Can we sum $\sum _{m=0}^{\infty }t^{m}J_{m}\left( k\right)$ this series?

Question

If it were to be $\sum _{m=-\infty}^{\infty }t^{m}J_{m}\left( k\right) $, it is known that it is equal to some power of exponentials. But for this case, i could only manage to write

$$\sum _{m=1}^{\infty }t^{m}J_{m}\left( k\right)+\sum _{m=1}^{\infty }t^{m}J_{-m}\left( k\right)+J_0(k)=\sum _{m=-\infty}^{\infty }t^{m}J_{m}\left( k\right) $$

which did not help.

Use $J_{-m}(k)=(-1)^m J_m(k)$. (Or something like that. It should be obvious from the series definition.) See also Wikipedia's page on the Jacobi-Anger expansion. — Semiclassical, Jun 06 '16 at 00:21

score 0 · Accepted Answer · answered Jun 06 '16 at 15:29

Due to the Jacobi-Anger expansion, $$ e^{ik\cos\theta} = \sum_{n\in\mathbb{Z}} (ie^{i\theta})^n J_n(k) \tag{1} $$ hence:

$$ \sum_{m\in\mathbb{Z}} t^m J_m(k) = \color{red}{\exp\left(k\cdot \frac{t^2-1}{2t}\right)}\tag{2} $$

under the assumption that the LHS is convergent.

Can we sum $\sum _{m=0}^{\infty }t^{m}J_{m}\left( k\right)$ this series?

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