ChapterZero

Weierstrass’ criterion

Filed under: General — Alex @ 7:20 pm 8/13/2007

Here’s ‘an extremely far-reaching criterion’ that can be used to determine the convergence of a series (a proof can be found in Knopp’s Theory and Applications of Infinite Series):

A series $\sum_{n=0}^\infty a_n$ of complex terms, for which

$\displaystyle \frac{a_{n+1}}{a_n} = 1 - \frac{\alpha}{n} - \frac{A_n}{n^\lambda}$

with bounded– where $\alpha$ is complex and arbitrary, and $\lambda > 1$ – is absolutely convergent if, and only if, $\Re{(\alpha)} > 1.$

For $\Re{(\alpha)} \leq 0$ the series is invariably divergent. If $0 < \Re{(\alpha)} \leq 1$ , both the series

$\displaystyle \sum_{n=0}^\infty |(a_n - a_{n+1})| \text{ and } \sum_{n=0}^\infty (-1)^n a_n$

are convergent.

Furthermore, if $\Re{(\alpha)} \leq 1$ , the series is divergent.

In itself, this is a useful result, but it is particularly useful for determining the convergence properties of a power series on its circle of convergence:

If, as in the preceding theorem,

$\displaymath \frac{a_{n+1}}{a_n} = 1 - \frac{\alpha}{n} - \frac{A_n}{n^\lambda}$

where $\alpha$ is arbitrary, $\lambda > 1$ , and is bounded, the series $\sum a_n z^n$ is absolutely convergent for , divergent for every , and for the points of the circumference , the series will

converge absolutely, if $\Re{(\alpha)} > 1$

converge conditionally, if $0 < \Re{(\alpha)} \leq 1$ except for where it diverges

diverge, if $\Re{(\alpha)} \leq 0.$

That’s a pretty powerful corollary. As an example, use it to show that the generalized harmonic series (my term for the series $\sum_{n=0}^\infty \frac{z^n}{n}$ ) diverges only at one point on the unit circle!

Weierstrass’ criterion

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