# Alternating Harmonic Series Spin-off

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 $$B_{1\ 000\ 000\ 000}=-2$$.

• My intuition agrees with you since $n$ is equidistributed $\mod \pi$. It would be a surprise for me that the series actually diverges. – mertunsal Jun 23 at 14:00
• 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 at 16:30
• @PepeSilvia I don't believe it's bounded. Please see my edited post. – Jaeseop Ahn Jun 23 at 16:33
• @Pepe Silvia If such a bound exists, since all the conditions of Dirichlet test is satisfied there won't be need of another grouping. – mertunsal Jun 23 at 16:34
• @JaeseopAhn did u see my answer below? – mathworker21 Jun 27 at 14:14

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 $$O(\frac{N}{\log^2 N})$$.

• 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 at 4:22
• @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 at 4:41
• Find my question here. Thank you! – Jaeseop Ahn Jun 28 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 at 20:59
• @samerivertwice I don't quite understand. What does this have to do with $\pi$ being normal? – Jaeseop Ahn Jul 7 at 19:10