3
$\begingroup$

Determine whether $\sum_{n=1}^{\infty}\dfrac{\sin (n^2)}{n}$ converges.

This question was proposed by Dr. Wolfgang Hintzeh here: Convergence of $\sum_{k=1}^\infty \frac{\sin(k(k-1))}{k}$

My attempt

I can see that all terms are bounded by $\frac{1}{n}$ but we cannot conclude that all terms are less than some $\frac{1}{n^p},p>1$. And because both Ratio Test and Comparison Test give no information about the convergence, I formed the corresponding integral, but since the Integral Test is not valid for this kind of series, I want to use it to check whether the series converges absolutely.

I constructed the graph of $\displaystyle\int_1^{\infty}\left|\dfrac{\sin (x^2)}{x}\right|dx$:

enter image description here

How is that possible for the integral to be less than $0$ on some intervals? What is wrong with my method? Can anybody give me some hints to prove or disprove its convergence?

| cite | improve this question | | | | |
$\endgroup$
  • 2
    $\begingroup$ It looks unlikely to converge absolutely, I'd expect $|\sin n^2|$ to be equidistributed in the interval $[0,1]$ and so the series to compare unfavourably with $\sum\frac1n$. Proving that seems hard though. $\endgroup$ – Angina Seng Dec 8 '19 at 2:53
  • $\begingroup$ The engine computes indefinite integral and definite integral in different ways. $\endgroup$ – xbh Dec 8 '19 at 3:06
  • $\begingroup$ @xbh Ok, then how to calculate the integral? it is hard to evaluate... $\endgroup$ – Kevin. S Dec 8 '19 at 3:08
  • $\begingroup$ This series has positive terms. I don't understand why you are distinguishing it converging form it converging absolutely. And either way, the Integral Test doesn't apply, because $\sin^2(x)/x$ is not a decreasing function. $\endgroup$ – alex.jordan Dec 8 '19 at 3:17
  • 1
    $\begingroup$ @KevinSong I might rewrite the summand as $\frac{\sin (n^2)}{n}$ (with parentheses) to avoid confusion with the more common expression $\sin^2 n = (\sin n)^2$. $\endgroup$ – Travis Willse Dec 8 '19 at 3:36
11
$\begingroup$

To answer the question in the title of the post, the series $$ \sum_{n=1}^{\infty}\frac{\sin(n^2)}{n} $$ converges conditionally - but the proof relies on some more advanced concepts.

Actually, this is a special case of a problem solved on mathoverflow. More precisely, on that page they show a more general result, which states that for a wide range of $x\in\mathbb R$ the series $$ \sum_{n=1}^{\infty}\frac{\sin(2\pi xn^2)}{n} $$ is conditionally convergent. Your question is the special case $x=1/2\pi$, which happens to satisfy the crucial condition on $x\in\mathbb R$ for the result to apply.

The exact condition on $x\in\mathbb R$ is that $x$ is not a Liouville number. Liouville numbers are an extremely rare class of transcendental numbers that have amazingly accurate rational approximations. There is a quantity called the irrationality measure which can be given to any irrational number, and if it takes the value $\infty$ then the number is Liouville (and conversely, if it is finite then the number is non-Liouville). This paper gives upper bounds for the irrationality measures of some common numbers, including $\pi$, which has the consequence that $\pi$ is non-Liouville. Since $x=\frac{1}{2\pi}$ is obtained from $\pi$ using rational operations, it is therefore non-Liouville as well.

| cite | improve this answer | | | | |
$\endgroup$
  • 1
    $\begingroup$ About that solution on mathoverflow... in it, they use the property of Liouville numbers that there is $n^a < q \leq n$ such that $|2x - p/q| \leq 1/nq$. This looks a little different from the other definitions of Liouville numbers that I have found online. Do you happen to know where I can find this formulation, or at least why ${1 \over \pi}$ satisfies the condition? $\endgroup$ – Zarrax Dec 8 '19 at 5:08
  • 1
    $\begingroup$ @Zarrax that's a good question, which might be even more valuable directly on the mathoverflow answer itself $\endgroup$ – pre-kidney Dec 8 '19 at 5:17
  • 1
    $\begingroup$ The answer is over three years old so I wasn't sure if it would get a response, but maybe I should still do that. I'm pretty sure ${1 \over \pi}$ satisfies an appropriate approximation property but it's been hard to pin down the right reference. $\endgroup$ – Zarrax Dec 8 '19 at 5:37
2
$\begingroup$

Proof for divergence of $$ \newcommand{\abs}[1]{\left\vert #1 \right\vert} \newcommand\rme{\mathrm e} \newcommand\imu{\mathrm i} \newcommand\diff{\,\mathrm d} \DeclareMathOperator\sgn{sgn} \renewcommand \epsilon \varepsilon \newcommand\trans{^{\mathsf T}} \newcommand\F {\mathbb F} \newcommand\Z{\mathbb Z} \newcommand\R{\Bbb R} \newcommand \N {\Bbb N} \renewcommand\geq\geqslant \renewcommand\leq\leqslant \int_1^{+\infty} \abs {\frac {\sin (x^2)}{x}} \mathrm dx. $$ Easy to see $\abs {\sin (x^2)} \geq\sin^2 (x^2) = (1 - \cos (x^2))/2$ since $0 \leq \abs {\sin (u)} \leq 1$. Then $$ \int_1^{ +\infty } \frac {\abs {\sin (x^2)}}x \geq \frac 12 \int_1^{+\infty} \frac 1x \diff x - \frac 12 \int_1^{+\infty} \frac {\cos (2x^2)}x \diff x =: \frac 12( I_1 + I_2). $$ Clearly $I_1$ diverges. For $I_2$, via the substitution $u = 2x^2$, or $x = \sqrt {u/2}$, $$ I_2 = \int_2^{+\infty} \frac {\cos u} {\sqrt {u/2}} \diff \sqrt {u/2} = \int_2^{+\infty} \frac {\cos u} {2u} \diff u, $$ and this is convergent by Dirichlet Test. Therefore the original integral is a sum of a divergent integral and a convergent integral, which shall be divergent as well.

Counterexample for integral test when the terms are not decreasing

Consider $$ f(x) = \frac {\pi x} {1 + (\pi x)^6 \sin^2 (\pi x)}, x \geq 0. $$ Then $f$ is nonnegative, and the limit $f(+\infty)$ does not exist, since $f (n) = \pi n \to +\infty$ as $n \to \infty$. Then the series $\sum_1^{+\infty} f(n)$ diverges since a necessary condition for convergence is that $\lim_{n \to +\infty} f(n) = 0$. But the integral $\int_0^{+\infty} f(x) \diff x$ converges since $$ \int_n^{n+1} f(x) \diff x = \int_{n\pi}^{(n+1)\pi} \frac x {1 + x^6 \sin^2 x} \diff x\leq \frac {2\pi^2}{n^2}, $$ for each $n \in \N^*$, then the integral is bounded from above by $\pi^4/3$.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ I thought the question is about the conditional convergence of a sum, not about the integral, but I think I have misread the question. It is very confusing, since the title of the question doesn't seem to match with what is asked in the body of the question. $\endgroup$ – pre-kidney Dec 8 '19 at 3:24
  • $\begingroup$ @pre-kidney Yes. I cannot guarantee this is what the OP wants since the question is not very clear to me. $\endgroup$ – xbh Dec 8 '19 at 3:28
  • $\begingroup$ Thanks for sharing this. $\endgroup$ – Kevin. S Dec 8 '19 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.