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Difference between revisions of "Ermakov convergence criterion"

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''for a series with positive numbers as terms''
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{{MSC|40A05}}
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{{TEX|done}}
  
Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036200/e0362001.png" /> be a positive decreasing function for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036200/e0362002.png" />. If the inequality
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A criterion for the convergence of a series $\sum_n f(n)$, where $f:[1, \infty[\to [0, \infty[$ is a monotone decreasing function, established by
  
<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/e/e036/e036200/e0362003.png" /></td> </tr></table>
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Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036200/e0362001.png" /> be a positive decreasing function for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036200/e0362002.png" />. If the inequality V.P. Ermakov in {{Cite|Er}}. If there is $\lambda< 1$ such that
 
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\[
holds for these values of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036200/e0362004.png" /> with a <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/e/e036/e036200/e0362005.png" />, then the series
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\frac{e^x f(e^x)}{f(x)} < \lambda
 
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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/e/e036/e036200/e0362006.png" /></td> </tr></table>
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for sufficiently large $x$, then the series $\sum_n f(n)$ converges. If instead
 
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\[
converges; if
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\frac{e^x f(e^x)}{f(x)}\geq 1
 
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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/e/e036/e036200/e0362007.png" /></td> </tr></table>
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for all sufficiently large $x$, then the series diverges. In particular the convergence or divergence of the series can be decided of the limit
 
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\[
then the series diverges. In particular, if the following limit exists and
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\lim_{x\to\infty} \frac{e^x f(e^x)}{f(x)}
 
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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/e/e036/e036200/e0362008.png" /></td> </tr></table>
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exists and differs from 1.
 
 
then the series converges (diverges). This criterion was established by V.P. Ermakov [[#References|[1]]].
 
 
 
====References====
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  V.P. Ermakov,  "A new criterion for convergence and divergence of infinite series of constant sign" , Kiev  (1872) (In Russian)</TD></TR></table>
 
 
 
 
 
 
 
====Comments====
 
  
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Ermakov's criterion can be derived from the [[Integral test|integral test]].
  
 
====References====
 
====References====
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  T.J. Bromwich,  "An introduction to the theory of infinite series" , Macmillan  (1947)</TD></TR></table>
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{|
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|-
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|valign="top"|{{Ref|Br}}|| T.J. Bromwich,  "An introduction to the theory of infinite series" , Macmillan  (1947)
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|-
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|valign="top"|{{Ref|Er}}|| V.P. Ermakov,  "A new criterion for convergence and divergence of infinite series of constant sign" , Kiev  (1872)  (In Russian)
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|-
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|}

Revision as of 13:31, 10 December 2013

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

A criterion for the convergence of a series $\sum_n f(n)$, where $f:[1, \infty[\to [0, \infty[$ is a monotone decreasing function, established by

Let be a positive decreasing function for . If the inequality V.P. Ermakov in [Er]. If there is $\lambda< 1$ such that \[ \frac{e^x f(e^x)}{f(x)} < \lambda \] for sufficiently large $x$, then the series $\sum_n f(n)$ converges. If instead \[ \frac{e^x f(e^x)}{f(x)}\geq 1 \] for all sufficiently large $x$, then the series diverges. In particular the convergence or divergence of the series can be decided of the limit \[ \lim_{x\to\infty} \frac{e^x f(e^x)}{f(x)} \] exists and differs from 1.

Ermakov's criterion can be derived from the integral test.

References

[Br] T.J. Bromwich, "An introduction to the theory of infinite series" , Macmillan (1947)
[Er] V.P. Ermakov, "A new criterion for convergence and divergence of infinite series of constant sign" , Kiev (1872) (In Russian)
How to Cite This Entry:
Ermakov convergence criterion. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Ermakov_convergence_criterion&oldid=30931
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article