Ratio test

From Wikipedia, the free encyclopedia

In mathematics, the ratio test is a test (or "criterion") for the convergence of a series

$\sum_{n=0}^\infty a_n$

whose terms are real or complex numbers. The test was first published by Jean le Rond d'Alembert and is sometimes known as d'Alembert's ratio test. The test makes use of the number

$L = \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$

where "lim" denotes the limit as n goes to infinity,

The ratio test states that:

if L < 1 then the series converges absolutely, and
if L > 1 then the series diverges.

If L = 1 or if the limit does not exist, then the test is inconclusive (there exist both convergent and divergent series that satisfy those cases).

1 Examples
2 L=1 and Raabe's test
3 See also
4 References

[edit] Examples

[edit] Converging

Consider the series:

$\sum_{n=1}^\infty\frac{n}{e^n}$

Putting this into the ratio test:

$\lim_{n\rightarrow\infty}\left\|\frac{a_{n+1}}{a_n}\right\|$	= $\lim_{n\rightarrow\infty}\left\|\frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|\frac{n+1}{e^{n+1}}\cdot\frac{e^n}{n}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|\frac{n+1}{n}\cdot\frac{e^n}{e^n\cdot e}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|(1+\frac{1}{n})\cdot\frac{1}{e}\right\|$
	= $1\cdot\frac{1}{e}$
	= $\frac{1}{e} (<1)$

Thus the series converges as $\frac{1}{e}$ is less than 1.

[edit] Diverging

Consider the series:

$\sum_{n=1}^\infty\frac{e^n}{n}$

Putting this into the ratio test:

$\lim_{n\rightarrow\infty}\left\|\frac{a_{n+1}}{a_n}\right\|$	= $\lim_{n\rightarrow\infty}\left\|\frac{\frac{e^{n+1}}{n+1}}{\frac{e^n}{n}}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|\frac{e^{n+1}}{n+1}\cdot\frac{n}{e^n}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|\frac{n}{n+1}\cdot\frac{e^n\cdot e}{e^n}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|(1-\frac{1}{n+1})\cdot e\right\|$
	= $1\cdot e$
	= $\!\, e (>1)$

Thus the series diverges as $e$ is greater than 1.

[edit] Inconclusive

If one has

$\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$

it is impossible to deduce from the ratio test if the series converges or diverges.

For example, the series

$\sum_{n=1}^\infty 1$

diverges, but

$\lim_{n\rightarrow\infty}\left|\frac{1}{1}\right| = 1.$

On the other hand,

$\sum_{n=1}^\infty \frac{1}{n^2}$

converges absolutely, but

$\lim_{n\rightarrow\infty}\left|\frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}}\right| = 1.$

Finally,

$\sum_{n=1}^\infty (-1)^n\frac{1}{n}$

converges conditionally but

$\lim_{n\rightarrow\infty}\left|\frac{\frac{(-1)^{n+1}}{(n+1)}}{\frac{(-1)^{n}}{n}}\right| = 1.$

[edit] L=1 and Raabe's test

As seen in the previous example, the ratio test is inconclusive when the limit of the ratio is 1. An extension of the ratio test due to Raabe sometimes allows one to deal with this case. Raabe's test states that if

$\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$

and if a positive number c exists such that

$\lim_{n\rightarrow\infty} \,n\left(\,\left|\frac{a_{n+1}}{a_n}\right|-1\right)=-1-c$

then the series will be absolutely convergent. d'Alembert's ratio test and Raabe's test are the first and second theorem in a hierarchy of such theorems due to Augustus De Morgan.

[edit] See also

[edit] References

Knopp, Konrad, "Infinite Sequences and Series", Dover publications, Inc., New York, 1956. (§ 3.3, 5.4) ISBN 0-486-60153-6

Whittaker, E. T., and Watson, G. N., A Course in Modern Analysis, fourth edition, Cambridge University Press, 1963. (§ 2.36, 2.37) ISBN 0-521-58807-3

Retrieved from "http://en.wikipedia.org../../../r/a/t/Ratio_test.html"

Categories: Mathematical series | Convergence tests

$\lim_{n\rightarrow\infty}\left\|\frac{a_{n+1}}{a_n}\right\|$	= $\lim_{n\rightarrow\infty}\left\|\frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|\frac{n+1}{e^{n+1}}\cdot\frac{e^n}{n}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|\frac{n+1}{n}\cdot\frac{e^n}{e^n\cdot e}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|(1+\frac{1}{n})\cdot\frac{1}{e}\right\|$
	= $1\cdot\frac{1}{e}$
	= $\frac{1}{e} (<1)$

$\lim_{n\rightarrow\infty}\left\|\frac{a_{n+1}}{a_n}\right\|$	= $\lim_{n\rightarrow\infty}\left\|\frac{\frac{e^{n+1}}{n+1}}{\frac{e^n}{n}}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|\frac{e^{n+1}}{n+1}\cdot\frac{n}{e^n}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|\frac{n}{n+1}\cdot\frac{e^n\cdot e}{e^n}\right\|$
	= $\lim_{n\rightarrow\infty}\left\|(1-\frac{1}{n+1})\cdot e\right\|$
	= $1\cdot e$
	= $\!\, e (>1)$

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Ratio test

From Wikipedia, the free encyclopedia

Contents

[edit] Examples

[edit] Converging

[edit] Diverging

[edit] Inconclusive

[edit] L=1 and Raabe's test

[edit] See also

[edit] References

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