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We know that the series $\sum (-1)^n/n$ converges, and clearly every other alternating harmonic series with the sign changing every two or more terms such as $$\left(1+\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)+\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)-\cdots$$ must converge. My question here is that does the series below also converge? $$\sum\frac{\textrm{sgn}(\sin(n))}{n}\quad\textrm{or}\quad\sum\frac{\sin(n)}{n|\sin(n)|}$$

Loosely speaking, the sign changes every $\pi$ terms. I'd be surprised if it doesn't converge. Wolfram Mathematica, after a couple of minutes of computing, concluded the series diverges but I can't really trust it. My first approach (assuming the series converges) was that if we bundle up terms with the same sign like the example above every bundle must have three or four terms, and since the first three terms of all bundles make an alternating series I was going to fiddle with the remaining fourth terms but they don't make an alternating series so I guess there's no point in this approach.

edit: I don't think we can use Dirichlet's test with $$b_n=\textrm{sgn}(\sin(n)).$$ The alternating cycle here is $\pi$ and I don't believe it would bound the series. For example if the cycle was a number very slightly smaller than $3+1/4$, then $B_n$ (sum of $b_n$) would get larger and larger every four bundles for some time. I believe this should happen for $\pi$ as well since it is irrational. I'm not entirely sure why but $|B_n|\leq3$ for most small $n$ though I guess it's because $\pi-3$ is slightly smaller than $1/7$? Anyway $B_{312\ 692}=4$, $B_{625\ 381}=5$, $B_{938\ 070}=6$, $B_{166\ 645\ 135}=-7$, and $B_{824\ 054\ 044}=8$. $|B_n|$ does not hit $9$ up to $n=1\ 000\ 000\ 000$ with $B_{1\ 000\ 000\ 000}=-2$.

Calum Gilhooley
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Jaeseop Ahn
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    I think the key to proving this is to show that $\sup_n\left\vert\sum_{i=1}^n\text{sgn}(\sin(i))\right\vert$ is bounded, i.e. that there are never more than a certain number of positive terms than negative terms and vice versa. With this you could likely form some alternate series style proof by grouping terms. – Pepe Silvia Jun 23 '20 at 16:30
  • @PepeSilvia I don't believe it's bounded. Please see my edited post. – Jaeseop Ahn Jun 23 '20 at 16:33
  • @JaeseopAhn did u see my answer below? – mathworker21 Jun 27 '20 at 14:14
  • @mathworker21 Yeah sorry, I've been occupied with something else and I'll accept your answer soon after reading it thoroughly. I've never heard of approximation exponent and if I understood it right your answer may not be applicable if the alternating cycle is not $\pi$, right? – Jaeseop Ahn Jun 27 '20 at 14:28
  • @JaeseopAhn I have to check. It might be the case that the bound I need does hold for any irrational number. I have to check the literature more closely; if I had to guess now, I would guess that my proof would work for any alternating cycle instead of $\pi$. – mathworker21 Jun 27 '20 at 14:30
  • I think so. For irrational $\alpha>1,$\begin{multline}B_n=\sum_{k=1}^n(-1)^{\lfloor k/\alpha\rfloor}=\left\lvert{k:1\leqslant k\leqslant n,\ (k/2\alpha)\in\left(0,\tfrac12\right)}\right\rvert\-\left\lvert{k:1\leqslant k\leqslant n,\ (k/2\alpha)\in\left(\tfrac12,1\right)}\right\rvert=O\left(n^{(\mu-2)/(\mu-1)+\epsilon}\right),\end{multline}where $\mu\geqslant2,$ so $B_n=O(n^p),$ $0<p<1.$ For $\pi,$ $\mu<7.11,$ whence $B_n=O(n^{0.84}).$ For almost all $\alpha,$ possibly including $\pi,$ $\mu=2,$ whence $B_n=O(n^\epsilon)$ for all $\epsilon>0.$ It's tempting to guess $B_n=O(\log\log n)!$ – Calum Gilhooley Jun 27 '20 at 15:32

1 Answers1

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Dirichlet's test is too weak. People should just forget it and just learn its proof.

Claim: If $\sum_{n=1}^N \text{sgn}(\sin(n)) = O(\frac{N}{\log^2 N})$, then $\sum_{n=1}^\infty \frac{\text{sgn}(\sin(n))}{n}$ converges.

Proof: $\sum_{n=1}^\infty \frac{\text{sgn}(\sin(n))}{n} = \lim_{N \to \infty} \left[\frac{\sum_{n=1}^N \text{sgn}(\sin(n))}{N}+\int_1^N \frac{\sum_{n \le t} \text{sgn}(\sin(n))}{t^2}dt\right]$ is obtained from summation by parts. The first term goes to $0$ by hypothesis of our claim, and $\int_1^\infty \frac{\sum_{n \le t} \text{sgn}(\sin(n))}{t^2} dt$ exists (and is finite) since the integrand is bounded above by $O(\frac{1}{t\log^2 t})$. $\square$

Note $\sum_{n=1}^N \text{sgn}(\sin(n)) = 2\#\{n \le N : \sin(n) > 0\}-N$, so, using that $\sin(n) > 0$ if and only if $\{\frac{n}{2\pi}\} \in (0,\frac{1}{2})$, we wish to show $\left|\#\{n \le N : \{\frac{n}{2\pi}\} \in (0,\frac{1}{2})\}-\frac{N}{2}\right| = O(\frac{N}{\log^2 N})$. Now, we use the fact that $\sup_{I \subseteq [0,1]} \left|\frac{\#\left\{n \le N : \{\frac{n}{2\pi}\} \in I\right\}}{N}-|I| \right| = O_\epsilon\left(N^{-\frac{1}{\mu-1}+\epsilon}\right)$ for any $\epsilon > 0$, where the supremum is over intervals $I$ and $\mu$ is the approximation exponent of $2\pi$. Since the approximation exponent of $2\pi$ is finite (since it is finite for $\pi$), we get $\left|\#\{n \le N : \{\frac{n}{2\pi}\} \in (0,\frac{1}{2})\}-\frac{N}{2}\right| = O(N^\alpha)$ for some $\alpha < 1$, which is obviously $O(\frac{N}{\log^2 N})$.

mathworker21
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  • Excellent, this is a kind of proof I never expected. So now I know $B_n$ is $O(N/\log^2(N))$ but do you know if it is actually unbounded? – Jaeseop Ahn Jun 28 '20 at 04:22
  • @JaeseopAhn I'm pretty sure it is unbounded. I would prove it's unbounded by showing that the pair correlations are small. If you want to see a full proof, ask this question in a separate post and I'll try to write a full proof. – mathworker21 Jun 28 '20 at 04:41
  • Find my question here. Thank you! – Jaeseop Ahn Jun 28 '20 at 11:40
  • Your proof is still not wrong but why did you go for $N/\log^2(N)$ out of all possible expressions? I just noticed that $\int_1^\infty\frac{1}{t\log^2(t)}dt$ does not converge. – Jaeseop Ahn Jun 28 '20 at 20:59
  • Also if I understood correctly about what "discrepancy" means and if the bound here is sharp, the series may actually not converge for an irrational cycle with $\mu=\infty$? That's what I'm starting to think. – Jaeseop Ahn Jun 28 '20 at 21:15
  • @JaeseopAhn I just said $\frac{N}{\log^2 N}$ cause that's kinda the largest thing that converges. You can write it as $\frac{N}{\log^2(N+1)}$ if it makes you feel better about that integral. The question I had in my mind was whether you still have a $O(\frac{N}{\log^2 N})$ bound for the discrepancy if $\mu = \infty$. This is the whole reason I said I don't know what the answer to your question is if we have something else instead of $\pi$. Obviously my argument goes through for anything with finite $\mu$ lol – mathworker21 Jun 28 '20 at 22:25
  • If I wasn't mistaken when planning a similar answer (only planning, because I didn't know about low-discrepancy sequences, and Theorem 445 of Hardy & Wright only gives $B_n = o(n),$ which isn't enough), summation by parts without integration, as in Theorem 8.27 of Apostol, Mathematical Analysis (2nd ed.) gives $$ \sum_{k=1}^n\frac{(-1)^{\left\lfloor{k/\pi}\right\rfloor}}{n} = \frac{B_n}{n+1} - \sum_{k=1}^n\frac{B_k}{k(k+1)}, $$ so it is enough to have $B_n = O(n/(\log{n})^2),$ because $\sum_{k=N}^\infty1/(k(\log{k})^2)$ converges, for all $N \geqslant 2.$ – Calum Gilhooley Jun 29 '20 at 00:01
  • @CalumGilhooley integration or no integration; it's all the same thing. i just remember the integration formula since there are fewer cross terms – mathworker21 Jun 29 '20 at 00:49
  • I can never remember either of them! $\ \ddot\frown$ – Calum Gilhooley Jun 29 '20 at 00:57
  • @CalumGilhooley $\sum_{n \le N} a_nf(n) = f(N)\sum_{n \le N} a_n - \int_1^N (\sum_{n \le t} a_n)f'(t)dt$ :) – mathworker21 Jun 29 '20 at 01:47
  • @mathworker21 Oh I just learned $\int^\infty\frac{1}{t\log^\alpha(t)}$ converges if and only if $\alpha>1$. Yeah $\frac{N}{\log^2(N)}$ seems to be the largest simple one. I was going to try Liouville's constant as the cycle instead of $\pi$ with Mathematica to see if the sum diverges but I quit because it would take forever to see it reach like 5 even if it actually diverges. – Jaeseop Ahn Jun 29 '20 at 13:05
  • @CalumGilhooley I don't remember either of them, either, so it took a couple of minutes for me to open Wikipedia and check the equality. – Jaeseop Ahn Jun 29 '20 at 13:07
  • Now just turn this into a proof that the binary digits of pi are normal and you have a proof to an open problem ;) – Robert Frost Jul 02 '20 at 10:16
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    @samerivertwice I don't quite understand. What does this have to do with $\pi$ being normal? – Jaeseop Ahn Jul 07 '20 at 19:10
  • @JaeseopAhn your "informally the sign changes every $\pi$ digits" can be mapped ${+,-}\to{0,1}$ then modulo a $\log_2$ morphism, it's a bit like $2^n\pi\pmod{2}$ which is the binary digits of $\pi$. But it was just a vague observation. I'm sure it won't stand up to rigour. – Robert Frost Jul 08 '20 at 06:55