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How to show that? $$\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^3=\frac{1}{2}\left(\pi-\frac{22}{7}\right)^3$$

I have no ideas to prove it, but it seems correct via Wolfram's calculator

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    $\begingroup$ Looks related to this one, also here. See if it helps you to start ... $\endgroup$ – rtybase May 27 at 21:18
  • $\begingroup$ This also seems to hold if you replace the pair $(22,7)$ with $(3,1)$ or $(355,113)$... $\endgroup$ – Carl Schildkraut May 27 at 22:50
  • $\begingroup$ @CarlSchildkraut Could be something to do with the $\pi$ continued fraction? $\endgroup$ – Anvit May 28 at 5:08
  • $\begingroup$ @Anvit That was my guess; it also seems to work for $(4,1)$ as well though. $\endgroup$ – Carl Schildkraut May 28 at 6:44
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    $\begingroup$ As well, something to do with Fourier series ? $\endgroup$ – Jean Marie May 28 at 8:03
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First some preliminary work that will be used later:

We have for $x\in(0,2\pi)$ $$\frac{\pi-x}{2}=\sum_{n=1}^\infty\frac{\sin(nx)}{n}$$ Shifting by $6\pi$ we have for $x\in (6\pi, 8\pi)$ $$\frac{7\pi-x}{2}=\sum_{n=1}\frac{\sin(nx)}{n} \tag1 $$ Or similarly for $x\in (20\pi,22\pi)$ $$\frac{21\pi-x}{2}=\sum_{n=1}\frac{\sin(nx)}{n}\tag 2$$

Integrating $(1)$ with respect to $x$ yields $$\sum_{n=1}\frac{\cos(nx)}{n^2}=\frac{(7\pi-x)^2}{4}+C$$ Now set $x=7\pi$ to get $C=-\frac{\pi^2}{12}$ and integrate again $$\sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=-\frac{(7\pi-x)^3}{12}-\frac{\pi^2}{12}x+K$$ And finally put $x=7\pi $ to get $K=7\pi\cdot \frac{\pi^2}{12}$. Thus for $x\in(6\pi,8\pi)$ we have $$\sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=-\frac{(7\pi-x)^3}{12}-\frac{\pi^2}{12}x+\frac{7\pi^3}{12}\tag3$$ Similarly things for $(2)$ yields for $x\in(20\pi,22\pi)$ $$\sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=-\frac{(21\pi-x)^3}{12}-\frac{\pi^2}{12}x+\frac{21\pi^3}{12}\tag4$$


Now back to the original sum. We have the formula $4\sin^3 x =3 \sin x-\sin(3x) $ so $$S=\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^3=\frac{1}{4\cdot 7^3}\left(3\sum_{n=1}^\infty \frac{\sin(22n)}{n^3}-\sum_{n=1}^\infty \frac{\sin(66n)}{n^3}\right)=\frac{1}{4\cdot 7^3}\left(3S_1-S_2\right)$$ Now things are easy because for $S_1$ we can set $x=22$ in $(3)$ and for $S_2$ we can set $x=66$ in $(4)$. $$ S_1=\sum_{n=1}^\infty \frac{\sin(22n)}{n^3}=-\frac{(7\pi-22)^3}{12}-\frac{22\pi^2}{12}+\frac{7\pi^3}{12}$$ $$S_2=\sum_{n=1}^\infty \frac{\sin(66n)}{n^3}=-\frac{(21\pi-66)^3}{12}-\frac{66\pi^2}{12}+\frac{21\pi^3}{12}$$ $$\Rightarrow S=\frac{1}{4\cdot 7^3}\left((7\pi-22)^3\left(-\frac3{12} +\frac{3^3}{12}\right)\right)=\frac{1}{2}\left(\pi-\frac{22} {7}\right)^3$$


Generalization. We have for $x\in\left((k-1)\pi,(k+1)\pi\right)$ $$\frac{k\pi-x}{2}=\sum_{n=1}^\infty \frac{\sin (nx)}{n}$$ $$\Rightarrow \sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=-\frac{(k\pi-x)^3}{12}-\frac{\pi^2}{12}x+\frac{k\pi^3}{12}$$ And for $x\in\left((3k-1)\pi,(3k+1)\pi\right)$ $$\sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=-\frac{(3k\pi-x)^3}{12}-\frac{\pi^2}{12}x+\frac{3k\pi^3}{12}$$ Here is where the magic happens: $$S(a,b)=\sum_{n=1}^\infty \frac{\sin^3(an)}{(bn)^3}=\frac{1}{4b^3}\left(3\sum_{n=1}^\infty \frac{\sin(an)}{n^3}-\sum_{n=1}^\infty \frac{\sin(3an)}{n^3}\right)$$ $$=\frac{1}{4b^3}\left(-3\frac{(k\pi-a)^3}{12}-\frac{3\pi^2}{12}a+\frac{3k\pi^3}{12}+\frac{(3k\pi-3a)^3}{12}+\frac{3\pi^2}{12}a-\frac{3k\pi^3}{12}\right)$$ $$=\frac{1}{4b^3}\left((k\pi-a)^3 \left(-\frac{3}{12}+\frac{27}{12}\right)\right)=\frac{1}{2b^3}(k\pi-a)^3$$ So for example a random series: $$S(123,321)=\sum_{n=1}^\infty \frac{\sin^3(123n)}{(321n)^3}=\frac{1}{2\cdot(321)^3}(39\pi-123)^3$$ If we set $b=k$ we get quite interesting things, mostly those combinations are found here, but the series is evaluable in an elementary form for any pair of numbers.

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    $\begingroup$ Probably you should mention that the "magic" (perfect cube) happens only for one third of integer numbers (see my comment to the answer of Claude Leibovici). Both $a=22$ and $a=123$ are such numbers (as well as other integer numbers close to an odd multiple of $\pi$). $\endgroup$ – user May 29 at 9:54
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    $\begingroup$ I was also thinking the generalization works only if both $a \in ((k-1)\pi, (k+1)\pi)$ and $3a \in ((3k - 1)\pi, (3k + 1)\pi)$, which means in fact $a \in ((k - 1/3)\pi, (k+1/3)\pi)$. This is a stronger condition, and is not satisfied for example by $a = 2$. But at least it does ensure uniqueness of the $k$, if it exists. $\endgroup$ – Tob Ernack May 29 at 16:05
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    $\begingroup$ For the cases not covered above, you can probably choose a different $k$ for the sum of $\frac{\sin(3an)}{n^3}$ than for the sum of $\frac{\sin(an)}{n^3}$ and a similar result should work. $\endgroup$ – Tob Ernack May 29 at 16:12
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    $\begingroup$ Damn, you're like Euler reincarnate. $\endgroup$ – clathratus Jun 27 at 21:47
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Too long for a comment.

I think that we could make amazing identities for the more general case of $$S_{a,b}=\sum_{n=1}^{\infty}\left(\frac{\sin(an)}{bn}\right)^3=\frac 1{b^3}\sum_{n=1}^{\infty}\left(\frac{\sin(an)}{n}\right)^3$$ Using first $$\sin ^3(a n)=\frac{3}{4} \sin (a n)-\frac{1}{4} \sin (3 a n)$$ $$S_{a,b}=\frac{3}{4b^3}\sum_{n=1}^{\infty}\frac{\sin(an)}{n^3}-\frac{1}{4b^3}\sum_{n=1}^{\infty}\frac{\sin(3an)}{n^3}$$ which is the imaginary part of $$T_{a,b}=\frac{3}{4b^3}\sum_{n=1}^{\infty}\frac{e^{ian}}{n^3}-\frac{1}{4b^3}\sum_{n=1}^{\infty}\frac{e^{3ian}}{n^3}$$ and now use the fact that $$\sum_{n=1}^{\infty}\frac{e^{ikn}}{n^3}=\text{Li}_3\left(e^{i k}\right)$$ As a result $$S_{a,b}=\frac{i}{8 b^3} \left(3 \text{Li}_3\left(e^{-i a}\right)-3 \text{Li}_3\left(e^{i a}\right)-\text{Li}_3\left(e^{-3 i a}\right)+\text{Li}_3\left(e^{3 i a}\right)\right)$$

Now, for the present case, $$i \left(\text{Li}_3\left(e^{-22 i}\right)-\text{Li}_3\left(e^{22 i}\right)\right)=-\frac{2}{3} (3 \pi -11) (4 \pi -11) (7 \pi -22)$$ $$i \left(\text{Li}_3\left(e^{-66 i}\right)-\text{Li}_3\left(e^{66 i}\right)\right)=-22 (\pi -3) (7 \pi -22) (10 \pi -33)$$ make $$S_{22,b}=\frac{(7 \pi -22)^3}{2 b^3}=\frac 12\left(\frac{7\pi}b-\frac {22} b \right)^3$$

In fact, exploring the cases where $$i\left(3 \text{Li}_3\left(e^{-i a}\right)-3 \text{Li}_3\left(e^{i a}\right)-\text{Li}_3\left(e^{-3 i a}\right)+\text{Li}_3\left(e^{3 i a}\right)\right)$$ is a multiple of a perfect cube, up to $a=100$ is found the sequence $$\{3,4,9,10,15,16,21,\color{red}{22},23,28,29,34,35,40,41,47,48,53,54,59,60,65,66,67,72,73,78,79,84, 85,91,92,97,98\}$$

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  • $\begingroup$ I assume that 16 should be added to the list. In fact it consists of the numbers satisfying the equation $$\left\lfloor\frac{3n}{2\pi}\right\rfloor-3\left\lfloor\frac{n}{2\pi}\right\rfloor=1.$$ $\endgroup$ – user May 29 at 8:57
  • $\begingroup$ @user. You are very correct ! I shall edit. Cheers :-) $\endgroup$ – Claude Leibovici May 29 at 9:04

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