When the Ratio Test Reaches Its Boundary
The Ratio Test is especially useful when consecutive terms have a simple relationship, as happens with factorials and exponential factors. But sometimes the calculation leads to the boundary value \(1\). That result is not a verdict: it does not tell us whether the series converges or diverges.
The important next step is to recognize the limit as inconclusive and find another property of the terms that can settle the question. In this tutorial, the same ratio limit will occur for series with opposite outcomes.
The Ratio Test and Its Conditions
Consider a series \(\sum_{n=1}^{\infty} a_n\). Suppose its terms are nonzero eventually, so the consecutive ratios are defined, and suppose the limit exists:
The Ratio Test gives a conclusion when this limit is strictly below or strictly above \(1\). When the limit equals \(1\), the test is inconclusive. For a series with both positive and negative terms, this test concerns absolute convergence through the ratios of absolute values.
The test requires a limit of the consecutive absolute ratios. If the limit does not exist, this version of the test does not supply a conclusion. In particular, a ratio limit of \(1\) is not the same as a conclusion that the series converges.
Why the Two Strict Conclusions Work
The mechanism behind the test is comparison with a geometric series. Here is a justification of the two conclusions that also shows why the boundary is different.
Choose a number \(q\) with \(L<q<1\). By the definition of the limit, there is an index \(N\) such that for every \(n\ge N\), \(\left|a_{n+1}/a_n\right|<q\).
Repeatedly applying the ratio inequality gives \(|a_{N+k}|\le |a_N|q^k\) for every integer \(k\ge 0\). The tail is bounded by a constant times a convergent geometric series, since \(0<q<1\). Thus the original series converges absolutely.
Choose \(q\) with \(1<q<L\). Eventually, \(|a_{n+1}/a_n|>q\). Consequently, the absolute values grow by a factor greater than \(q\) at each step, so the terms do not tend to zero. A series can converge only if its terms tend to zero; therefore it diverges.
At \(L=1\), neither argument works. There is no \(q<1\) that eventually bounds the ratios from above based on that limit, and no \(q>1\) that eventually bounds them from below. The absence of either bound is why the test cannot decide the outcome.
Worked Example: The Harmonic Series Diverges
Worked Example: The Harmonic Series Diverges
Consider \(\sum_{n=1}^{\infty}1/n\). First calculate the ratio of consecutive terms:
As \(n\to\infty\), \(1/n\to0\), so the ratio limit is \(1/(1+0)=1\). The Ratio Test is inconclusive.
Use the Integral Test instead. The function \(f(x)=1/x\) is positive, continuous, and decreasing for \(x\ge1\). Its improper integral is
By the Integral Test, the harmonic series diverges. Check: the terms \(1/n\) do tend to zero, so the nonzero-term test does not decide this series; the divergent integral supplies the needed conclusion. The ratio limit \(1\) itself did not.
Worked Example: A \(p\)-Series Can Converge at the Same Boundary
Worked Example: A \(p\)-Series Can Converge at the Same Boundary
Now consider \(\sum_{n=1}^{\infty}1/n^2\). Its consecutive-term ratio is
Again, the Ratio Test is inconclusive. The terms are values of \(f(x)=1/x^2\), which is positive, continuous, and decreasing for \(x\ge1\). The Integral Test applies, and
This improper integral is finite, so the series converges. Check: differentiating the antiderivative \(-1/x\) gives \(1/x^2\), and the integral's limit is indeed \(1\). Thus the Ratio Test gives the same boundary value as in the harmonic example, but the Integral Test distinguishes their outcomes.
Worked Example: Alternating Harmonic Terms
Worked Example: Alternating Harmonic Terms
Consider \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\). Taking absolute values in the ratio gives
The Ratio Test is inconclusive. To test the original series, write it as \(\sum(-1)^{n+1}b_n\), where \(b_n=1/n\). The terms \(b_n\) are positive, decrease because \(1/(n+1)<1/n\), and approach zero. The Alternating Series Test therefore shows that the original series converges.
It does not converge absolutely: its series of absolute values is \(\sum 1/n\), which diverges by the harmonic-series result. So the alternating series converges conditionally. Check: the absolute-ratio calculation uses the magnitudes \(1/n\), while the alternating sign is handled by the Alternating Series Test. The two tests address different features of the series.
What the Examples Show
All three examples have a ratio limit of \(1\), but the outcomes differ. The positive harmonic series diverges, the positive series with terms \(1/n^2\) converges, and the alternating harmonic series converges conditionally. The limit alone cannot distinguish these cases.
| Series | Ratio limit | Useful next test | Conclusion |
|---|---|---|---|
| \(\sum 1/n\) | \(1\) | Integral Test | Diverges |
| \(\sum 1/n^2\) | \(1\) | Integral Test | Converges |
| \(\sum (-1)^{n+1}/n\) | \(1\), using absolute ratios | Alternating Series Test | Converges conditionally |
Once the ratio test reaches this boundary, use information the ratio limit does not capture. For positive terms resembling \(1/n^p\), the Integral Test or the known p-series criterion can be useful. For alternating terms, check the conditions of the Alternating Series Test. For other structures, comparison may be more informative. The choice depends on the actual terms and on the hypotheses of the alternative test.
A Common Mistake
A frequent error is to treat \(L=1\) as if it were a convergence result. For example, after finding \(n/(n+1)\to1\) for the harmonic series, someone might write:
This implication is false. The Ratio Test makes no claim about convergence when \(L=1\), and the harmonic series actually diverges.
Another related error is to stop after the ratio calculation without stating its meaning. Always report both the limit and the test's conclusion: \(L=1\), so the Ratio Test does not decide the series. Then give the separate argument that resolves it.
Check Your Understanding
For each question, distinguish the ratio-test result from the conclusion of any other test you use.
- For \(\sum_{n=1}^{\infty}1/n^3\), calculate the consecutive-term ratio limit. Does the Ratio Test decide convergence?
- For \(\sum_{n=1}^{\infty}1/n\), what test can establish divergence after the ratio limit is found to be \(1\)?
- What two conditions on \(b_n\) are needed, along with alternating signs, to apply the Alternating Series Test?
- Can a series with absolute ratio limit \(1\) converge conditionally? Give the example from this tutorial and explain why.
- Why does the proof of the Ratio Test use a number \(q<1\) when \(L<1\), but cannot make that same choice when \(L=1\)?
