# Flint Hills series $\sum_{n=1}^{\infty}\frac{1}{n^3\sin^2 n}$

The well known Flint Hills series,

$$\sum_{n=1}^{\infty}\frac{1}{n^3\sin^2 n}$$Since I'm aware of the fact that the series convergence issue is still unknown or falls under unsolved problem however, making check on WA, the infinite sum has been approximated $$\sim 4.80$$.

Similarly, I came here where the proposer, Tobi Ope claims that

$$\sum_{n=1}^{\infty}\frac{1}{n^3\sin^2 n} =3\zeta(5)-\frac{2}{\pi^2}\zeta(2)\zeta(3)-2\sum_{k=1}^{\infty}\frac{\cot k}{k^4}\tag{1}$$ which doesn't meet the approximation done by Wolfram alpha.

Whether the series converges or not which is still a big unsolved issue so I believe the equality $$(1)$$ done is incorrect?

Similarly, what is wrong with Wolfram alpha? Is it possible to approximate or find the possible closed form for any such infinite series whose convergence is unknown?

Thank you

• Actually if it isn't convergent the sum is sufficient by itself. But it is actually a very interesting problem to wonder if it converges or not.
– EDX
Aug 10, 2020 at 11:21
• @EDX, If the sum isn't sufficient, it should imply the convergence? How can it be possible if problem isnot convergent? Aug 10, 2020 at 11:55
• I mean that finding a closed form if it isn't convergent is not meaningful. That means the expression of the sum is largely sufficient.
– EDX
Aug 10, 2020 at 12:22
• Aww, I get it now, so does I think. Thank you:) Aug 10, 2020 at 12:31
• You're welcome :)
– EDX
Aug 10, 2020 at 12:42