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I want to find $\displaystyle\sum_{n=1}^\infty\frac{\sin(n)}{n^3}$ by using method from this link : http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf

Since $f(z)=\frac{\sin{z}}{z^3}$ has a pole at $z=0$,

I get that $\displaystyle\sum_{n=-\infty ,n\neq0}^\infty\frac{\sin(n)}{n^3}=-Res_{z=0}\left(\frac{\pi\sin{z}\cot{(\pi z)}}{z^3}\right)=\frac{2\pi^2+1}{6}$ .

That is $\displaystyle\sum_{n=1}^\infty\frac{\sin(n)}{n^3}=\frac{2\pi^2+1}{12}$.

But wolfram compute that $\displaystyle\sum_{n=1}^\infty\frac{\sin(n)}{n^3}=\frac{(\pi-1)(2\pi-1)}{12}$. https://www.wolframalpha.com/input/?i=sum+sin(n)%2Fn%5E3,+n%3D1+to+inf

How am I wrong?

Thank you.

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  • $\begingroup$ the pole $z=0$ is double. $\endgroup$ – G Cab Jan 3 '18 at 10:02
  • $\begingroup$ @GCab It doesn't really matter: I did the Laurent series and got the same residue as the OP. $\endgroup$ – DonAntonio Jan 3 '18 at 10:03
  • $\begingroup$ @GCab : You mean has order 3? $$\frac{\pi\sin z\cos(\pi z)}{z^3\,\sin(\pi z)}$$ $\endgroup$ – LutzL Jan 3 '18 at 10:03
  • $\begingroup$ @Ble In the link you give there's only a result with the polylogarithm. Where did you get that number from? $\endgroup$ – DonAntonio Jan 3 '18 at 10:06
  • $\begingroup$ @DonAntonio : It is in "Alternate forms" which is in the 6th box of the WA answer. $\endgroup$ – LutzL Jan 3 '18 at 10:09
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You can not apply the approach from that paper directly as the key assumption of theorem 3.2 $$ |f(z)|\le \frac{M}{|z|^k} $$ for some $k>1$ and all large $z$, $|z|>R$ for some large $R$, is not satisfied for $f(z)=\frac{\sin z}{z^3}$ as the sine function is exponential in the imaginary directions, $$ \sin(z)=\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y) $$

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The best way to compute such series is using periodic functions. Suppose f is periodic with period $2\pi$ and is equal to ${x^{3}\over3}-{{\pi x^{2}\over 2}}+{{\pi ^{2}x}\over 6}$ in the interval $[-\pi,\pi]$. The Fourier series of this function is $\Sigma_{n=1}^{\infty} {\sin(2nx)\over n^{3}}$. Therefore replacing x=0.5 the result jumps out!

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  • $\begingroup$ very cute approach $\endgroup$ – G Cab Jan 3 '18 at 10:34
  • $\begingroup$ That's kind of you! $\endgroup$ – Mostafa Ayaz Jan 3 '18 at 10:39

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