# How to evaluate $\sum J_0(\alpha n) z^{-n}$ in closed form?

I need to evaluate $\sum_{n = -\infty}^{\infty} J_0(\alpha n) z^{-n}$ in closed form, where $z$ is complex variable and $J_0()$ is the zeroth order Bessel function of the first kind. How do I evaluate this summation?

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Explanation. In "signal processing" they say "Z transform" but mathematicians don't use that terminology. The question may be translated to:evaluate $$\sum_{n=-\infty}^{\infty}J_0(\alpha n)z^{-n}$$ is closed form. –  GEdgar Mar 24 '12 at 16:27
Since $J_0(x)$ is asymptotically proportional to $x^{-1/2} \cos x$, wouldn't the series diverge for all $z$? –  Steven Stadnicki Mar 28 '12 at 16:56
@StevenStadnicki Since $Z = r e^{j\theta}$, for large enough value of $r$ the sum $\sum |J_0(\alpha n) r^{-n}|$converges? –  sauravrt Mar 28 '12 at 17:07
@sauravrt But you're summing over positive and negative $n$, so if the terms go to $0$ on one side (e.g., if $|r|< 1$ then the terms go to $0$ for $n\rightarrow\infty$) then they grow unboundedly on the other (e.g., $n\rightarrow -\infty$). The harmonic term tweaks this behavior slightly, but not enough to get any kind of convergence. –  Steven Stadnicki Mar 28 '12 at 17:24

You can prove (or maybe already know) that in fact $J_0(-z)=J_0(z)$. So we can split up the sum as $$\underset{n=-\infty}{\overset{\infty}{\sum}}J_0(an)z^{-n}=\underset{n=-\infty}{\overset{-1}{\sum}}J_0(an)z^{-n}+\underset{n=0}{\overset{\infty}{\sum}}J_0(an)z^{-n} = \underset{n=1}{\overset{\infty}{\sum}}J_0(an)z^{n}+\underset{n=0}{\overset{\infty}{\sum}}J_0(an)z^{-n}$$ $$= J_0(0)+\underset{n=1}{\overset{\infty}{\sum}}J_0(an)(z^n+z^{-n}) = 1+\underset{n=1}{\overset{\infty}{\sum}}J_0(an)(z^n+z^{-n})$$