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Limit Comparison and Equivalents · Tutorial 625 of 998

Limit Comparison With Zero or Infinite Limits

Use the direction of a term ratio to decide when limit comparison proves convergence or divergence—and recognize when it proves neither.

Intermediate 11 min read

What You'll Learn

  • State the convergence and divergence implications when the ratio of positive terms tends to zero.
  • State the corresponding implications when the ratio tends to infinity.
  • Justify each implication using an eventual termwise inequality.
  • Identify the two situations in which these limit comparisons give no conclusion.
  • Avoid reversing a valid implication or treating an inconclusive case as a result.

Why the Direction of the Ratio Matters

When two series have positive terms, the ratio of their terms measures which sequence is smaller in the long run. If \(a_n/b_n\) tends to zero, then \(a_n\) eventually becomes smaller than \(b_n\). If the ratio tends to infinity, then \(a_n\) eventually becomes larger than \(b_n\). These inequalities can transfer convergence or divergence—but only in specific directions.

The limit comparison test with a finite positive ratio has a two-way conclusion. The zero and infinite cases are different: each gives a one-way implication, and each has a reverse situation in which the comparison does not decide the result.

Setup: Let \(a_n>0\) and \(b_n>0\) for every \(n\). We compare the series \(\sum a_n\) and \(\sum b_n\) using the limit of \(a_n/b_n\). Changing or ignoring finitely many initial terms does not affect whether a series converges or diverges.

The Two Valid Implications

If \(a_n/b_n\to 0\), then eventually \(a_n<b_n\). This lets convergence of the larger series \(\sum b_n\) imply convergence of the smaller series \(\sum a_n\). It does not let divergence of \(\sum b_n\) determine whether \(\sum a_n\) converges.

If \(a_n/b_n\to\infty\), then eventually \(a_n>b_n\). This lets divergence of the smaller series \(\sum b_n\) imply divergence of the larger series \(\sum a_n\). It does not let convergence of \(\sum b_n\) determine whether \(\sum a_n\) converges.

Theorem: Suppose \(a_n,b_n>0\).
  • If \(\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=0\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
  • If \(\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=\infty\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
In the other pairings—ratio tending to zero with \(\sum b_n\) divergent, or ratio tending to infinity with \(\sum b_n\) convergent—the test alone gives no conclusion.

Why the Implications Work

The proof uses the definition of the limits to obtain an eventual inequality, then applies the direct comparison principle for positive series. For a zero limit, it is enough to choose a ratio bound of \(1\). For an infinite limit, the same bound works in the opposite direction.

1
Ratio tends to zero.
By the definition of the limit, there is an index \(N\) such that \(n\ge N\) implies \(0<a_n/b_n<1\). Since \(b_n>0\), this gives \(0<a_n<b_n\).
2
Use convergence of the comparison series.
If \(\sum b_n\) converges, then its tail sums are bounded. For every \(m\ge N\), positivity and \(a_n<b_n\) give \(\sum_{n=N}^{m}a_n\le\sum_{n=N}^{m}b_n\). Thus the increasing partial sums of \(\sum a_n\) are bounded and converge. Adding the finite initial sum does not change convergence.
3
Ratio tends to infinity.
By the definition of divergence to infinity, there is an index \(N\) such that \(n\ge N\) implies \(a_n/b_n>1\), so \(a_n>b_n\). If \(\sum b_n\) diverges, its partial sums are unbounded. The partial sums of \(\sum a_n\) dominate the corresponding tail partial sums of \(\sum b_n\), so they too are unbounded. Therefore \(\sum a_n\) diverges.

This proves both claims. The direction of the inequality explains why the hypotheses differ: a smaller positive series is forced to converge by a convergent upper bound, while a larger positive series is forced to diverge by a divergent lower bound.

Worked Example: A Zero Ratio Gives Convergence

Worked Example: A Zero Ratio Gives Convergence

Determine the behavior of \(\sum 1/n^3\) by comparing it with \(\sum 1/n^2\). Set \(a_n=1/n^3\) and \(b_n=1/n^2\), which are positive for all \(n\in\mathbb N\). Compute the ratio:

$$ \frac{a_n}{b_n} = \frac{1/n^3}{1/n^2} = \frac{1}{n} \longrightarrow 0. $$

The comparison series \(\sum 1/n^2\) is a \(p\)-series with \(p=2>1\), so it converges. The zero-limit implication therefore gives convergence of \(\sum 1/n^3\).

Check: The ratio does tend to zero, and the comparison series is the larger convergent series. Thus the conclusion follows in the required direction.

Worked Example: An Infinite Ratio Gives Divergence

Worked Example: An Infinite Ratio Gives Divergence

Determine the behavior of \(\sum 1/\sqrt n\) by comparing it with the harmonic series \(\sum 1/n\). Let \(a_n=1/\sqrt n\) and \(b_n=1/n\). Both sequences are positive, and

$$ \frac{a_n}{b_n} = \frac{1/\sqrt n}{1/n} = \frac{n}{\sqrt n} = \sqrt n \longrightarrow\infty. $$

The comparison series \(\sum 1/n\) is the harmonic series and diverges. Since the ratio tends to infinity, \(\sum a_n\) diverges as well.

Check: For every \(n>1\), \(\sqrt n>1\), so \(a_n>b_n\). A positive series whose terms eventually exceed the terms of a divergent positive series must diverge.

When a Zero Ratio Does Not Decide the Result

A zero ratio says that \(a_n\) is eventually smaller than \(b_n\); it does not say that \(a_n\) is small enough for its series to converge. A divergent series can have smaller positive terms than another divergent series and still diverge. It can also have terms large enough to form a convergent series. Both outcomes are possible.

Worked Example: A Zero Ratio with a Divergent Comparison Series

Use \(b_n=1/n\), so \(\sum b_n\) diverges. First choose \(a_n=1/n^2\). Then

$$ \frac{a_n}{b_n} = \frac{1/n^2}{1/n} = \frac{1}{n} \longrightarrow 0. $$

The series \(\sum 1/n^2\) converges. In this case, the ratio tends to zero but the comparison series diverges, and the smaller series converges.

Now instead choose \(c_n=1/(n\ln n)\) for \(n\ge2\), still comparing with \(b_n=1/n\). The ratio is

$$ \frac{c_n}{b_n} = \frac{1/(n\ln n)}{1/n} = \frac{1}{\ln n} \longrightarrow 0. $$

Yet \(\sum_{n=2}^{\infty}1/(n\ln n)\) diverges. Indeed, \(f(x)=1/(x\ln x)\) is positive and decreasing for \(x\ge2\), and

$$ \int_2^R\frac{dx}{x\ln x} = \ln(\ln R)-\ln(\ln 2) \longrightarrow\infty \quad\text{as }R\to\infty. $$

The integral test therefore gives divergence. Check: Both choices have ratio tending to zero against the same divergent harmonic comparison series, but one series converges and the other diverges. Thus this case has no conclusion from the ratio test alone.

When an Infinite Ratio Does Not Decide the Result

An infinite ratio says that \(a_n\) is eventually larger than \(b_n\). If \(\sum b_n\) converges, being larger than \(b_n\) does not guarantee either convergence or divergence: \(a_n\) might still be small enough for its series to converge, or it might not be.

Worked Example: An Infinite Ratio with a Convergent Comparison Series

Take \(b_n=1/n^2\), whose series converges. One choice is \(a_n=1/n^{3/2}\). Its ratio to \(b_n\) is

$$ \frac{a_n}{b_n} = \frac{1/n^{3/2}}{1/n^2} = \sqrt n \longrightarrow\infty. $$

Nevertheless, \(\sum 1/n^{3/2}\) converges, since it is a \(p\)-series with \(p=3/2>1\).

For a different choice, let \(c_n=1/n\), with the same \(b_n=1/n^2\). Then

$$ \frac{c_n}{b_n} = \frac{1/n}{1/n^2} = n \longrightarrow\infty. $$

But \(\sum 1/n\) diverges. Check: In both cases the ratio tends to infinity and the comparison series converges, yet one series converges and the other diverges. The test gives no conclusion in this pairing.

A Common Mistake: Reversing the Implications

A frequent error is to remember that a zero ratio is associated with convergence and an infinite ratio with divergence, then apply those labels without checking the comparison series. The ratio alone does not establish either conclusion. The convergence or divergence of \(\sum b_n\) must match the direction of the applicable implication.

Worked Example: Correcting an Invalid Conclusion

Suppose \(a_n=1/(n\ln n)\) and \(b_n=1/n\), for \(n\ge2\). An incorrect argument is: “Since \(a_n/b_n\to0\), the series \(\sum a_n\) converges.” The ratio calculation is correct:

$$ \frac{a_n}{b_n}=\frac{1}{\ln n}\longrightarrow 0. $$

But the proposed conclusion does not follow: the comparison series \(\sum b_n\) diverges, which is precisely the inconclusive case for a zero ratio. In fact, as shown by the integral calculation above, \(\sum_{n=2}^{\infty}1/(n\ln n)\) diverges.

Correction: A zero ratio proves convergence only when the comparison series \(\sum b_n\) converges. Here it diverges, so the ratio test gives no conclusion; an additional argument, such as the integral test, is needed.

How to Apply the Test Reliably

Before drawing a conclusion, keep track of which sequence appears in the numerator and which appears in the denominator. The theorem is stated for \(a_n/b_n\): changing the order changes whether the relevant limit is zero or infinity. Then check that the known behavior of the comparison series matches the valid implication.

Limit of \(a_n/b_n\)Behavior of \(\sum b_n\)Conclusion about \(\sum a_n\)
ZeroConvergesConverges
ZeroDivergesNo conclusion from this test
InfinityDivergesDiverges
InfinityConvergesNo conclusion from this test

“No conclusion” does not mean that the series has no behavior to determine. Every series either converges or diverges; it means only that these facts about the ratio and the comparison series do not settle which one occurs. Another comparison or a different convergence test may still resolve the question.

Takeaway: With positive terms, a zero ratio makes \(\sum a_n\) smaller and transfers convergence from \(\sum b_n\). An infinite ratio makes \(\sum a_n\) larger and transfers divergence from \(\sum b_n\). In the opposite pairings, the comparison is inconclusive.

Check Your Understanding

For each question, pay attention both to the ratio and to the behavior of the comparison series.

  1. If \(a_n,b_n>0\), \(a_n/b_n\to0\), and \(\sum b_n\) converges, what can be concluded about \(\sum a_n\)? Why?
  2. If \(a_n/b_n\to\infty\) and \(\sum b_n\) diverges, what can be concluded about \(\sum a_n\)?
  3. Suppose \(a_n/b_n\to0\) but \(\sum b_n\) diverges. Does the limit comparison argument decide whether \(\sum a_n\) converges? Explain.
  4. Give the eventual termwise inequality implied by \(a_n/b_n\to\infty\), assuming \(a_n,b_n>0\).
  5. Why does an infinite ratio together with convergence of \(\sum b_n\) not determine the behavior of \(\sum a_n\)?

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