ratio test of d’Alembert
A lighter version of the ratio test is the
Ratio test of d’Alembert. Let a1+a2+… be a series with positive terms.
1∘. If there exists a number q such that 0<q<1 and
an+1an≤q for alln≥n0, | (1) |
then the series converges.
2∘. If there exists a number n0 such that
an+1an≥ 1 for alln≥n0, | (2) |
then the series diverges.
Proof. 1∘. By the condition (1), we have an+1≤anq; thus we get the estimations
an0+1≤an0q, |
an0+2≤an0+1q≤an0q2, |
⋯ ⋯ ⋯ |
an0+p≤an0+p-1q≤…≤an0qp, |
⋯ ⋯ ⋯ |
Because an0q+an0q2+…+an0qp+… is a convergent geometric series
, those inequalities
and the comparison test
imply that the series
an0+1+an0+2+…+an0+p+… |
and as well the whole series a1+a2+… is convergent.
2∘. The condition (2) yields
an0+1≥an0,an0+2≥an0+1≥an0,… |
and since an0 is positive, the limit of an as n tends to infinity cannot be 0. Hence the given series does not fulfil the necessary condition of convergence.
Example. If the variable x in the power series
∞∑n=0n!xn |
is distinct from zero, we have
|(n+1)!xn+1||n!xn|=(n+1)|x|≥ 1 for alln≥n0. |
Then the series does not converge absolutely (http://planetmath.org/AbsoluteConvergence). The known theorem of Abel says that the series diverges for all
x≠0. It means that the radius of convergence is 0.
References
- 1 Л. Д. Кудрявцев: Математический анализ. I том. Издательство ‘‘Высшая школа’’. Москва (1970).
Title | ratio test of d’Alembert |
---|---|
Canonical name | RatioTestOfDAlembert |
Date of creation | 2013-03-22 19:12:28 |
Last modified on | 2013-03-22 19:12:28 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 9 |
Author | pahio (2872) |
Entry type | Theorem |
Classification | msc 40A05 |
Related topic | FiniteChangesInConvergentSeries |