Why the Ratio Limit Is Not the Whole Decision
A limit comparison begins by placing the terms of a positive-term series beside those of a benchmark series whose behavior is known. The ratio of the terms measures their relative size. But the ratio's limit does not, by itself, always determine whether the first series converges.
The crucial question is not just whether the ratio is small or large. It is also whether the benchmark converges or diverges, and which direction of comparison that information supports. A finite positive ratio limit gives a two-way comparison. A zero or infinite limit gives only a one-way comparison, so the benchmark's behavior matters especially strongly.
All statements below concern positive terms, at least from some index onward. This condition matters: the comparison arguments use inequalities between terms and the corresponding partial sums.
The Three Ratio-Limit Cases
If the ratio tends to a finite positive number, the two terms eventually differ by at most constant factors. Consequently, the two series have the same convergence behavior.
If the ratio tends to zero, the terms \(a_n\) eventually become smaller than \(b_n\). Thus convergence of the benchmark forces convergence of \(\sum a_n\). But if the benchmark diverges, knowing that \(a_n\) is smaller does not settle whether \(\sum a_n\) converges.
If the ratio tends to infinity, the terms \(a_n\) eventually become larger than \(b_n\). Thus divergence of the benchmark forces divergence of \(\sum a_n\). But if the benchmark converges, a larger positive-term series might either converge or diverge.
- If \(0<L<\infty\), \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge.
- If \(L=0\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
- If \(L=\infty\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
Why These Conclusions Follow
The reasoning is a direct application of inequalities and the comparison test for positive-term series. If one positive-term series is eventually no larger than a convergent series, its partial sums are bounded above, so it converges. If it is eventually no smaller than a divergent positive-term series, its partial sums are unbounded, so it diverges.
For \(0<L<\infty\), the definition of the limit gives, for all sufficiently large \(n\), \(L/2<a_n/b_n<3L/2\). Multiplying by \(b_n>0\) gives \((L/2)b_n<a_n<(3L/2)b_n\).
The lower bound shows that divergence of \(\sum b_n\) forces divergence of \(\sum a_n\). The upper bound shows that convergence of \(\sum b_n\) forces convergence of \(\sum a_n\). Hence their behavior agrees.
If \(a_n/b_n\to0\), then eventually \(a_n/b_n<1\), so \(a_n<b_n\). This upper bound transfers convergence from \(b_n\) to \(a_n\), but it supplies no lower bound that could transfer divergence.
If \(a_n/b_n\to\infty\), then eventually \(a_n/b_n>1\), so \(a_n>b_n\). This lower bound transfers divergence from \(b_n\) to \(a_n\), but it supplies no upper bound that could transfer convergence.
For example, in the finite-limit case, the inequalities can be summed from a sufficiently large index \(N\) through \(m\). The resulting bounds compare the tails of the two series by fixed positive constants. Adding or removing the finite initial sums cannot affect convergence, which completes the justification.
A Finite Positive Limit: The Test Decides Both Ways
Worked Example: A Ratio Limit of Three
Consider \(\sum_{n=1}^{\infty} a_n\), where \(a_n=(3n+2)/(n^3+4n)\). Compare it with \(b_n=1/n^2\). Both terms are positive for \(n\ge1\), and the benchmark \(\sum 1/n^2\) converges by the \(p\)-series test.
Compute the ratio, keeping the algebra visible:
Divide numerator and denominator by \(n^3\):
The limit is finite and positive. Therefore the two series have the same behavior. Since \(\sum 1/n^2\) converges, \(\sum (3n+2)/(n^3+4n)\) converges. The check is the limit calculation: the numerator tends to \(3\), the denominator tends to \(1\), and their quotient tends to \(3\), as required for the test.
A Zero Limit: Convergence Can Transfer
Worked Example: Smaller Than a Convergent Benchmark
Take \(a_n=1/n^3\) and the benchmark \(b_n=1/n^2\). Both terms are positive for \(n\ge1\), and the benchmark series converges.
Because the ratio tends to zero and \(\sum b_n=\sum 1/n^2\) converges, the test concludes that \(\sum a_n=\sum 1/n^3\) converges. In this particular case the term inequality can also be checked directly: \(1/n^3\le 1/n^2\) for every \(n\ge1\). The comparison has the direction needed to transfer convergence.
A Zero Limit Against a Divergent Benchmark: No Verdict
When \(a_n/b_n\to0\) and \(\sum b_n\) diverges, the fact that \(a_n\) is eventually smaller is not enough to determine the behavior of its series. The following two choices use the same divergent benchmark but lead to opposite outcomes.
Worked Example: One Zero-Limit Ratio, Two Outcomes
Use \(b_n=1/\sqrt{n}\), whose series diverges by the \(p\)-series test because its exponent is \(1/2\le1\).
First choose \(a_n=1/n\). Then
Nevertheless, \(\sum a_n=\sum 1/n\) diverges: it is the harmonic series. Thus a zero ratio against a divergent benchmark does not prove convergence.
Now keep the same benchmark and choose \(c_n=1/n^2\). Its ratio is
But \(\sum c_n=\sum 1/n^2\) converges. The two ratios both tend to zero while the corresponding series have different behavior. This confirms that the limit and the divergent benchmark alone do not decide the result. Each series behavior stated here agrees with the \(p\)-series test, with the harmonic series as the exponent-one case.
An Infinite Limit: Divergence Can Transfer
Worked Example: Larger Than a Divergent Benchmark
Take \(a_n=1/\sqrt{n}\) and \(b_n=1/n\). The benchmark is the harmonic series, so it diverges. The terms are positive, and the ratio is
For every \(n\ge1\), \(\sqrt{n}\ge1\), so \(a_n\ge b_n\). The divergence conclusion follows: \(\sum 1/\sqrt{n}\) diverges. More generally, the infinite ratio guarantees this inequality eventually, and a termwise lower bound by a divergent positive-term series forces divergence.
An Infinite Limit Against a Convergent Benchmark: No Verdict
An infinite ratio says \(a_n\) is eventually larger than \(b_n\). If \(\sum b_n\) converges, that lower bound does not decide whether \(\sum a_n\) converges: a larger series could still converge, or it could diverge.
Worked Example: Two Infinite-Limit Ratios, Two Outcomes
Use \(b_n=1/n^2\), a convergent benchmark. First let \(a_n=1/n^{3/2}\). Then
Even though the ratio tends to infinity, \(\sum a_n=\sum 1/n^{3/2}\) converges, since its \(p\)-series exponent is greater than \(1\).
Now choose \(c_n=1/n\), with the same benchmark. This time,
Yet \(\sum c_n=\sum 1/n\) diverges. Both examples have an infinite ratio and a convergent benchmark, but their series behave differently. Therefore this combination is inconclusive; it does not mean that either series must converge or that it must diverge.
Diagnosing an Inconclusive Result
“Inconclusive” is a statement about the information supplied by this comparison, not a claim that the series has no answer. If a ratio limit lands in one of the two undecided cases, you can still settle the original series by choosing a different benchmark or using another valid comparison. The earlier material on comparison tests and \(p\)-series provides familiar tools for doing so.
A related error is to read \(a_n/b_n\to\infty\) as an automatic proof of divergence. For \(a_n=1/n^{3/2}\) and \(b_n=1/n^2\), the ratio is \(\sqrt n\to\infty\), but \(\sum a_n\) converges. An infinite ratio proves divergence only when the benchmark diverges.
A Reliable Decision Routine
Before writing a conclusion, record both the ratio limit and the benchmark behavior. Then identify whether the available inequality points in the direction needed for convergence or divergence.
| Ratio limit | Benchmark behavior | What follows for \(\sum a_n\)? |
|---|---|---|
| \(0<L<\infty\) | Convergent | Convergent |
| \(0<L<\infty\) | Divergent | Divergent |
| \(0\) | Convergent | Convergent |
| \(0\) | Divergent | No conclusion from this comparison |
| \(\infty\) | Divergent | Divergent |
| \(\infty\) | Convergent | No conclusion from this comparison |
The table is a decision guide, not a shortcut around checking the hypotheses. Make sure the terms are positive eventually, compute the ratio correctly, determine its limit, and know the benchmark's behavior. Only then does the conclusion follow.
Check Your Understanding
For each question, state what the limit comparison test establishes and whether the information is conclusive.
- If \(a_n,b_n>0\), \(a_n/b_n\to 5\), and \(\sum b_n\) diverges, what can you conclude about \(\sum a_n\)?
- If \(a_n/b_n\to0\) and \(\sum b_n\) converges, what can you conclude about \(\sum a_n\)? Which direction of comparison supports the answer?
- If \(a_n/b_n\to0\) while \(\sum b_n\) diverges, does the test prove that \(\sum a_n\) converges? Explain why or why not.
- If \(a_n/b_n\to\infty\) and \(\sum b_n\) diverges, what follows about \(\sum a_n\)?
- Give the reason an infinite ratio with a convergent benchmark does not, by itself, decide the behavior of \(\sum a_n\).
