In mathematical analysis, the ratio test is a test (or "criterion") for the convergence of a series
∑ n = 1 ∞ a n , {\displaystyle \sum _{n=1}^{\infty }a_{n},}
where each term is a real or complex number and all but finitely many terms are non-zero. 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.
The test
The usual form of the test makes use of the limit
The ratio test states that:
if L < 1 then the series converges absolutely; if L > 1 then the series diverges; 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 | a n + 1 a n | {\displaystyle R=\lim \sup \left|{\frac {a_{n+1}}{a_{n}}}\right|}
r = lim inf | a n + 1 a n | {\displaystyle r=\lim \inf \left|{\frac {a_{n+1}}{a_{n}}}\right|} . Then the ratio test states that:
if R < 1, the series converges absolutely; if r > 1, the series diverges; or equivalently if | a n + 1 a n | > 1 {\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|>1} for all large n (regardless of the value of r), the series also diverges; this is because | a n | {\displaystyle |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.
Examples
Convergent because L < 1 Consider the series
∑ n = 1 ∞ n e n {\displaystyle \sum _{n=1}^{\infty }{\frac {n}{e^{n}}}}
Applying the ratio test, one computes the limit
L = lim n → ∞ | a n + 1 a n | = lim n → ∞ | n + 1 e n + 1 n e n | = 1 e < 1. {\displaystyle 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.}
Since this limit is less than 1, the series converges.
Divergent because L > 1 Consider the series
∑ n = 1 ∞ e n n . {\displaystyle \sum _{n=1}^{\infty }{\frac {e^{n}}{n}}.}
Putting this into the ratio test:
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