Ratio test

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 non-zero 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

$L=\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|$ (1)

in the cases where this limit exists.

The ratio test states that:

• if L < 1 then the series converges absolutely;
• if L > 1 then the series does not converge;
• if L = 1 or the limit fails to exist, then the test is inconclusive (there exist both convergent and divergent series that satisfy that case).

In cases where the limit fails to exist, it is possible to generalize the test using a limit superior. Let

$L=\lim \sup \left|{\frac {a_{n+1}}{a_{n}}}\right|$

Then the ratio test states that:

• if L < 1, then the series converges absolutely, and
• if the inequality $\left|{\frac {a_{n+1}}{a_{n}}}\right|>1$ holds for all but finitely many n, then the series diverges.

The test is inconclusive otherwise. Note that if the divergence criterion holds then the absolute value of the series is increasing for large enough n, so $\lim a_{n}\neq 0$, which implies divergence. A slightly weaker version of the same divergence criteria can be stated as follows using the limit inferior:

• if $\ell =\lim \inf \left|{\frac {a_{n+1}}{a_{n}}}\right|>1,$ then the series diverges.

The test is inconclusive if ℓ ≤ 1 ≤ L.

If the limit in (1) exists, then it is equal to the limit superior and limit inferior, and so the original version of the ratio test is seen as a special case of the latter.

Examples

Converging

Consider the series:

$\sum _{n=1}^{\infty }{\frac {n}{e^{n}}}$

Putting this into the ratio test:

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {\frac {n+1}{e^{n+1}}}{\frac {n}{e^{n}}}}\right|\\&={\frac {1}{e}}<1.\end{aligned}}$

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:

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {\frac {e^{n+1}}{n+1}}{\frac {e^{n}}{n}}}\right|\\&=e>1.\end{aligned}}$

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

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

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

Proof

Suppose that $L=\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|<1$. We can then show that the series converges absolutely by showing that its terms eventually become less than those of a geometric series with r < 1. To do this, let $r={\frac {L+1}{2}}$. Then r is strictly between L and 1, and $|a_{n+1}|<r|a_{n}|$ for sufficiently large n (say, n greater than N). Hence $|a_{n+k}|<r^{k}|a_{n}|$ for each n > N and k > 0, and so

$\sum _{i=1}^{\infty }|a_{i}|=\sum _{i=1}^{N}|a_{i}|+\sum _{i=N+1}^{\infty }|a_{i}|=P_{N}+\sum _{i=1}^{\infty }|a_{N+i}|<P_{N}+\sum _{i=1}^{\infty }r^{i}|a_{N+1}|=P_{N}+|a_{N+1}|\sum _{i=1}^{\infty }r^{i}=P_{N}+|a_{N+1}|{\frac {r}{1-r}}<\infty$

where $P_{N}$ is the sum of the first N terms $\left|a_{n}\right|$. Hence the series converges absolutely.

On the other hand, if L > 1, then $|a_{n+1}|>|a_{n}|$ for sufficiently large n, so that the limit of the summands is non-zero. Hence the series diverges.

L=1

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 Joseph Ludwig 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

$\lim _{n\rightarrow \infty }\,n\left(\,\left|{\frac {a_{n+1}}{a_{n}}}\right|-1\right)<-1$

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.

Higher order tests

The next cases in de Morgan's hierarchy are Bertrand's and Gauss's test. Each test involves slightly different higher order asymptotics. If

$\left|{\frac {a_{n}}{a_{n+1}}}\right|=1+{\frac {1}{n}}+{\frac {\rho _{n}}{n\ln n}}$

then the series converges if lim inf ρ_n > 1, and diverges if lim sup ρ_n < 1. This is Bertrand's test.

If

$\left|{\frac {a_{n}}{a_{n+1}}}\right|=1+{\frac {h}{n}}+{\frac {C_{n}}{n^{r}}}$

where r > 1 and C_n is bounded, then the series converges if h > 1 and diverges if h ≤ 1. This is Gauss's test.

These are both special cases of Kummer's test for the convergence of the series Σa_n. Let ζ_n be an auxiliary sequence of positive constants. Let

$\rho =\lim _{n\to \infty }\left(\zeta _{n}{\frac {a_{n}}{a_{n+1}}}-\zeta _{n+1}\right).$

Then if ρ > 0, the series converges. If ρ < 0 and Σ1/ζ_n diverges, then the series diverges. Otherwise the test is inconclusive.

See also

• Root test
• Radius of convergence

References

• d'Alembert, J. (1768), Opuscules, vol. V, pp. 171–183.

• Apostol, Tom M. (1974), Mathematical analysis (2nd ed.), Addison-Wesley, ISBN 978-0-201-00288-1: §8.14.

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

• Rudin, Walter (1976), Principles of Mathematical Analysis (3rd ed.), New York: McGraw-Hill, Inc., ISBN 0-07-054235-X: §3.34.

• Weisstein, Eric W. "Bertrand's Test". MathWorld.

• Weisstein, Eric W. "Gauss's Test". MathWorld.

• Weisstein, Eric W. "Kummer's Test". MathWorld.

• Watson, G. N.; Whittaker, E. T. (1963), A Course in Modern Analysis (4th ed.), Cambridge University Press, ISBN 0-521-58807-3: §2.36, 2.37.