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I've got a function whose general form is

$f(x) = \frac{1}{x^\alpha}$

where $x > 0$ and $\alpha > 0$. I would like to show that if $0< \alpha < 1$ $f(x)$ has slow decay and if $\alpha > 1$ the $f(x)$ has rapid decay. (I've already verified these properties of $f(x)$ using a graphing application.)

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    Well, what definitions of "rapid" and "slow decay" are you using?2010-10-27
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    Slow/fast relative to what?2010-10-27
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    I'm trying to understand why $\sum_{x = 0}^\infty f(x) = \infty$ if $\alpha > 1$ and why $\sum_{x = 0}^\infty f(x) < \infty$ if $0 < \alpha < 1$. (If it matters, this is in relation to long range dependence -- http://en.wikipedia.org/wiki/Long_range_dependence.)2010-10-27
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    If it is the convergence/divergence of the series you are looking for then you should read about the "integral test".2010-10-27
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    Thanks. A search for the integral test led me to Paul Dawkins' notes (http://tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx) which I am currently studying. But I'd like to ask if (in the absence of growth) divergence implies slow decay in some sense.2010-10-27

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