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Difference between revisions of "Raabe criterion"

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\frac{|a_{n+1}|}{|a_n|} \leq 1 - \frac{R}{n}
 
\frac{|a_{n+1}|}{|a_n|} \leq 1 - \frac{R}{n}
 
\end{equation}
 
\end{equation}
holds, then $\sum_n a_n$ converges absolutely. If instead there is $R\leq 1$ such that
+
holds, then $\sum_n a_n$ converges absolutely. If instead there is $N$ such that
 
\[
 
\[
\frac{|a_{n+1}|}{|a_n|} \geq 1 - \frac{R}{n}\,  
+
\frac{|a_{n+1}|}{|a_n|} \geq 1 - \frac{1}{n} \qquad \forall n \geq N\, ,
 
\]
 
\]
for sufficiently large $n$, then the series $\sum_n |a_n|$ diverges. However, the series itself might still converge, as can be seen taking  
+
then the series $\sum_n |a_n|$ diverges, which can be easily shown comparing it to the [[Harmonic series|harmonic series]]. However, the series itself might still converge, as can be seen taking  
 
\[
 
\[
 
\sum_n (-1)^n \frac{1}{\sqrt{n}}\, .
 
\sum_n (-1)^n \frac{1}{\sqrt{n}}\, .
 
\]
 
\]
 
+
The number $R$ is related to the limit
Observe moreover that the [[Harmonic series|harmonic series]] $\sum \frac{1}{n}$ (which diverges) and the series $\sum_n \frac{1}{n^2}$ (which converges) have both the property that
+
\[
 +
\lim_{n\to \infty} n \left(1-\frac{|a_n|}{|a_{n+1}|}\right)
 +
\]
 +
and the criterion can therefore be compared to [[Gauss criterion|Gauss' criterion]]. Observe however that the [[Harmonic series|harmonic series]] $\sum \frac{1}{n}$ (which diverges) and the series $\sum \frac{1}{n (\log n)^2}$ (which converges) have both the property that
 
\[
 
\[
 
\lim_{n\to \infty} n \left(1-\frac{a_n}{a_{n+1}}\right) = 1\, .
 
\lim_{n\to \infty} n \left(1-\frac{a_n}{a_{n+1}}\right) = 1\, .

Latest revision as of 10:49, 10 December 2013

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

on the convergence of a series of complex numbers

A criterion for the convergence of series of complex numbers $\sum_n a_n$, proved by J. Raabe. If $a_n \neq 0$ and there is a number $R>1$ such that for sufficiently large $n$ the inequality \begin{equation} \frac{|a_{n+1}|}{|a_n|} \leq 1 - \frac{R}{n} \end{equation} holds, then $\sum_n a_n$ converges absolutely. If instead there is $N$ such that \[ \frac{|a_{n+1}|}{|a_n|} \geq 1 - \frac{1}{n} \qquad \forall n \geq N\, , \] then the series $\sum_n |a_n|$ diverges, which can be easily shown comparing it to the harmonic series. However, the series itself might still converge, as can be seen taking \[ \sum_n (-1)^n \frac{1}{\sqrt{n}}\, . \] The number $R$ is related to the limit \[ \lim_{n\to \infty} n \left(1-\frac{|a_n|}{|a_{n+1}|}\right) \] and the criterion can therefore be compared to Gauss' criterion. Observe however that the harmonic series $\sum \frac{1}{n}$ (which diverges) and the series $\sum \frac{1}{n (\log n)^2}$ (which converges) have both the property that \[ \lim_{n\to \infty} n \left(1-\frac{a_n}{a_{n+1}}\right) = 1\, . \]

References

[Kn] K. Knopp, "Theorie und Anwendung der unendlichen Reihen" , Springer (1964) (English translation: Blackie, 1951 & Dover, reprint, 1990)
How to Cite This Entry:
Raabe criterion. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Raabe_criterion&oldid=30919
This article was adapted from an original article by E.G. Sobolevskaya (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article