Why the Ratio Test Needs Careful Interpretation
A series can be difficult to classify by looking at its terms one at a time. The Ratio Test examines how the absolute values of consecutive terms compare. It is especially useful when terms contain factorials or powers, because taking a ratio often cancels much of the expression.
The test has three distinct outcomes: a ratio limit below one proves absolute convergence; a limit above one proves divergence; and a limit equal to one gives no conclusion. A correct justification must state which outcome applies and must not claim more than the test establishes.
Set Up the Test
Consider a series \(\sum_{n=1}^{\infty} a_n\). Its absolute-value series is \(\sum_{n=1}^{\infty}|a_n|\). The original series is absolutely convergent if this absolute-value series converges. Absolute convergence also guarantees convergence of the original series.
The Ratio Test examines the limit of \(\frac{|a_{n+1}|}{|a_n|}\). This ratio is defined when \(a_n\ne 0\). In the standard form of the test, the terms are nonzero for all sufficiently large \(n\), and the ratio has a limit. A finite number of initial terms does not affect whether a series converges or diverges.
- If \(0\le L<1\), then \(\sum |a_n|\) converges, so \(\sum a_n\) converges absolutely.
- If \(1<L\le\infty\), then \(\sum a_n\) diverges.
- If \(L=1\), the test is inconclusive: it does not establish convergence or divergence.
The numerator is the absolute value of the next term, and the denominator is the absolute value of the current term. Using absolute values matters: the test assesses the sizes of terms, not their signs.
Why a Limit Below One Proves Absolute Convergence
Suppose \(L<1\). Choose a number \(r\) such that \(L<r<1\). By the definition of a limit, the ratios are eventually less than \(r\). Thus, for some index \(N\), \(\frac{|a_{n+1}|}{|a_n|}<r\) whenever \(n\ge N\). Equivalently, \(|a_{n+1}|<r|a_n|\) for those indices.
Repeatedly applying this inequality bounds the tail terms by a geometric sequence. The first bound, at \(k=0\), is equality; for later terms the inequalities are strict:
Since \(0<r<1\), the geometric series \(\sum_{k=0}^{\infty}|a_N|r^k\) converges. The absolute-value tail is bounded term by term by this convergent geometric series, so it converges as well. Adding the finite number of terms before \(N\) does not change convergence. Therefore \(\sum |a_n|\) converges.
Why a Limit Above One Proves Divergence
Suppose \(1<L<\infty\). Choose \(r\) with \(1<r<L\). Eventually the consecutive ratios exceed \(r\), so \(|a_{n+1}|>r|a_n|\) for all sufficiently large \(n\). If \(L=\infty\), the ratios eventually exceed any chosen fixed number greater than one, so the same argument applies.
For some \(N\), iterating the inequality gives \(|a_{N+k}|\ge |a_N|r^k\) for \(k\ge 0\), with strict inequality for \(k\ge 1\). The \(k=0\) case is equality, not a strict inequality. Because \(r>1\) and \(|a_N|>0\), these term magnitudes do not approach zero. A necessary condition for a series to converge is that its terms approach zero. Therefore the series diverges.
Worked Example: A Factorial Series That Diverges
Worked Example: A Factorial Series That Diverges
Determine whether \(\sum_{n=1}^{\infty}\frac{n!}{3^n}\) converges or diverges using the Ratio Test. Let \(a_n=\frac{n!}{3^n}\). The terms are positive, so \(|a_n|=a_n\).
First calculate the next term: \[ a_{n+1}=\frac{(n+1)!}{3^{n+1}}. \] Now form the ratio and simplify: \[ \frac{|a_{n+1}|}{|a_n|} = \frac{(n+1)!}{3^{n+1}}\cdot\frac{3^n}{n!} = \frac{n+1}{3}. \] Taking the limit gives \[ \lim_{n\to\infty}\frac{n+1}{3}=\infty. \] For a direct check of the divergence conclusion, if \(n\ge 6\), then \(\frac{n+1}{3}\ge\frac{7}{3}>2\). Thus \(a_{n+1}>2a_n\) for every \(n\ge6\). In particular, for \(k\ge0\), \(a_{6+k}\ge a_6 2^k\), with equality at \(k=0\) and strict inequality when \(k\ge1\). Since \(a_6=\frac{6!}{3^6}=\frac{720}{729}=\frac{80}{81}>0\), the terms do not approach zero. The series diverges, as the Ratio Test predicts.
Worked Example: A Factorial Series That Converges Absolutely
Worked Example: A Factorial Series That Converges Absolutely
Determine whether \(\sum_{n=1}^{\infty}\frac{5^n}{n!}\) converges. Set \(a_n=\frac{5^n}{n!}\); again the terms are positive, so their absolute values are unchanged.
The next term is \(a_{n+1}=\frac{5^{n+1}}{(n+1)!}\), and the ratio simplifies as follows: \[ \frac{|a_{n+1}|}{|a_n|} = \frac{5^{n+1}}{(n+1)!}\cdot\frac{n!}{5^n} = \frac{5}{n+1}. \] Therefore, \[ \lim_{n\to\infty}\frac{5}{n+1}=0. \] Since \(0<1\), the Ratio Test establishes absolute convergence. To see the geometric comparison behind the result, for \(n\ge9\) we have \(\frac{5}{n+1}\le\frac{5}{10}=\frac12\). Hence \(a_{n+1}\le\frac12a_n\) from that point onward, and the tail is bounded by a convergent geometric series. The ratio limit and the comparison agree: the series converges absolutely.
What a Limit of One Does—and Does Not—Tell You
When \(L=1\), the Ratio Test is inconclusive. This means only that this test does not decide the series. It does not mean the series diverges, and it does not mean it converges. Other tests or known series classifications may still settle the question.
Worked Example: Two Series with Ratio Limit One
Compare \(\sum_{n=1}^{\infty}\frac1n\) and \(\sum_{n=1}^{\infty}\frac1{n^2}\). For the first series, set \(a_n=\frac1n\). Then \[ \frac{|a_{n+1}|}{|a_n|} = \frac{1/(n+1)}{1/n} = \frac{n}{n+1}, \qquad \lim_{n\to\infty}\frac{n}{n+1}=1. \] The Ratio Test is inconclusive. The series is the harmonic series, which diverges by the p-series classification.
For the second series, set \(b_n=\frac1{n^2}\). Its ratio is \[ \frac{|b_{n+1}|}{|b_n|} = \frac{1/(n+1)^2}{1/n^2} = \frac{n^2}{(n+1)^2} = \left(\frac{n}{n+1}\right)^2, \qquad \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^2=1. \] The Ratio Test is again inconclusive. This series converges by the p-series classification, since its exponent is \(2>1\). Thus both ratio limits really are one, but the series have different outcomes. The calculations confirm that a limit of one cannot decide convergence.
Writing the Justification Clearly
A strong written justification names the test, shows the ratio calculation, evaluates the limit, and states exactly the conclusion supported by the result. For example, when the limit is below one, say that the series converges absolutely. When it is above one or infinite, state that the series diverges. When it equals one, explicitly say that the Ratio Test is inconclusive and use another valid argument if a final classification is required.
Write \(a_n\) for the general term and check that the ratio is defined for all sufficiently large \(n\).
Find \(\frac{|a_{n+1}|}{|a_n|}\) and simplify before taking the limit.
Compare the result with one, keeping the three possible outcomes distinct.
Use absolute convergence for a limit below one, divergence for a limit above one, and “inconclusive” when the limit equals one.
Common Pitfall: Treating One as a Divergence Result
A common incorrect argument is: “For \(\sum 1/n\), the ratio limit is one, so the series diverges by the Ratio Test.” The limit calculation is correct, but the conclusion is not: the Ratio Test gives no conclusion when the limit equals one. The harmonic series does diverge, but its divergence must be justified by another result, such as the p-series classification.
The same correction prevents the opposite mistake. For \(\sum 1/n^2\), the ratio limit is also one, so the Ratio Test does not prove convergence either. That series converges by the p-series classification. A test’s inconclusive outcome is not a classification; it is a signal to choose another valid method.
Also avoid omitting absolute values when terms can be negative. The Ratio Test is based on \(\frac{|a_{n+1}|}{|a_n|}\). A signed ratio can behave differently because signs may alternate, while the test’s conclusions concern the magnitudes of terms.
Check Your Understanding
For each question, distinguish the ratio calculation from the conclusion it permits.
- If \(\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=\frac34\), what does the Ratio Test establish?
- If the limit is \(2\), what does the Ratio Test establish, and what necessary condition for convergence helps explain the result?
- If the limit is \(1\), what may you conclude from the Ratio Test alone?
- Why is \(\frac{|a_{n+1}|}{|a_n|}\) used instead of the signed ratio \(\frac{a_{n+1}}{a_n}\)?
- For \(\sum_{n=1}^{\infty}\frac{4^n}{n!}\), calculate the consecutive absolute-term ratio and its limit, then state the test’s conclusion.
