7
$\begingroup$

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

$\endgroup$
8
  • 1
    $\begingroup$ 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. $\endgroup$
    – EDX
    Aug 10, 2020 at 11:21
  • $\begingroup$ @EDX, If the sum isn't sufficient, it should imply the convergence? How can it be possible if problem isnot convergent? $\endgroup$
    – Naren
    Aug 10, 2020 at 11:55
  • $\begingroup$ 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. $\endgroup$
    – EDX
    Aug 10, 2020 at 12:22
  • $\begingroup$ Aww, I get it now, so does I think. Thank you:) $\endgroup$
    – Naren
    Aug 10, 2020 at 12:31
  • $\begingroup$ You're welcome :) $\endgroup$
    – EDX
    Aug 10, 2020 at 12:42

1 Answer 1

Reset to default
1
$\begingroup$

This is changed now, it says the series diverges. Now, that is just a limitation of Wolfram Alpha, indeed, in Wolfram Mathematica 13.2.1 (Alpha is still 12 version, indeed, see https://www.wolframalpha.com/input?i=Integrate%5Bx%2FSqrt%5Bx%5E4+%2B+10+x%5E2+-+96+x+-+71%5D%2C+x%5D) it just returns SumConvergence[Csc[n]^2/n^3, n], because that cannot be solved. In fact it cannot even solve the series that depend on Pi measure that we know Pi has already, last example here: https://mathematica.stackexchange.com/a/258979/82985 proved by genius Terence Tao here: https://mathoverflow.net/a/282290/160207

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .