Why Compare Consecutive Terms?
Some series are difficult to classify by comparing their terms directly with familiar benchmarks. Factorials can grow rapidly, while exponential factors can either amplify or offset that growth. The Ratio Test turns this competition into a calculation with consecutive terms: factorials often cancel, and an exponential factor contributes a constant multiplier.
The result is useful only when its conditions are met. A limiting ratio below one guarantees convergence; a limit above one guarantees divergence. A limit equal to one does not decide the question.
The Ratio Test
Suppose the terms \(a_n\) are nonzero for all sufficiently large \(n\), and the limit
exists, allowing \(L=+\infty\). The Ratio Test gives these conclusions:
- If \(L<1\), then \(\sum a_n\) converges absolutely.
- If \(L>1\), including \(L=+\infty\), then \(\sum a_n\) diverges.
- If \(L=1\), the test gives no conclusion.
Absolute convergence means that the series of magnitudes, \(\sum |a_n|\), converges. If terms are zero from some point onward, the series instead has only finitely many nonzero terms and converges; in that case the consecutive-term ratio need not be used.
Why the Test Works
The central reasoning is a comparison with a geometric series. When the ratio is eventually less than a fixed number below one, each successive term is bounded by a geometric step. When the ratio is eventually greater than one, the magnitudes of terms cannot approach zero.
Choose a number \(r\) such that \(L<r<1\). By the definition of the limit, there is an index \(N\) such that \(\left|\frac{a_{n+1}}{a_n}\right|<r\) for every \(n\ge N\).
Repeated multiplication gives \(|a_{N+k}|\le |a_N|r^k\) for every integer \(k\ge0\); equality holds when \(k=0\), and the inequality is strict when \(k\ge1\). Thus the tail is bounded term by term by a constant multiple of a convergent geometric series.
For \(0\le r<1\), the geometric tail beginning at \(k=m\) has sum \(r^m/(1-r)\), which tends to zero as \(m\to\infty\). The comparison shows the tails of \(\sum |a_n|\) tend to zero, so that series converges.
Choose \(c\) with \(1<c<L\) when \(L\) is finite; if \(L=+\infty\), choose any \(c>1\). Eventually \(|a_{n+1}|>c|a_n|\), so the magnitudes grow at least geometrically from that point and do not tend to zero. Since convergence of a series requires its terms to tend to zero, the series diverges.
This argument explains why the test concerns consecutive terms and why the two decisive cases are ratios below or above one. A ratio below one eventually forces geometric decrease; a ratio above one eventually forces growth. The test does not require every early ratio to have the same behavior as the limiting ratio.
Factorials and Exponentials in the Ratio
To apply the test, form \(a_{n+1}/a_n\) before taking a limit. Use the factorial identity \((n+1)!=(n+1)n!\), and its repeated version, to cancel matching factors. For an exponential factor, \(q^{n+1}/q^n=q\). These exact cancellations usually expose a constant or a simple expression whose limit is easy to find.
Do not estimate factorial growth from memory when direct cancellation is available. For instance, a factor \((2n)!\) changes to \((2n+2)!\) when \(n\) increases by one, adding two factors, not one.
Worked Example: A Factorial Ratio with an Exponential Factor
Determine whether the series
converges. Set \(a_n=\frac{n!\,3^n}{(2n)!}\). All terms are positive, so the ratio in the test is \(a_{n+1}/a_n\). Write out both terms:
Cancel \(n!\) and \(3^n\), then expand the two new factors in the denominator:
Therefore,
Since \(0<1\), the Ratio Test proves that the series converges absolutely. Checking the cancellation, increasing \(n\) by one adds the factor \(n+1\) in the numerator but adds both \(2n+2\) and \(2n+1\) in the factorial denominator; the simplified ratio is therefore consistent.
Worked Example: A Finite Ratio Limit Below One
Test the series
Let \(a_n=\frac{(2n)!}{(n!)^2\,9^n}\). The next term divided by this term is
Expand the new factorial factors and cancel:
Dividing numerator and denominator by \(n\), or comparing their leading terms, gives
Because \(\frac49<1\), the series converges absolutely. The limit is not zero: the factorial ratio tends to \(4\), while the factor \(9^{-n}\) contributes \(1/9\) to the consecutive-term ratio. Their product is \(4/9\), confirming the computed limit.
Worked Example: A Ratio Limit Above One
Determine the behavior of
Set \(a_n=\frac{3^n n!}{(n+2)!}\). Consecutive terms give
Thus
Since \(3>1\), the Ratio Test says that the series diverges. The actual ratio also verifies the reason: for \(n\ge4\), \(3(n+1)/(n+3)>2\), because \(3n+3>2n+6\) is equivalent to \(n>3\). Hence the terms eventually more than double at each step and cannot tend to zero. The divergence conclusion agrees with the necessary condition that the terms of a convergent series approach zero.
What the Limit Tells You—and What It Does Not
For factorial-and-exponential terms, the ratio limit summarizes their long-term competition. A factorial may leave a factor that grows with \(n\), while an exponential contributes a fixed factor. After exact cancellation, the resulting limit determines whether the Ratio Test applies decisively.
| Ratio limit | Conclusion | Reason |
|---|---|---|
| \(L<1\) | Absolute convergence | The tail is bounded by a convergent geometric series. |
| \(L>1\) | Divergence | Terms eventually fail to approach zero. |
| \(L=1\) | No conclusion from this test | The consecutive-term ratio alone does not settle convergence. |
A common mistake is to treat \(L=1\) as if it meant convergence. For example, the terms \(a_n=1/n\) have \(a_{n+1}/a_n=n/(n+1)\), which tends to \(1\), but the series \(\sum 1/n\) diverges. The limit \(1\) is not a verdict either way; it means only that this test has not decided the series. The Ratio Test at the boundary is a separate issue.
A Common Factorial-Cancellation Error
When the denominator contains \((2n)!\), increasing the index produces \((2n+2)!\), and therefore two new factors:
A tempting but incorrect cancellation is to write this quotient as just \(2n+2\). That drops the factor \(2n+1\), changing the ratio and potentially changing the limit. For example, in the first worked example the correct ratio is
not \(\frac{3(n+1)}{2n+2}=\frac32\). The correct limit is \(0\), not \(\frac32\). A reliable check is to expand each factorial only far enough to display every factor added when the index increases.
Using the Test Efficiently
For a series with factorials, exponentials, or both, organize the calculation in this order:
- Name the general term \(a_n\).
- Form \(\left|a_{n+1}/a_n\right|\), keeping all factors visible.
- Cancel factorials and exponential powers exactly.
- Evaluate the limit only after simplifying.
- Compare the limit with one and state the Ratio Test conclusion, including whether convergence is absolute.
This method keeps the algebra separate from the conclusion. It also makes it easier to catch missing factors before they affect the limit. The Ratio Test is especially effective when consecutive terms share most of their factorial and exponential structure; it is less informative when the limiting ratio is one.
Check Your Understanding
For each question, form the ratio of consecutive term magnitudes before deciding what the Ratio Test establishes.
- For \(a_n=\frac{n!}{5^n}\), calculate \(\left|\frac{a_{n+1}}{a_n}\right|\) and determine the series behavior.
- For \(a_n=\frac{4^n}{(2n)!}\), simplify the consecutive-term ratio and find its limit.
- For \(a_n=\frac{(2n)!}{(n!)^2\,5^n}\), determine the ratio limit and state what the test concludes.
- What conclusion does the Ratio Test give if the limiting ratio is \(1\)?
- Why must \((2n+2)!/(2n)!\) contain two factors?
