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Does the sequence $|a^{-n} J_0(n)|$ converge? I used the approximation $J_0(n) \approx \sqrt{\frac{2}{\pi n}} \cos(n - \pi/4)$ and assume that $a > 1$. Since the sample sequence of $J_0(n)$ is weighted by a decaying exponential $a^{-n}$, I suspect this will converge, but I'm unable to confirm this.

I want to know if

$\sum_{n=1}^\infty a^{-n}|J_0(n)|$

converges for above approximation?

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1 Answer 1

up vote 1 down vote accepted

Since $a^{-n}|J_0(n)|\sim a^{-n}\sqrt{\frac 2{\pi n}}|\cos(n-\pi /4)|$, the series $\sum_{n\geq 0}a^{-n}|J_0(n)|$converges if and only if $\sum_{n\geq 1}a^{-n}\sqrt{\frac 2{\pi n}}|\cos(n-\pi /4)|$ converges. But for $n\geq 1$: $$0\leq a^{-n}\sqrt{\frac 2{\pi n}}|\cos(n-\pi /4)|\leq a^{-n}$$ so the series $\sum_{n\geq 1}a^{-n}\sqrt{\frac 2{\pi n}}|\cos(n-\pi /4)|$ converges. In particular, $\lim_{n\to\infty}\sqrt{\frac 2{\pi n}}\cos(n-\pi /4)=0$

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So I understand that the sequence $a^{−n}|J_0(n)|$ is bounded by $a_{-n}$ and since the sequence $a_{-n}$ is converges for $n \geq 1$, the sequence $a^{−n}|J_0(n)|$ also converges. Is my understanding correct? –  sauravrt Mar 22 '12 at 16:47
    
Yes, it is correct. –  Davide Giraudo Mar 22 '12 at 20:20

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