Conjecture: $\lim_{x\rightarrow 0_+} \sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n} = \frac{\pi}{4}$

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?

Solution 1:

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.

Solution 2:

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.