This problem was posed by a colleague in China.

Yes. It is Expected to. And the sum is approximately 4.80672 as per Mathematica for fewer terms.

The above answer is from Mathematica, More discussions are available in stackexchange and a paper at these two links:

https://mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1

https://arxiv.org/pdf/1104.5100.pdf

Dear Panchatcharam Mariappan, thank you for useful links. However, we have to be careful with the conclusion - for larger values of n, the sum significantly exceeds the value ~4.80672 (see enclosed figure from WolframAlpha).

Whether it is finite or not (?) I think it is further open, but an expected answer is positive

Dear Viera,

Yes. It exceeds and the provided link says it is expected to converge. The approximation is only for first few terms, mathematica did not help me to find the value this time. After seeing the picture from mathematica, I was confused as well and hence I provided the source of discussion and the article. Now corrected the answer as "Yes" to "Yes, it is expected to"; Thank you for your suggestions.

Dear All, 

since  obviously $sin(n) \neq 0, n \in \mathbb N$, the series is positive and possesses a finite upper bound equal to $\frac{\zeta(3)}{\min_{n \in \mathbb N} \sin^2(n)}$. 

On the other hand, there is the series expansion of the $cosecans^2(n)$ which could (?) help in finding some closed form expression for the posed sum. So we have

\[ S = \sum_{n \geq 1} \frac1{n^3 \sin^2(n)} = \sum_{n \geq 1} \frac1{n^3} \sum_{k \in \mathbb Z} \frac1{(k\pi + n)^2} =  \sum_{k \in \mathbb Z} \sum_{n \geq 1} \frac1{n^3 (k\pi + n)^2}.\]

Now, transforming this expression, we get 

\[ S = \zeta(5) + 2 \sum_{k, n \geq 1} \frac{n^2+k^2\pi^2}{(n^2-k^2\pi^2)^2 n^3} .\]

Now, "only" the final closed form expression remains.

Kind regards,

Tibor

A note: $ \sup{ \sin^{-2}(n) : n \ge 1 } = \infty$

Dear Joachim,

you've right. Just ignore the first part of my answer. May be the double series calculation will lead to some result.

Rgrds,

Tibor

Thank Dr. Mariappan very much for your providing two links below.

(1) https://mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1

(2) https://arxiv.org/pdf/1104.5100.pdf

http://mathworld.wolfram.com/FlintHillsSeries.html

As far as I know  it is not known whether the Flint Hills series. At first glance it seems  that it is not easy.

Do it numerically, my guess is convergent-

First 10^6 terms give

30.31453940238829279...

See the  article

Max.A.Alekseyev

April 28, 2011

One convergence of the the Flint Hills series

I believe that Messahel Abdelkader means the following arXiv preprint:

https://arxiv.org/abs/1104.5100

Till now, we know that the following three links are useful for understanding this problem:

(1) https://mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1 

(2) https://arxiv.org/abs/1104.5100 

(3) http://mathworld.wolfram.com/FlintHillsSeries.html

Thank everybody who provided the above references.

We have 1/((sin n)^2)=1+1/((tg n)^2).

So 1/(n^3)=1/[(n^3)((sin n)^2)]-1/[(n^3)((tg n)^2)].

We obtain SUM{1/(n^3)}=SUM{1/[(n^3)((sin n)^2)]}-SUM{1/[(n^3)((tg n)^2)]}.

SUM{1/(n^3)}=1.20205......(the Apery's constant, i.e. the Riemann zeta function at the point s=3).

Perhaps the above mentioned facts can be useful.