Is Bessel function $J_0(n)$ absolutely summable?

Question

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.

Answer 1

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 $\sum_{n=0}^{+\infty}|J_0(n)|$.