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Is the Bessel function $J_0(n)$ absolutely summable i.e $\sum_{n=0}^{\infty}|J_0(n)| < \rm C$? Since $\lim\limits_{n \to\infty} J_0(n) = 0$, I'd assume the absolute sum converges to a constant value too.

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up vote 2 down vote accepted

We use the asymptotic equivalent $J_0(n)\sim\sqrt{\frac 2{n\pi}}\cos(n-\pi/4)$, so $|J_0(n)|\sim\sqrt{\frac 2{n\pi}}|\cos(n-\pi/4)|=:a_n$. Let $b_n:=\sqrt{\frac{\pi}2}a_n$. Then $$2b_n\geq\frac 2{\sqrt n}\cos^2\left(n-\frac{\pi}4\right)=\frac 1{\sqrt n}\left(1+\cos\left(2n-\frac{\pi}2\right)\right)=\frac{1-\sin(2n)}{\sqrt n}\geq 0.$$ Since $\sum_{n>0}\frac 1{\sqrt n}$ is divergent and $\sum_{n>0}\frac{\sin (2n)}{\sqrt n}$ is convergent (use Abel transform), the series $\sum_{n>0}b_n$ is divergent and so is the series $\sum_{n=0}^{+\infty}|J_0(n)|$.

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From your description, it seems that divergence is mainly due to divergence of $\frac{1}{\sqrt{n}}$. Then in that case for $r > 1$, following should hold $\sum_{n>0} |r^{-n}J_0(n)| < \infty$. Or in other words I'm saying that Bessel function will be absolutely summable for some range of $r$. The idea here is similar with the concepts of Z-transform or Laplace transform which include wider class of functions compared to Fourier transform. –  sauravrt Mar 21 '12 at 1:10
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