# How to show that a function has slow or rapid decay

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? –  Mariano Suárez-Alvarez Oct 27 '10 at 17:28
Slow/fast relative to what? –  Ｊ. Ｍ. Oct 27 '10 at 17:29
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 -- en.wikipedia.org/wiki/Long_range_dependence.) –  Olumide Oct 27 '10 at 17:43
If it is the convergence/divergence of the series you are looking for then you should read about the "integral test". –  AD. Oct 27 '10 at 18:18
Thanks. A search for the integral test led me to Paul Dawkins' notes (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. –  Olumide Oct 27 '10 at 22:15

If you start at 1 (your sum is not defined at x=0), you can bound the sums as integrals: $\int_{x=2}^{\infty}x^{-\alpha} dx \lt \sum_{x = 1}^\infty x^{-\alpha} \lt \int_{x=1}^{\infty}x^{-\alpha} dx$. One will solve it for $\alpha \le 1$ and one for $\alpha \gt 1$

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Another approach is to use Cauchy's condensation test which states that a series $\sum a_n$, where the terms $a_n$ are positive and decreasing, is convergent if and only if $\sum 2^m a_{2^m}$ is convergent.

In this example the line $\mathbf{Re}(s)=1$ is the edge of the region of ansolute convergence of the usual formula $\sum_{n=1}^\infty 1/n^s$ for the Riemann zeta function.

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According to the Wikipedia article, the terms $a_{n}$ should be non-increasing (which, of course, is not necessarily 'decreasing'). –  Robert Smith Oct 28 '10 at 0:58

I'm probably talking nonsense but

$\int_1^\infty \frac{1}{x^\alpha} = \left. \frac{x^{1 - \alpha}}{1 - \alpha} \right|_1^\infty$

The integral diverges if $1 - \alpha > 0$ i.e. if $\alpha < 1$, but converges if $1 - \alpha < 0$ i.e. $\alpha > 1$.

My difficulty now is in showing that the integral is a bound on the sum, which from what I've studied can only be done if the terms of the sum are taken as follows