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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?

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  • $\begingroup$ You can try, maybe (I dont tested) express $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ $\endgroup$
    – Masacroso
    Jul 20, 2015 at 17:18
  • $\begingroup$ My first though would be to see if the series is the fourier series of a known function, and from there use Parseval $\endgroup$ Jul 20, 2015 at 17:18
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    $\begingroup$ ??? This is just calculus... $\int_0^x \sin(n^2y)\,dy=(1-\cos(n^2x))/n^2$. $\endgroup$ Jul 20, 2015 at 17:37
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    $\begingroup$ Aargh. Never mind, sorry. Ignore that hand-slapping-forhead sound... $\endgroup$ Jul 20, 2015 at 17:43
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    $\begingroup$ I have the answer, just give me a minute to write it up. $\endgroup$
    – amcalde
    Jul 20, 2015 at 17:50

3 Answers 3

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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.

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  • $\begingroup$ 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? $\endgroup$ Jul 20, 2015 at 19:52
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    $\begingroup$ @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 $(1)$. $\endgroup$ Jul 20, 2015 at 19:54
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    $\begingroup$ @JohnBarber: anyway, between $(1)$ and $(2)$ I replaced $m$ with $2m^2$ to get a familiar polynomial, namely $n^4+4m^4 = (n^2-2mn+2m^2)(n^2+2mn+2m^2)$. $\endgroup$ Jul 20, 2015 at 19:59
  • $\begingroup$ 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. $\endgroup$ Jul 20, 2015 at 20:01
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    $\begingroup$ @JohnBarber: oh, I'm so sorry, that was just a typo, now fixed. $\endgroup$ Jul 20, 2015 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.

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    $\begingroup$ 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. $\endgroup$ Jul 20, 2015 at 19:42
  • $\begingroup$ Why do you call it "weakly convergent"? $\endgroup$ Jul 20, 2015 at 20:06
  • $\begingroup$ @YoTengoUnLCD: because we have pointwise convergence but it is really slow. Moreover, the limit function is quite irregular. $\endgroup$ Jul 20, 2015 at 20:08
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    $\begingroup$ This answer is not full of rigor, but that was announced, so I don't think it really deserves a downvote. (+1) back. $\endgroup$ Jul 20, 2015 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 $$ \lim_{x\rightarrow 0_+} \sum_{n=1}^{\infty} \frac{\sin(n x)}{n} = \frac{\pi}{2} $$

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    $\begingroup$ Unfortunately, this has little to share with the present question. $\endgroup$ Jul 20, 2015 at 19:30
  • $\begingroup$ But this problem can be generalized for sin(x*n^k)/n, now we have the limits for k=1 and k=2. $\endgroup$ Jul 20, 2015 at 19:40
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    $\begingroup$ However, the case $k=1$ is well-known and your answer does not really deal with the case $k=2$. What happens for $k=3$? $\endgroup$ Jul 20, 2015 at 19:44
  • $\begingroup$ My conjecture is that in general the limit is equal to Pi / (2k) $\endgroup$ Jul 20, 2015 at 19:53

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