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If someone is curious I have solved it here: https://math.stackexchange.com/a/3242204/647013

This question is related to this post https://math.stackexchange.com/q/3241994/647013, but I am fairly sure this is a computer job to disprove it. The following result is given: $$ \sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^3=\frac{1}{2}\left(\pi-\frac{22}{7}\right)^3$$ It can be rewritten as: $$S=\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^3=\frac{3}{4\cdot 7^3}\sum_{n=1}^\infty \frac{\sin(22n)}{n^3}-\frac{1}{4\cdot 7^3}\sum_{n=1}^\infty \frac{\sin(66n)}{n^3}$$ $$=\frac{1}{1372}\left(3\text{Cl}_3(22)-\text{Cl}_3(66)\right)$$ Where $\text{Cl}$ is the Clausen function of order $3$: https://en.wikipedia.org/wiki/Clausen_function.

Can someone with a more advanced computer check if this result matches?

I could only verify up to $100$ decimal places.

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    $\begingroup$ Your "alternative of writing into Mathematica" doesn't seem to make sense with Mathematica syntax or known built-in functions. Can you write it in a form that's actually possible to copy-paste? Put it in between ` ` tick marks, too, if possible. That'll make it look like code. $\endgroup$
    – b3m2a1
    May 27, 2019 at 22:24
  • $\begingroup$ I think you should use the Sl function, not Cl, according to the Wikipedia: $S=[3Sl_3(22)-Sl_3(66)]/1372$. Clausen functions are real and imaginary parts of PolyLog in Mathematica. $\endgroup$
    – Roman
    May 28, 2019 at 10:53

1 Answer 1

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Are you sure the result is false?

sum = Sum[(Sin[22 n]/(7 n))^3, {n, 1, ∞}]
Block[{$MaxExtraPrecision = 10000},
     N[sum - 1/2 (π - 22/7)^3, 50]
]

$$ \left|\sum a_n-x\right|<10^{-10000.} $$

On a second thought,

FullSimplify[sum]

yields

1/686 (-22 + 7 π)^3

and so the sum is correct, and MMA is able to confirm it. Neat!

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