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We know that a sum of the form $\sum_{n=0}^{\infty} \left|\frac{sin(a\pi n)}{a\pi n}\right|$ where $a$ is not an integer, is unbounded and tends to infinity. But what about the expression $\sum_{n=0}^{\infty} \left|\frac{sin(a\pi n)}{a\pi n}\right|^p$ where $1<p<2$. Would anyone please provide some insight on this?

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$\operatorname{sin}(\pi n) = 0$ for any $n\in \mathbf{N}$. Is there any ambiguity to be revised in your description? –  Ansel Apr 18 '11 at 12:34
    
Thank you for pointing this out Ansel. There were alot of thoughts going through my head when i jotted this down! –  Ahmed Apr 18 '11 at 13:18
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1 Answer

up vote 1 down vote accepted

Since $|\sin(a\pi n)|\le1$ and the potential series $\sum_{n=1}^{\infty} n^{-p}$ converges if and only if $p>1$, the comparison principle implies that your series converges for all $p>1$.

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Thank you Julian. Would you know of any way to get a closed form of the expression? –  Ahmed Apr 18 '11 at 13:44
    
Using Fourier series you can check that $\sum_{n=0}^\infty\frac{\sin(a\pi n)}{a\pi n}=1+\frac{1-a}{2a}$. For the sums with absolute value, I have no idea of how to get a closed expression for its value. –  Julián Aguirre Apr 18 '11 at 22:04
    
About my previous comment: the value of the sum is valid for $0< a<2$. –  Julián Aguirre Apr 19 '11 at 7:47
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