Ratio test

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In mathematics, the ratio test is a criterion for convergence or divergence of a series whose terms are real or complex numbers. It considers the ratio of successive terms of the series; if the ratio tends to a limit less than 1 for terms further along the series, then it converges absolutely. The test was first published by Jean le Rond d'Alembert and is sometimes known as d'Alembert's ratio test.

Formally, the ratio test states that if

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

then the series

\sum_{n=1}^\infty a_n

converges absolutely, and if

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

then the series diverges. In particular, if

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

exists, then the series converges absolutely if that limit is < 1 and diverges if it is > 1. There exist both convergent and divergent series for which the limit is exactly 1, and consequently the test is inconclusive in that case.

An extension of the ratio test due to Raabe allows one to sometimes deal with the case when the limit is exactly 1. 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 test and Raabe's test are the first and second theorem in a hierarchy of such theorems due to Augustus De Morgan.

Contents

Examples

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.

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.

References

  • Knopp, Konrad, "Infinite Sequences and Series", Dover publications, Inc., New York, 1956. (§ 3.3, 5.4) ISBN 0486601536

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