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 $1 . Would anyone please provide some insight on this?
Summability of a sinc function power 'p', where 1
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sequences-and-series
signal-processing
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3$\operatorname{sin}(\pi n) = 0$ for any $n\in \mathbf{N}$. Is there any ambiguity to be revised in your description? – 2011-04-18
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0Thank you for pointing this out Ansel. There were alot of thoughts going through my head when i jotted this down! – 2011-04-18
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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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0Thank you Julian. Would you know of any way to get a closed form of the expression? – 2011-04-18
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0Using 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. – 2011-04-18
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0About my previous comment: the value of the sum is valid for $0< a<2$. – 2011-04-19