? I just checked it, and I don't see the error. – John Barber Jul 20 '15 at 17:26
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    ??? This is just calculus... $\int_0^x \sin(n^2y),dy=(1-\cos(n^2x))/n^2$. – David C. Ullrich Jul 20 '15 at 17:37
  • @DavidC.Ullrich Yes, but $1 - \cos(x) = 2\sin^2(x/2)$. – John Barber Jul 20 '15 at 17:41
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    Aargh. Never mind, sorry. Ignore that hand-slapping-forhead sound... – David C. Ullrich Jul 20 '15 at 17:43
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    I have the answer, just give me a minute to write it up. – amcalde Jul 20 '15 at 17:50
  • @amcalde Great. It's not even clear to me that the original series converges... – David C. Ullrich Jul 20 '15 at 17:58
  • @DavidC.Ullrich: it is pointwise convergent by partial summation and Weyl's bound. – Jack D'Aurizio Jul 20 '15 at 19:40
  • @JackD'Aurizio I think Weyl's bound works only for numbers which partial quotients don't grow too fast, otherwise the convergence problem is open (and maybe for some Liouville numbers it could be divergent). – CuriousGuest Jul 21 '15 at 09:43
  • @CuriousGuest: Weyl's bound proves that $\sum_{n\geq 1}\frac{\sin(n^k x)}{n}$ is pointwise convergent for any $k$, there is nothing "open" here. – Jack D'Aurizio Jul 21 '15 at 14:31
  • @JackD'Aurizio No, it does not. The Weyl's bound depends on numbers $q$ in the representation of $x/\pi$ as $\frac{a}{q}+\frac{\theta}{q^2}$. For some numbers it's impossible to find $q$ in the interval $[N,N^k]$ where the bound is nontrivial. In fact, even before Weyl's bound Hardy and Littlewood proved that there are points where this series diverge (see Hardy G.H., Littlewood J.E. Some problems of Diophantine approximation II. Acta Mathematica, 37 (1914), Theorem 2.20, p. 222). – CuriousGuest Jul 21 '15 at 15:00
  • @CuriousGuest: yes, I know. But the point is through Weyl's bound it is not difficult to prove that every $\sum\frac{\sin(n^k x)}{n}$ is pointwise convergent. As an alternative, you may integrate against an approximate identity like I did below, and prove it this way. Digamma-related series arise. – Jack D'Aurizio Jul 21 '15 at 15:30
  • @JackD'Aurizio The statement about pointwise convergence of this series is simply not true, as Hardy and Littlewood showed in 1914, so you can't prove it either through Weyl's bound or any other way. The series converge almost everywhere but not pointwise. – CuriousGuest Jul 21 '15 at 16:49
  • @CuriousGuest: converges almost everywhere in which sense, if it is not pointwise convergent? – Jack D'Aurizio Jul 21 '15 at 17:02
  • I think it doesn't converge at $x = \pi / 2$, for example. – John Barber Jul 21 '15 at 17:02
  • Sure, so it is almost everywhere pointwise convergent. – Jack D'Aurizio Jul 21 '15 at 17:03
  • Could the set of points where it doesn't converge be dense? – John Barber Jul 21 '15 at 17:05
  • $\sin\left(n^{2}x\right) = {1 \over \sqrt{\pi}},\Im\left[\sqrt{-,{\mathrm{i} \over x}}\int_{-\infty}^{\infty}\exp\left({y^{2} \over x\mathrm{i}} + 2ny\right),\mathrm{d}y\right],,\quad\Im\left(1 \over x\right) > 0$. I don't know whether it's useful but, at least, $n$ appears as a linear factor. – Felix Marin Nov 11 '16 at 08:09

3 Answers3

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An interesting trick is to consider that the integral of $f(x)$ against the approximate identity $m e^{-mx}$ is given by:

$$ \int_{0}^{+\infty}f(x)\,me^{-mx}\,dx=\sum_{n\geq 1}\frac{mn}{m^2+n^4} \tag{1}$$ However: $$ \frac{2m^2 n}{4m^4+n^4} = \frac{m}{2}\left(\frac{1}{2m^2-2mn+n^2}-\frac{1}{2m^2+2mn+n^2}\right)\tag{2}$$ hence: $$ \sum_{n\geq 1}\frac{2m^2 n}{4m^4+n^4}=\frac{i}{4}\left(H_{-m(i+1)}-H_{m(-1+i)}-H_{m(1-i)}+H_{m(1+i)}\right)\tag{3}$$ and by letting $m\to +\infty$ we get: $$\begin{eqnarray*} \lim_{x\to 0^+}f(x) &=& \lim_{m\to +\infty}\sum_{n\geq 1}\frac{2m^2 n}{4m^4+n^4}\\&=&\frac{i}{4}\left(\log(-1-i)-\log(-1+i)-\log(1-i)+\log(1+i)\right)\\&=&\color{red}{\frac{\pi}{4}}\tag{4}\end{eqnarray*}$$ proving your conjecture through the unlimited power of the digamma function.

The same approach leads to:

$$\begin{eqnarray*} \lim_{x\to 0^+}\sum_{n\geq 1}\frac{\sin(n^3 x)}{n}&=&\lim_{m\to +\infty}\sum_{n\geq 1}\frac{m^3 n^2}{m^6+n^6}\\&=&\lim_{m\to +\infty}\frac{\pi}{6}\left(-\coth(m\pi)+\coth\left(\frac{1+i\sqrt{3}}{2}\,m\pi\right)+\coth\left(\frac{1-i\sqrt{3}}{2}\,m\pi\right)\right)\\&=&\color{red}{\frac{\pi}{6}}.\tag{5}\end{eqnarray*}$$

The reasonable conjecture:

$$\forall k\in\mathbb{Z}^+,\quad \lim_{x\to 0^+}\sum_{n\geq 1}\frac{\sin(n^{k}x)}{n}=\frac{\pi}{2k}\tag{6}$$

is left as object of further investigations.

Jack D'Aurizio
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  • OK, I understand step (1). $m e^{-m x}$ becomes a Dirac delta function at $0_+$ as $m\rightarrow \infty$. I understand step (2).You lose me in step (3). Are those Harmonic numbers with complex arguments? I'm not familiar with those, but I can accept that they exist. I don't see where they came from, though. Between steps (3) and (4) it looks like you made an implicit substitution $m\rightarrow 2 m^2$, but did you miss a factor of 4 somewhere? And finally, where do all those $\log$s come from? – John Barber Jul 20 '15 at 19:52
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    @JohnBarber: they come out when applying: $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}. $$ The logarithmic derivative of the $\Gamma$ function is the main tool for computing series like the one appearing in the RHS of $(1)$. – Jack D'Aurizio Jul 20 '15 at 19:54
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    @JohnBarber: anyway, between $(1)$ and $(2)$ I replaced $m$ with $2m^2$ to get a familiar polynomial, namely $n^4+4m^4 = (n^2-2mn+2m^2)(n^2+2mn+2m^2)$. – Jack D'Aurizio Jul 20 '15 at 19:59
  • At last, if you compute $$\sum_{n\geq 1}\frac{1}{n^2+1}$$ with any reasonable technique (just perform a search on MSE if you are lazy), you'll see where the logarithms and hyperbolic cotangents come from. – Jack D'Aurizio Jul 20 '15 at 20:01
  • So, in step (2) you've got a $4 m^2$ on the bottom, and then in the last sequence of steps it's become simply $m^4$. Is there a dropped 4 somewhere? – John Barber Jul 20 '15 at 20:04
  • @JohnBarber: there is nothing dropped. To go from the RHS of $(1)$ to the LHS of $(2)$ I replaced $m$ with $2m^2$. Then the LHS of $(2)$ equals the RHS of $(2)$ since $(2m^2+2mn+n^2)(2m^2-2mn+n^2)=4m^4+n^4$ and $\frac{m}{2}(2mn+2mn)=2m^2 n.$ – Jack D'Aurizio Jul 20 '15 at 20:07
  • I understand that part. What I'm wondering is why you've got $\sum_{n\ge 1} \frac{2 m^2n}{m^4 + n^4}$ just above Eq. (4). What happened to the 4 in the denominator? – John Barber Jul 20 '15 at 20:11
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    @JohnBarber: oh, I'm so sorry, that was just a typo, now fixed. – Jack D'Aurizio Jul 20 '15 at 20:12
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OK, I think I've got an answer to this using my non-rigorous "physicist's math".

Using simple algebra, we can write $$ \frac{\sin(n^2 x)}{n} = \frac{\sin(n^2 x)}{2\sqrt{n^2 x}\left(\sqrt{n^2 x} - \frac{1}{2}\sqrt{x}\right)}\Delta_n\, , $$ where $\Delta_n = n^2 x - {(n-1)}^2 x$. So our limit becomes: $$ \lim_{x\rightarrow 0_+} f(x) = \lim_{x\rightarrow 0_+} \sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{2\sqrt{n^2 x}\left(\sqrt{n^2 x} - \frac{1}{2}\sqrt{x}\right)} \Delta_n\, . $$ This is a kind of "warped" Riemann sum over the function $$ g(z) = \frac{\sin(z)}{2\sqrt{z}\left(\sqrt{z} \,-\, \frac{1}{2}\sqrt{x}\right)}\, , $$ where $x$ is a fixed positive number: $$ \lim_{x\rightarrow 0_+} f(x) = \lim_{x\rightarrow 0_+} \sum_{n=1}^{\infty} g(z_n)\, \Delta_n\, . $$

I use the word "warped" here because the "sampling points" $z_n = n^2 x$ are not evenly-spaced, but grow closer and closer together as $x\rightarrow 0_+$. $\Delta_n = z_n - z_{n-1}$ is the width of the $n^{\mathrm{th}}$ "column" of area under the graph of $g(z)$. Thus: \begin{align} \lim_{x\rightarrow 0_+} f(x) &= \lim_{x\rightarrow 0_+} \int_{0}^{\infty} dz\, \frac{\sin(z)}{2\sqrt{z}\left(\sqrt{z} \,-\, \frac{1}{2}\sqrt{x}\right)}\\ &= \int_{0}^{\infty} dz\, \frac{\sin(z)}{2z}\\ &= \frac{\pi}{4}\, . \end{align}

Obviously I did some non-rigorous things there, such as exchanging limits and integrals, so if anyone has an alternative demonstration, I'd be curious to see it.

John Barber
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    It is dangerous to perform such "wild" manipulations with a series that is so weakly convergent, but the idea to use something similar to partial summation to resort to the Dirichlet integral is interesting. – Jack D'Aurizio Jul 20 '15 at 19:42
  • Why do you call it "weakly convergent"? – YoTengoUnLCD Jul 20 '15 at 20:06
  • @YoTengoUnLCD: because we have pointwise convergence but it is really slow. Moreover, the limit function is quite irregular. – Jack D'Aurizio Jul 20 '15 at 20:08
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    This answer is not full of rigor, but that was announced, so I don't think it really deserves a downvote. (+1) back. – Jack D'Aurizio Jul 20 '15 at 20:11
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I have a solution for similar problem $$ \sum_{n=1}^{\infty} \frac{\sin(n x)}{n}\ =\frac{1}{2} i \left(\log \left(1-e^{i x}\right)-\log \left(1-e^{-i x}\right)\right)=\arctan\left(\frac{\sin (x)}{1-\cos (x)}\right) $$ and $$ \lim_{x\rightarrow 0_+} \sum_{n=1}^{\infty} \frac{\sin(n x)}{n} = \frac{\pi}{2} $$

Vaclav Kotesovec
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