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Difference between revisions of "Logarithmic convergence criterion"

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A series of numbers <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l060/l060580/l0605801.png" />, where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l060/l060580/l0605802.png" />, converges if there is a number <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l060/l060580/l0605803.png" /> such that
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{{MSC|40A05}}
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{{TEX|done}}
  
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<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l060/l060580/l0605804.png" /></td> </tr></table>
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A criterion for the convergence of series $\sum a_n$ of positive real numbers. If there are $\alpha > 0$ and $N$ such that
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\begin{equation}\label{e:compare1}
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and diverges if
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\frac{\ln (a_n)^{-1}}{\ln n} \geq 1 + \alpha \qquad \forall n\geq N
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\end{equation}
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<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l060/l060580/l0605805.png" /></td> </tr></table>
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then the series converges. If there is $N$ such that
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\begin{equation}\label{e:compare2}
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\frac{\ln (a_n)^{-1}}{\ln n} \leq 1 \qquad \forall n \geq N
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\end{equation}
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then the series diverges. Indeed \eqref{e:compare1} implies that $\sum_n a_n$ is dominated by $\sum \frac{1}{n^{1+\alpha}}$ (namely that
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\[
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\left.a_n \leq \frac{1}{n^{1+\alpha}}\right)\, ,
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\]
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whereas \eqref{e:compare2} implies that $\sum_n a_n$ dominates the [[Harmonic series|harmonic series]].

Latest revision as of 10:35, 10 December 2013

2020 Mathematics Subject Classification: Primary: 40A05 [MSN][ZBL]

A criterion for the convergence of series $\sum a_n$ of positive real numbers. If there are $\alpha > 0$ and $N$ such that \begin{equation}\label{e:compare1} \frac{\ln (a_n)^{-1}}{\ln n} \geq 1 + \alpha \qquad \forall n\geq N \end{equation} then the series converges. If there is $N$ such that \begin{equation}\label{e:compare2} \frac{\ln (a_n)^{-1}}{\ln n} \leq 1 \qquad \forall n \geq N \end{equation} then the series diverges. Indeed \eqref{e:compare1} implies that $\sum_n a_n$ is dominated by $\sum \frac{1}{n^{1+\alpha}}$ (namely that \[ \left.a_n \leq \frac{1}{n^{1+\alpha}}\right)\, , \] whereas \eqref{e:compare2} implies that $\sum_n a_n$ dominates the harmonic series.

How to Cite This Entry:
Logarithmic convergence criterion. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Logarithmic_convergence_criterion&oldid=14834
This article was adapted from an original article by V.I. Bityutskov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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