Limit comparison test

The limit comparison test is a fundamental tool in mathematical analysis for assessing the convergence or divergence of an infinite series \sum_{n=1}^\infty a_n with nonnegative terms a_n \geq 0 by comparing it to another series \sum_{n=1}^\infty b_n with positive terms b_n > 0, whose convergence is already known, through the evaluation of the limit L = \lim_{n \to \infty} \frac{a_n}{b_n}. If this limit exists and satisfies $0 < L < \infty, then the two series either both converge or both diverge. This test is particularly useful when direct comparison is inconclusive, as it leverages the asymptotic behavior of the terms for large n rather than requiring strict inequalities throughout.^[1] The test assumes that both series have nonnegative terms to ensure the limit is meaningful and the comparison aligns with the monotone nature of partial sums.^[2] Commonly, b_n is chosen as a p-series \sum \frac{1}{n^p} (which converges for p > 1 and diverges for p \leq 1) or a geometric series \sum r^n (converging for |r| < 1), allowing the test to resolve series where terms grow or decay similarly at infinity.^[3] In practice, the limit is often computed using techniques like L'Hôpital's rule when \frac{a_n}{b_n} yields an indeterminate form \frac{\infty}{\infty} or \frac{0}{0}. The proof of the test relies on the comparison test itself: for large n, a_n is bounded by multiples of b_n (specifically, \frac{L}{2} b_n < a_n < 2L b_n), so the partial sums of \sum a_n behave comparably to those of \sum b_n. This method extends the direct comparison test and is a staple in calculus for handling series beyond elementary forms, such as those involving rational functions of n.^[2]

Standard Limit Comparison Test

Formal Statement

The limit comparison test is a criterion used to determine the convergence or divergence of an infinite series by comparing it to another series whose behavior is known. Specifically, for two series \sum_{n=1}^\infty a_n and \sum_{n=1}^\infty b_n where a_n \geq 0 and b_n > 0 for all n, if the limit

\lim_{n \to \infty} \frac{a_n}{b_n} = L

exists and satisfies $0 < L < \infty, then \sum a_n and \sum b_n either both converge or both diverge.^[1]^[2]^[4] The assumption of non-negative terms (a_n \geq 0) and strictly positive terms (b_n > 0) is essential, as it ensures that the ratio a_n / b_n is well-defined and non-negative for sufficiently large n, allowing the asymptotic comparison to align with the direct comparison test without complications from alternating signs or negative values.^[1]^[4] In practice, the positivity condition needs to hold only for all n greater than some fixed N, since the convergence of a series depends solely on the tail of the terms.^[4] This test extends the direct comparison test by focusing on the limiting ratio rather than term-by-term inequalities.^[2]

Proof

To prove the convergence case of the standard limit comparison test, suppose \sum b_n converges where b_n > 0 for all n, and \lim_{n \to \infty} \frac{a_n}{b_n} = L with $0 < L < \infty. By the definition of the limit, for any \epsilon > 0, there exists N \in \mathbb{N} such that for all n \geq N, \left| \frac{a_n}{b_n} - L \right| < \epsilon. Choosing $0 < \epsilon < L ensures L - \epsilon > 0, and thus \frac{a_n}{b_n} < L + \epsilon, or equivalently, a_n < (L + \epsilon) b_n. Since \sum b_n converges, the constant multiple \sum (L + \epsilon) b_n also converges, and by the direct comparison test, \sum a_n converges.^[5] For the divergence case, again assume b_n > 0 for all n and \lim_{n \to \infty} \frac{a_n}{b_n} = L with $0 < L < \infty. Using the same \epsilon- N definition with $0 < \epsilon < L, for n \geq N, \frac{a_n}{b_n} > L - \epsilon, so a_n > (L - \epsilon) b_n. If \sum b_n diverges, then \sum (L - \epsilon) b_n diverges as a constant multiple of a divergent series. By the direct comparison test, \sum a_n must diverge.^[5]

Examples

Consider the series \sum_{n=1}^\infty \frac{1}{n^2 + 1}. Compare it to the p-series \sum_{n=1}^\infty \frac{1}{n^2}, which converges since p = 2 > 1. Compute the limit:

\lim_{n \to \infty} \frac{\frac{1}{n^2 + 1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n^2 + 1} = 1.

Since the limit is 1 (positive and finite) and the comparison series converges, the original series \sum_{n=1}^\infty \frac{1}{n^2 + 1} converges.^[1] For divergence, consider the series \sum_{n=1}^\infty \frac{n}{n^2 + 1}. Compare it to the harmonic series \sum_{n=1}^\infty \frac{1}{n}, which diverges. The limit is:

\lim_{n \to \infty} \frac{\frac{n}{n^2 + 1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^2}{n^2 + 1} = 1.

Since the limit is 1 (positive and finite) and the comparison series diverges, \sum_{n=1}^\infty \frac{n}{n^2 + 1} diverges.^[1]

One-Sided Limit Comparison Tests

Statements

The one-sided limit comparison tests extend the standard limit comparison test by addressing cases where the limit of the ratio of terms is 0 or \infty, providing one-way implications for convergence or divergence under specific conditions.^[6] Assume that a_n > 0 and b_n > 0 for all sufficiently large n.^[6] If \lim_{n \to \infty} \frac{a_n}{b_n} = 0 and \sum b_n converges, then \sum a_n converges; in this case, a_n is asymptotically smaller than b_n.^[6]^[7] If \lim_{n \to \infty} \frac{a_n}{b_n} = \infty and \sum b_n diverges, then \sum a_n diverges; here, a_n is asymptotically larger than b_n.^[6]^[7] These statements differ from the standard limit comparison test, which requires a finite positive limit to conclude that both series converge or diverge together.^[6]

Proofs

The one-sided limit comparison test for convergence addresses the case where \lim_{n \to \infty} \frac{a_n}{b_n} = 0 with a_n > 0, b_n > 0, and \sum b_n converges. By the definition of the limit, for \varepsilon = 1 > 0, there exists N \in \mathbb{N} such that for all n \geq N, \frac{a_n}{b_n} < 1, which implies a_n < b_n. The partial sums of \sum_{n=N}^\infty a_n are therefore bounded above by the tail of the convergent series \sum_{n=N}^\infty b_n, so \sum_{n=N}^\infty a_n converges by the direct comparison test. Thus, \sum a_n converges.^[8] The one-sided limit comparison test for divergence addresses the case where \lim_{n \to \infty} \frac{a_n}{b_n} = \infty with a_n > 0, b_n > 0, and \sum b_n diverges. By the definition of the limit, for M = 1 > 0, there exists N \in \mathbb{N} such that for all n \geq N, \frac{a_n}{b_n} > 1, which implies a_n > b_n. The partial sums of \sum_{n=N}^\infty a_n therefore exceed those of the divergent series \sum_{n=N}^\infty b_n, so \sum_{n=N}^\infty a_n diverges by the direct comparison test. Thus, \sum a_n diverges.^[8]

Examples

Consider the series \sum_{n=3}^{\infty} \frac{1}{n^2 \log n}. To determine its convergence using the one-sided limit comparison test with L = 0, compare it to the p-series \sum_{n=1}^{\infty} \frac{1}{n^2}, which converges since p = 2 > 1. Compute the limit:

\lim_{n \to \infty} \frac{\frac{1}{n^2 \log n}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{1}{\log n} = 0.

Since the limit is 0 and the comparison series converges, the original series \sum_{n=3}^{\infty} \frac{1}{n^2 \log n} converges. A simpler illustration is the series \sum_{n=1}^{\infty} \frac{1}{n^3}, compared to \sum_{n=1}^{\infty} \frac{1}{n^2}, which converges as noted above. The limit is:

\lim_{n \to \infty} \frac{\frac{1}{n^3}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{1}{n} = 0.

Thus, \sum_{n=1}^{\infty} \frac{1}{n^3} converges by the one-sided limit comparison test. For divergence with L = \infty, consider \sum_{n=2}^{\infty} \frac{n}{\log n}. Compare it to the harmonic series \sum_{n=1}^{\infty} \frac{1}{n}, which diverges. The limit is:

\lim_{n \to \infty} \frac{\frac{n}{\log n}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^2}{\log n} = \infty.

Since the limit is \infty and the comparison series diverges, \sum_{n=2}^{\infty} \frac{n}{\log n} diverges.

Converse and Limitations

Converse Statements

The converse statements for the one-sided limit comparison tests provide the flipped implications under the assumption of the specified limit values, again requiring that the terms a_n > 0 and b_n > 0 for all sufficiently large n.^[8] For the case where \lim_{n \to \infty} \frac{a_n}{b_n} = 0, if \sum a_n diverges, then \sum b_n diverges. This follows from the fact that the limit being zero implies a_n < b_n eventually, allowing the direct comparison test to establish that the larger series \sum b_n must diverge if the smaller one does.^[8] For the case where \lim_{n \to \infty} \frac{a_n}{b_n} = \infty, if \sum a_n converges, then \sum b_n converges. Here, the infinite limit implies b_n < a_n eventually, so the direct comparison test shows that the smaller series \sum b_n converges if the larger one does; this scenario is unusual, as it typically arises when analyzing a potentially larger series whose convergence would force the smaller to converge as well.^[8] These converse forms are the logical counterparts to the original one-sided statements, offering utility when the behavior of the "smaller" series (in the L=0 case) or the "larger" series (in the L=\infty case) is known, thereby allowing inference about the other series.^[8]

Proofs of the Converse

The converse statements for the one-sided limit comparison tests establish the reverse implications under the assumption of positive terms a_n > 0 and b_n > 0 for sufficiently large n. Specifically, for the case where \lim_{n \to \infty} \frac{a_n}{b_n} = 0, the converse asserts that if \sum a_n diverges, then \sum b_n diverges. Similarly, for \lim_{n \to \infty} \frac{a_n}{b_n} = \infty, the converse asserts that if \sum a_n converges, then \sum b_n converges. These converses complete the bidirectional relationship, making the one-sided tests equivalences in the presence of positivity.^[8] To prove the converse for L = 0, proceed by contradiction. Assume \lim_{n \to \infty} \frac{a_n}{b_n} = 0 and \sum a_n diverges, but suppose \sum b_n converges. By the original one-sided limit comparison test, since the limit is 0 and \sum b_n converges, it follows that \sum a_n converges, which contradicts the assumption that \sum a_n diverges. Therefore, \sum b_n must diverge.^[8] An alternative direct proof uses reciprocal bounds. Since \lim_{n \to \infty} \frac{a_n}{b_n} = 0, it follows that \lim_{n \to \infty} \frac{b_n}{a_n} = \infty. Thus, for any M > 0, there exists N such that for all n > N, \frac{b_n}{a_n} > M, or equivalently, b_n > M a_n. The tail of the series then satisfies

\sum_{n=N+1}^\infty b_n > M \sum_{n=N+1}^\infty a_n.

If \sum a_n diverges, the tail \sum_{n=N+1}^\infty a_n diverges, so \sum_{n=N+1}^\infty b_n diverges for any M > 0, implying that \sum b_n diverges by the comparison test.^[8] For the converse when L = \infty, again use contradiction. Assume \lim_{n \to \infty} \frac{a_n}{b_n} = \infty and \sum a_n converges, but suppose \sum b_n diverges. By the original one-sided test for infinity, since the limit is \infty and \sum b_n diverges, \sum a_n diverges, contradicting the assumption. Hence, \sum b_n converges. This case is symmetric to the L = 0 converse by interchanging the roles of a_n and b_n, as \lim_{n \to \infty} \frac{b_n}{a_n} = 0.^[8] These converses hold due to the if-and-only-if structure inherent in the original one-sided implications, reinforced by the positivity of terms, which ensures that bounding arguments via the limit apply bidirectionally without sign issues.^[8]

Limitations and Counterexamples

The limit comparison test requires that the limit \lim_{n \to \infty} \frac{a_n}{b_n} exists and equals a positive finite number for its conclusion to hold; if the limit does not exist, the test cannot be applied. For instance, consider the series \sum a_n where a_n = \frac{1}{n} if n is even and a_n = \frac{1}{n^2} if n is odd, compared to b_n = \frac{1}{n}. Along even indices, \frac{a_n}{b_n} = 1, while along odd indices, \frac{a_n}{b_n} = \frac{1}{n} \to 0, so the limit oscillates and does not exist, rendering the test inapplicable despite \sum b_n diverging. Similarly, for \sum \frac{1}{n \sin^2 n} compared to \sum \frac{1}{n}, the ratio \frac{1/\ (n \sin^2 n)}{1/n} = \frac{1}{\sin^2 n} oscillates without limit due to the dense distribution of \sin n in [-1, 1], preventing application of the test.^[1]^[9] Even when the limit exists, the test may be inconclusive if it equals 0 or \infty, providing no information about convergence. A key limitation arises when \lim_{n \to \infty} \frac{a_n}{b_n} = 0 and \sum b_n diverges: the behavior of \sum a_n remains undetermined, as it could converge or diverge depending on how much smaller a_n is than b_n. For example, comparing \sum \frac{1}{n^2} (which converges) to \sum \frac{1}{n} (which diverges) gives \lim_{n \to \infty} \frac{1/n^2}{1/n} = \lim_{n \to \infty} \frac{1}{n} = 0, yet \sum \frac{1}{n^2} converges, showing the test yields no conclusion. Conversely, comparing \sum \frac{1}{n \log n} (which diverges for n \geq 2) to \sum \frac{1}{n} gives \lim_{n \to \infty} \frac{1/(n \log n)}{1/n} = \lim_{n \to \infty} \frac{1}{\log n} = 0, again inconclusive under the limit comparison test, though other methods like the integral test confirm divergence. This highlights the need for careful benchmark selection; an inappropriate b_n can lead to inconclusive results even if a suitable one exists.^[1]^[10] The one-sided implications provide partial guidance but underscore further limitations: if the limit is 0 and \sum b_n converges, then \sum a_n converges, but the converse fails when \sum b_n diverges, as illustrated above. Misapplying a "converse" in such cases—e.g., incorrectly assuming \lim = 0 and \sum b_n diverges implies \sum a_n diverges—leads to errors, as the p-series counterexample demonstrates. In practice, alternatives like the ratio test or root test may be necessary when the limit does not exist or is extreme. The test, a refinement of earlier comparison methods developed by 19th-century analysts, remains a core tool but requires verification of its prerequisites to avoid these pitfalls.^[1]^[10]

Limit comparison test

Standard Limit Comparison Test

Formal Statement

Proof

Examples

One-Sided Limit Comparison Tests

Statements

Proofs

Examples

Converse and Limitations

Converse Statements

Proofs of the Converse

Limitations and Counterexamples

References