Consider the series: $$\sum_{m=1}^{\infty}\frac{\sin(2\pi n x^{1/m})}{ m}\;\;\;\;n\in\mathbb{N}$$

Other than formal manipulation of the Taylor series of the $\sin$ function, is there a way to expand it in terms of any functions besides the roots of $x$ ? A truncated version looks like this. It seems to me that a Fourier-Neumann expansion could be obtained, but i wasn't able to prove that.

The series converges -at least heuristically- for all positive-real values of $x$. How can we prove that rigorously ?

  • $\begingroup$ Have you tried using Dirichlet's test? $\endgroup$ – Ayesha Jan 5 '14 at 18:31

We have $\sin(2\pi n x^{1/m}) = \sin(2 \pi n \{ x^{1/m} \})$, where $\{x\}$ takes the fractional part of $x$. Since $x^{1/m}$ approaches $1$ monotonically decreasing if $x>1$, then eventually $\{x^{1/m}\}=x^{1/m}-1$ and $\{x^{1/m}\} \to 0$ as $m \to \infty$. If $0<x<1$ then $\{x^{1/m}\} = x^{1/m}$.

In the first case we get $\sin(2\pi n x^{1/m}) = (2\pi n) \; O(x^{1/m}-1)$. But $x^{1/m}-1= \frac{1}{m} \ln(x) + \cdots = O(1/m)$. Thus, for $x>1$, because $\sin(2\pi n x^{1/m}) = O(1/m)$, the series does converge.

When $0<x<1$, we note the symmetry $\sin x = - \sin x$ and its periodicity. Hence, we can write $\sin(2 \pi n x^{1/m}) = \sin(2 \pi n \{1-x^{1/m}\})$ do the same thing. Again, for $0<x<1$, the series does converge.

(For $x=1$ or $x=0$ the convergence is obvious and $x<0$ will not work.)


We showed in the process that for appropriately small $x$ that $\sin(2\pi n x^{1/m}) = (2\pi n) (x^{1/m}-1) + \cdots = 2\pi n \frac{1}{m} \ln(x) +\cdots.$

So we get that the sum is $$S(x) = \frac{n\pi^3}{3} \ln(x) +O(ln(x)^2)$$

  • $\begingroup$ nice indeed .... what about an expansion ? $\endgroup$ – Mohammad Al Jamal Jan 5 '14 at 18:38

Note that $$ x^{1/m}=1+\frac{\ln x}{m}+\mathcal O\left(\frac{1}{m^2}\right). $$ Thus by Mean Value Theorem, there is a $\xi$, such that $$ \sin \left(2\pi nx^{1/m}\right)-\sin(2\pi n)=\cos (\xi)\left(\frac{\ln x}{m}+\mathcal O\left(\frac{1}{m^2}\right)\right), $$ amd hence $$ \left|\sin \left(2\pi nx^{1/m}\right)\right|\le \frac{c}{m}, $$ for some $c>0$. This makes the series absolutely convergent, using for example the comparison test.

  • $\begingroup$ nice indeed .... what about an expansion ? $\endgroup$ – Mohammad Al Jamal Jan 5 '14 at 18:39
  • $\begingroup$ Nice! This is a slciker proof of convergence than mine. Just a slightly nitpicking note, $\xi = \xi(m)$, which is important generally, but goes away since $\cos(\xi)$ is bounded. $\endgroup$ – abnry Jan 5 '14 at 18:41

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