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The sum is $$\sum_{n>0} \mathrm{i}^n\frac{J_n(x)}{n}\sin\frac{2\pi n}{3}\stackrel{?}{=}0$$ and $$\sum_{n>0} (\mathrm{-i})^n\frac{J_n(x)}{n}\sin\frac{2\pi n}{3}\stackrel{?}{=}0$$

I suspect they are zero because I am working on a project which from symmetry consideration they should be zero.

If anyone can point out they cannot be zero, or they cannot identically be zero, that will be great, too.

I've asked a question about Bessel function before, see Does this Infinite summation of Bessel function has a closed form? I hope the reference it contains may do some help.

Community
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an offer can't refuse
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2 Answers2

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For $x = 1$, numerical evaluation of the first sum in Maple gives me $$0.0502908663091560136488560173842+ 0.381051541033830840946600100117\, i$$ so I can be pretty confident that they are not $0$.

Robert Israel
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By using the generating function \begin{align} e^{i x \cos\phi} = J_{0}(x) + 2 \sum_{n=1}^{\infty} i^{n} \, J_{n}(x) \, \cos(n \phi) \end{align} integration with respect to $\phi$leads to \begin{align} \sum_{n=1}^{\infty} \frac{i^{n}}{n} \, J_{n}(x) \, \sin\left(\frac{2 n \pi}{3}\right) = \frac{\pi}{3} \, J_{0}(x) + \frac{1}{2} \int_{0}^{2\pi/3} e^{i x \cos\phi} \, d\phi. \end{align} The integral is not equal to $- \frac{\pi}{3} \, J_{0}(x)$, which leads to a non-zero result. For the second summation letting $x \rightarrow - x$ leads to \begin{align} \sum_{n=1}^{\infty} \frac{(-i)^{n}}{n} \, J_{n}(x) \, \sin\left(\frac{2 n \pi}{3}\right) = \frac{\pi}{3} \, J_{0}(x) + \frac{1}{2} \int_{0}^{2\pi/3} e^{-i x \cos\phi} \, d\phi. \end{align}

Leucippus
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