Signed variant of the Flint Hills series

Question

I asked my Calculus 2 students to come up with a series the convergence of which they are unable to decide. One of the students, Denis Zelent, invented a very interesting one: $$ \sum_{n = 1}^\infty \frac{1}{n^2 \sin n} \, . \tag{1} $$

Question (short version): Has convergence of this series been studied in literature?

My immediate answer was that this must have something to do with the irrationality measure $\mu$ of $\pi$. Obviously, $\mu \geqslant 2$, and the best currently known upper bound for $\mu$ is $\mu \leqslant 7.6063\!\ldots\,$, due to Salikhov; see [V. Kh. Salikhov, On the Irrationality Measure of $\pi$. Russ. Math. Surv 63(3):570–572, 2008]. It is widely believed that $\mu = 2$.

In fact, the inequality $\mu < 3$ is equivalent to convergence of $1 / (n^2 \sin n)$ to zero. Thus if we knew that $\mu \geqslant 3$, the series (1) would diverge. On the other hand, if we had $\mu < 2$ (which is of course absurd), then one could easily show that $1 / (n^2 \sin n) = O(n^{-1 - \varepsilon})$ for some $\varepsilon > 0$, which would imply absolute convergence of the series (1).

My student searched the web and realized that his question is related to the well-known open problem, asking whether the Flint Hills series $$ \sum_{n = 1}^\infty \frac{1}{n^3 \sin^2 n} \tag{2} $$ converges. An extension of this problem asks for convergence of a more general series $$ \sum_{n = 1}^\infty \frac{1}{n^p |\sin n|^q} , \tag{3} $$ which is equivalent to absolute convergence of the series (1) when $p = 2$ and $q = 1$. For more details, see [Max. A. Alexeyev, On convergence of the Flint Hills series, arXiv:1104.5100, 2011].

To summarise, this is what we have found so far:

• lack of convergence of $1 / (n^2 \sin n)$ to zero would imply that $\mu \geqslant 3$, which is very unlikely;

• convergence (in particular: absolute convergence) of the series (1) would imply $\mu \leqslant 3$, which means it is certainly an open problem;

• lack of absolute convergence of the series (1) would not have any consequences for the estimates of $\mu$.

Question (long version)

1. Does absolute convergence of the series (1) imply any tighter bounds on the estimates of the irrationality measure $\mu$ of $\pi$?
2. Vice versa: Assuming that $\mu$ is known, can one tell whether the series (1) converges absolutely?
3. Same questions with absolute convergence changed into convergence. In other words: are cancellations of any help here?
4. Does the series (1) has a fancy name, similar to Flint Hills series and Cookson Hills series? (And if not: can Denis choose an appropriate mountain range?)

Edited: I just noticed David Simmons's answer to an MO question on the Flint Hills series, which reduces the question of its convergence to a similar question for a series involving convergents of $\pi$. The same argument should work for the absolute convergence of the series (1), but I do not see right away if it leads to an answer to question 1.

Answer 1

You mentioned my argument from the other thread, which can be used to show that if $\mu(\pi) < 3$, then your series converges. But this also means you can't get any further information about $\mu(\pi)$ from knowing that your series converges -- it would converge if $\pi$ were any of the numbers with exponent of irrationality $<3$, so you cannot rule out $\pi$ being any of those numbers (I know I am speaking of $\pi$ as though it were a variable, but this is necessary in order for questions like "does A imply B?" to make sense when A and B are fixed) In other words, the answer to question 1 is "No".