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

\( \sum_{n=1}^\infty a_n, \)

where each term is a real or complex number and an is nonzero when n is large. The test was first published by Jean le Rond d'Alembert and is sometimes known as d'Alembert's ratio test or as the Cauchy ratio test.[1]


Given the following geometric series:

\( \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \)

The quotient

\( a_{n+1}/a_n = (1/2)^{n+1}/(1/2)^n \)

of any two adjacent terms is 1/2. The sum of the first m terms is given by:

\( 1 - \frac{1}{2^m}. \)

As m increases, this converges to 1, so the sum of the series is 1. On the other hand given this geometric series:

\( \sum_{n=1}^\infty 2^n = 2 + 4 + 8 + \cdots \)

The quotient \( a_{n+1}/a_n \) of any two adjacent terms is 2. The sum of the first m terms is given by

\( 2^{m+1} - 2, \)

which increases without bound as m increases, so this series diverges. More generally, the sum of the first m terms of the geometric series is given by:

\( \sum_{n=1}^{m} r^n = \frac{r}{r-1} (r^m - 1). \)

Whether this converges or diverges as m increases depends on whether r, the quotient of any two adjacent terms, is less than or greater than 1. Now consider the series:

\sum_{n=1}^\infty \frac{n+1}{n} \left(\frac{1}{2}\right)^n = \frac{2}{1} \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{4} + \frac{4}{3} \cdot \frac{1}{8} + \cdots \)

This is similar to the first convergent sequence above, except that now the ratio of two terms is not fixed at exactly 1/2:

\( \left(\frac{n+1}{n} \left(\frac{1}{2}\right)^n\right)/\left(\frac{n}{n-1} \left(\frac{1}{2}\right)^{n-1}\right) = \frac{n^2-1}{2n^2} = \frac{1}{2} - \frac{1}{2n^2}. \)

However, as n increases, the ratio still tends in the limit towards the same constant 1/2. The ratio test generalizes the simple test for geometric series to more complex series like this one where the quotient of two terms is not fixed, but in the limit tends towards a fixed value. The rules are similar: if the quotient approaches a value less than one, the series converges, whereas if it approaches a value greater than one, the series diverges.

The test

The usual form of the test makes use of the limit

\( L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|. \) (1)

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, because there exist both convergent and divergent series that satisfy this case.

It is possible to make the ratio test applicable to certain cases where the limit L fails to exist, if limit superior and limit inferior are used. The test criteria can also be refined so that the test is sometimes conclusive even when L = 1. More specifically, let

\( R = \lim\sup \left|\frac{a_{n+1}}{a_n}\right| \)
\( r = \lim\inf \left|\frac{a_{n+1}}{a_n}\right|. \)

Then the ratio test states that:[2][3]

if R < 1, the series converges absolutely;
if r > 1, the series diverges;
if \( \left|\frac{a_{n+1}}{a_n}\right|\ge 1 \) for all large n (regardless of the value of r), the series also diverges; this is because |a_n| is nonzero and increasing and hence an does not approach zero;
the test is otherwise inconclusive.

If the limit L in (1) exists, we must have L=R=r. So the original ratio test is a weaker version of the refined one.

Convergent because L<1

Consider the series

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

Putting this into the ratio test:

\( L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right| = \frac{1}{e} < 1. \)

As every term is positive, the series converges.
Divergent because L>1

Consider the series

\( \sum_{n=1}^\infty\frac{e^n}{n}. \)

Putting this into the ratio test:

\( L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{e^{n+1}}{n+1}}{\frac{e^n}{n}} \right| = e > 1. \)

Thus the series diverges.
Inconclusive because L=1

Consider the three series

\( \sum_{n=1}^\infty 1, \)
\( \sum_{n=1}^\infty \frac{1}{n^2}, \)
\( \sum_{n=1}^\infty (-1)^n\frac{1}{n}. \)

The first series diverges, the second one converges absolutely and the third one converges conditionally. However, the term-by-term magnitude ratios \( \left|\frac{a_{n+1}}{a_n}\right| \)of the three series are respectively \( 1, \frac{n^2}{(n+1)^2} \) and \( \frac{n}{n+1} \). So, in all three cases, we have \( \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1 \). This illustrates that when L=1, the series may converge or diverge and hence the original ratio test is inconclusive. For the first series \sum_{n=1}^\infty 1, however, as the term-by-term magnitude ratio \( \left|\frac{a_{n+1}}{a_n}\right|=1 \) for all n, we can apply the third criterion in the refined version of the ratio test to conclude that the series diverges.

In this example, the ratio of adjacent terms in the blue sequence converges to L=1/2. We choose r = (L+1)/2 = 3/4. Then the blue sequence is dominated by the red sequence rk for all n ≥ 2. The red sequence converges, so the blue sequence does as well.

Below is a proof of the validity of the original ratio test.

Suppose that \( L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_{n}}\right| < 1 \). We can then show that the series converges absolutely by showing that its terms will eventually become less than those of a certain convergent geometric series. 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 \( |a_{n+i}| < r^i|a_{n}| \) for each n > N and i > 0, and so

\( \sum_{i=N+1}^{\infty}|a_i| = \sum_{i=1}^{\infty} \left |a_{N+i} \right | < \sum_{i=1}^{\infty}r^{i}|a_{N+1}| = |a_{N+1}| \sum_{i=1}^{\infty} r^{i} = |a_{N+1}|\frac{r}{1 - r} < \infty. \)

That is, 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.

Extensions for L=1

As seen in the previous example, the ratio test may be inconclusive when the limit of the ratio is 1. Extensions to ratio test, however, sometimes allows one to deal with this case. For instance, the aforementioned refined version of the test handles the case

\( \left|\frac{a_{n+1}}{a_n}\right|\ge 1. \)

Below are some other extensions.
Raabe's test

This extension is due to Joseph Ludwig Raabe. It states that if

\( \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=1, \)
\( \lim_{n\to\infty} n\left(\left|\frac{a_n}{a_{n+1}}\right|-1\right)=R, \)

then the series will be absolutely convergent if R>1 and divergent if R<1.[4] 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.

Proof of Raabe's test

Let R > p > 1. Then:

\( \lim_{n\to \infty} n\left(\left|\frac {a_n}{a_{n+1}}\right|-1\right) = R > p =\lim_{ n\to \infty} n\left(\left(1+\frac {1}{n}\right)^p-1\right) \)
\( \Rightarrow \quad n\left(\left|\frac{a_n}{a_{n+1}}\right|-1\right) > n\left(\left(1+\frac {1}{n}\right)^p-1\right) \quad \text{ for large } n
\Rightarrow \quad \left|\frac {a_n}{a_{n+1}}\right| > \frac {(n+1)^p}{n^p} \)

and since the p-series \( \sum n^{-p} \) converges for p>1 the series \( \sum { { a }_{ n } } \) converges for R>1. The divergence for R<1 follows by comparing with the harmonic series.[5]
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 ρn > 1, and diverges if lim sup ρn < 1. This is Bertrand's test.[6]


\( \left|\frac{a_n}{a_{n+1}}\right| = 1+ \frac{h}{n} + \frac{C_n}{n^r} \)

where r > 1 and Cn is bounded, then the series converges if h > 1 and diverges if h ≤ 1. This is Gauss's test.[7]

These are both special cases of Kummer's test for the convergence of the series Σan. Let ζn 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/ζn diverges, then the series diverges. Otherwise the test is inconclusive.[8]

See also

Root test
Radius of convergence

Rudin 1976, §3.34
Apostol 1974, §8.14


d'Alembert, J. (1768), Opuscules 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.
Hazewinkel, Michiel, ed. (2001), "Bertrand criterion", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
Hazewinkel, Michiel, ed. (2001), "Gauss criterion", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
Hazewinkel, Michiel, ed. (2001), "Kummer criterion", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
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.

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