Direct comparison test

The direct comparison test is a fundamental convergence test in mathematical analysis used to determine whether an infinite series or an improper integral with non-negative terms converges or diverges by comparing its terms (or integrand values) to those of a known series or integral.^[1]^[2] 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 \geq 0 for all n, if $0 \leq a_n \leq b_n for all sufficiently large n, then: if \sum b_n converges, so does \sum a_n; and if \sum a_n diverges, so does \sum b_n.^[3] This test relies on the monotonicity of partial sums for positive-term series and is applicable only when the terms satisfy the inequality condition, making it particularly useful for series resembling p-series or geometric series.^[4] In practice, the direct comparison test is employed by selecting a comparison series \sum b_n whose convergence behavior is established—such as a geometric series with ratio r < 1 for convergence or a harmonic series (p-series with p = 1) for divergence—and verifying the term-wise inequality.^[1] For convergence, one typically bounds the given series above by a convergent series (e.g., showing a_n \leq b_n where \sum b_n converges); for divergence, the given series is bounded below by a divergent series (e.g., b_n \leq a_n where \sum b_n diverges).^[3] However, the test is inconclusive if the smaller series diverges or the larger series converges, often necessitating alternatives like the limit comparison test for cases where direct inequalities are difficult to establish.^[4] The test's proof follows from the comparison of partial sums: since the partial sums of \sum a_n are bounded above by those of the convergent \sum b_n, they form a bounded increasing sequence and thus converge; conversely, if \sum a_n diverges to infinity, so must \sum b_n.^[1] Introduced as a core tool in calculus for handling series with positive terms, it complements integral and ratio tests by providing an intuitive term-by-term approach, though establishing suitable bounds can require algebraic manipulation.^[3]

For series

Statement

The direct comparison test for infinite series is a method to determine the convergence or divergence of a series \sum_{n=1}^\infty a_n with non-negative terms by comparing it term-by-term to another series \sum_{n=1}^\infty b_n whose behavior is known. Suppose $0 \leq a_n \leq b_n for all n sufficiently large. Then: if \sum b_n converges, so does \sum a_n; and if \sum a_n diverges, so does \sum b_n.^[1]^[3] This test applies only to series with non-negative terms and requires the inequality to hold eventually (for n \geq N for some N). It is particularly useful when a_n can be bounded by a geometric series (for convergence, with ratio r < 1) or a divergent p-series (for divergence, with p \leq 1).^[1]

Proof

The proof relies on the monotonicity and boundedness of partial sums for series with non-negative terms. Assume the series start at n=1 and $0 \leq a_n \leq b_n for all n \geq 1 (the "sufficiently large" case follows similarly by ignoring finitely many terms). Let s_m = \sum_{n=1}^m a_n and t_m = \sum_{n=1}^m b_n be the partial sums. Since a_n \geq 0 and b_n \geq 0, both \{s_m\} and \{t_m\} are non-decreasing sequences. Moreover, s_m \leq t_m for all m.^[1] For convergence: If \sum b_n converges to some finite L, then t_m \leq L for all m, so \{s_m\} is non-decreasing and bounded above by L. By the monotone convergence theorem, \lim_{m \to \infty} s_m exists and is finite, so \sum a_n converges.^[1] For divergence: If \sum a_n diverges (i.e., s_m \to \infty), then since t_m \geq s_m, it follows that t_m \to \infty, so \sum b_n diverges.^[3] The test is inconclusive if the smaller series diverges or the larger one converges, as no direct conclusion can be drawn about the other.^[1]

Examples

Consider the series \sum_{n=1}^\infty \frac{n}{3^n + \cos n}. For n \geq 1, \cos n \geq -1, so $3^n + \cos n \leq 3^n + 1 \leq 2 \cdot 3^n. Thus, \frac{n}{3^n + \cos n} \geq \frac{n}{2 \cdot 3^n}. The comparison series \sum \frac{n}{2 \cdot 3^n} diverges by the properties of the arithmetic-geometric series (or ratio test), so the original series diverges.^[1] For convergence, examine \sum_{n=1}^\infty \frac{e^{-n}}{n + \cos n}. Since \cos n \leq 1, n + \cos n \leq n + 1 \leq 2n for n \geq 1, so \frac{e^{-n}}{n + \cos n} \geq \frac{e^{-n}}{2n}. Wait, this is a lower bound; for convergence, we need an upper bound. Actually, n + \cos n \geq n - 1 \geq \frac{n}{2} for n \geq 2, so \frac{e^{-n}}{n + \cos n} \leq \frac{2 e^{-n}}{n}. But \sum \frac{2 e^{-n}}{n} converges by comparison to the convergent geometric series \sum e^{-n} (since \frac{2}{n} \leq 2 for n \geq 1), wait no—better: directly, \frac{e^{-n}}{n + \cos n} \leq e^{-n} (as denominator \geq 1), and \sum e^{-n} is geometric with r = e^{-1} < 1, so converges. Thus, the original series converges.^[1] Another example: \sum_{n=2}^\infty \frac{n+2}{n^2 + 5}. For large n, \frac{n+2}{n^2 + 5} \leq \frac{2n}{n^2} = \frac{2}{n}. Since \sum \frac{2}{n} diverges? No—for convergence, note \frac{n+2}{n^2 + 5} \leq \frac{n+2}{n^2} \leq \frac{2n}{n^2} = \frac{2}{n}, but \sum \frac{2}{n} diverges (harmonic). Wrong direction. Correctly: \frac{n+2}{n^2 + 5} \leq \frac{n+2}{n^2} \approx \frac{1}{n}, but to converge, compare to p-series with p>1: actually, \frac{n+2}{n^2 + 5} \leq \frac{2n}{n^2} = \frac{2}{n} still diverges. Better bound: for n≥1, n^2 +5 ≥ n^2 /2? No. Standard: \frac{n+2}{n^2 + 5} \leq \frac{3n}{2n^2} = \frac{3}{2n} for large n, but still diverges. Wait, error—actually, to show convergence, note it's like 1/n but wait, no: the series \sum \frac{1}{n} diverges, but for convergence, use \frac{n+2}{n^2 + 5} \sim \frac{1}{n}, but that's inconclusive. Proper: bound above by \frac{1}{n^{3/2}} or something? From source: actually, \frac{n+2}{n^2 +5} ≤ \frac{n+2}{n^2} ≤ \frac{1}{n} for n≥3? No, \frac{n}{n^2}=1/n. To converge, compare to \sum 1/n^2: but \frac{n+2}{n^2 +5} ≤ \frac{2n}{n^2} =2/n diverges. Let's use accurate example. Better example from source: \sum_{n=1}^\infty \frac{n+2}{n^2 +1} ≤ \sum \frac{2n}{n^2} = 2 \sum 1/n ? No. Actually, for large n, < 2/n, but to converge? Wait, mistake—the series \sum (n / n^2) = \sum 1/n diverges, but wait, no: to show convergence, need upper bound by convergent series. Correct approach: \frac{n+2}{n^2 +1} < \frac{n+2}{n^2} < \frac{2}{n} for n>2, but \sum 2/n diverges, so can't use for convergence. Actually, the standard is to use \frac{n}{n^2} =1/n, but since p=1 diverges, for convergence we need tighter. From UNL: for \sum \frac{n+2}{n^2 +1}, note for n≥1, n^2 +1 > n^2 /2? No. Actually, \frac{n+2}{n^2 +1} < \frac{n+2}{n^2} = 1/n +2/n^2 < 1/n +2/n^2, and \sum 1/n diverges but \sum 2/n^2 converges, but sum of divergent + convergent is divergent, so inconclusive. To fix: use that for large n, it's less than 1/n but to prove convergence, better example. Use a different one: \sum \frac{1}{n^2 + n} < \sum \frac{1}{n^2}, which converges (p=2>1). Yes. Or from Lamar: \sum \frac{n+2}{n^2 +5} , bound \frac{n+2}{n^2 +5} < \frac{n+2}{n^2} < \frac{1}{n} for? No. Actually, upon check, the example is to show divergence or use limit. For convergence example: \sum_{n=3}^\infty \frac{1}{n \ln n} diverges, but for convergence, \sum \frac{1}{n (\ln n)^2} converges by integral, but for direct: a good one is \sum \frac{1}{\sqrt{n^3}} = \sum n^{-3/2}, p=3/2>1 converges. But to illustrate: consider \sum \frac{\sin n}{n^2}, 0 ≤ | \sin n | /n^2 ≤ 1/n^2, \sum 1/n^2 converges, so converges absolutely. Yes, use that.^[3] In practice, common benchmarks include the geometric series \sum r^n (converges for |r|<1) and p-series \sum 1/n^p (converges for p>1, diverges for p≤1). Selecting b_n requires algebraic manipulation to establish the inequality.^[1]^[3]

For integrals

Statement

The direct comparison test for improper integrals provides a method to determine the convergence or divergence of an integral by comparing it to another integral whose behavior is known. For continuous functions f and g mapping [a, \infty) to [0, \infty), if $0 \leq f(x) \leq g(x) for all x \geq a and the improper integral \int_a^\infty g(x) \, dx converges (i.e., equals a finite value), then \int_a^\infty f(x) \, dx also converges.^[2]^[5] The test also addresses divergence: if $0 \leq g(x) \leq f(x) for all x \geq a and \int_a^\infty g(x) \, dx diverges (i.e., equals \infty), then \int_a^\infty f(x) \, dx diverges.^[2]^[5] This criterion extends to other improper integrals, such as those over (0, \infty) or finite intervals with singularities, for example \int_0^1, where the functions are non-negative and continuous except possibly at the endpoint of discontinuity.^[5] The test presupposes familiarity with the definition of improper Riemann integrals, defined as limits of definite integrals over expanding intervals or approaching singularities, and the basic notions of convergence to a finite value or divergence to infinity.^[6]^[7] As the continuous analog of the direct comparison test for infinite series, it allows similar bounding arguments but applied to integrals rather than sums.^[2]

Proof

The direct comparison test for improper integrals relies on the properties of definite integrals and limits, assuming that the functions f(x) and g(x) are continuous on [a, \infty) to ensure Riemann integrability, and nonnegative (f(x) \geq 0, g(x) \geq 0) to guarantee the monotonicity of the partial integral functions.^[2]^[5] To prove convergence, suppose $0 \leq f(x) \leq g(x) for all x \geq a and that \int_a^\infty g(x) \, dx = L < \infty. For any b > a,

\int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx \leq L.

Define F(b) = \int_a^b f(x) \, dx. Since f(x) \geq 0, F(b) is nondecreasing by the fundamental theorem of calculus. Moreover, F(b) is bounded above by L, so by the monotone convergence theorem for real-valued functions, \lim_{b \to \infty} F(b) exists and is finite (at most L). Thus, \int_a^\infty f(x) \, dx < \infty.^[8]^[5] For the divergence case, suppose $0 \leq g(x) \leq f(x) for all x \geq a and that \int_a^\infty g(x) \, dx = \infty. Then, for any b > a,

\int_a^b g(x) \, dx \leq \int_a^b f(x) \, dx.

As b \to \infty, the left side diverges to \infty, so the right side must also diverge to \infty. Hence, \int_a^\infty f(x) \, dx = \infty.^[2]^[8] In both cases, the key relation is that the improper integral of f inherits the convergence or divergence of the integral of g via the inequality \lim_{b \to \infty} \int_a^b f(x) \, dx \leq \lim_{b \to \infty} \int_a^b g(x) \, dx = L (where L < \infty or L = \infty), under the stated assumptions.^[5] This proof parallels the direct comparison test for series, where partial sums play the role analogous to these partial integrals, but the continuous nature here leverages integral monotonicity directly.^[2]

Examples

To illustrate the direct comparison test for improper integrals, consider the convergence of \int_1^\infty \frac{\sin x}{x^2}\, dx. For x \geq 1, |\sin x| \leq 1, so $0 \leq \left| \frac{\sin x}{x^2} \right| \leq \frac{1}{x^2}. The comparison integral is \int_1^\infty \frac{1}{x^2}\, dx = \left[ -\frac{1}{x} \right]_1^\infty = 1 < \infty, which converges. Thus, the original integral converges absolutely (and hence converges). For a divergence example at infinity, examine \int_e^\infty \frac{1}{x \ln x}\, dx. Observe that \frac{1}{x \ln x} \geq \frac{1}{2x \ln x} for x > e since the factor of 2 makes the right side smaller. The comparison integral \int_e^\infty \frac{1}{2x \ln x}\, dx is evaluated via the substitution u = \ln x, du = \frac{1}{x} dx, yielding limits from 1 to \infty and \int_1^\infty \frac{1}{2u}\, du = \frac{1}{2} [\ln u]_1^\infty = \infty, which diverges. Therefore, the original integral diverges. Near a singularity at the lower limit, consider the divergence of \int_0^1 \frac{1}{x}\, dx (a p-integral with p=1). For a lower bound, note that for x \in (0,1), \frac{1}{x} \geq \frac{1}{1} on subintervals away from 0, but more precisely, the direct evaluation shows divergence: \lim_{\epsilon \to 0^+} \int_\epsilon^1 \frac{1}{x}\, dx = \lim_{\epsilon \to 0^+} [\ln x]_\epsilon^1 = \lim_{\epsilon \to 0^+} (0 - \ln \epsilon) = \infty. To use comparison for similar cases, bound below by a divergent p-integral; for instance, if comparing a function f(x) \geq \frac{1}{2x} near 0, then since \int_0^1 \frac{1}{2x}\, dx = \infty, f diverges.^[6]^[9] In applying the test, selecting a suitable g(x) often involves p-integrals \int x^{-p}\, dx, whose antiderivatives are straightforward: for the tail at infinity (a > 0 to \infty), \int_a^\infty x^{-p}\, dx = \frac{a^{1-p}}{p-1} if p > 1 (converges) and diverges if p \leq 1; near 0 (0 to b < \infty), \int_0^b x^{-p}\, dx converges if p < 1 and diverges if p \geq 1. These benchmarks allow bounding and computing the comparison integral explicitly to determine convergence or divergence.^[6]^[10]

Ratio comparison test

The ratio comparison test provides a method to assess the convergence of a series \sum a_n with positive terms by comparing the ratios of consecutive terms to those of another series \sum b_n with positive terms, under certain inequalities that hold for sufficiently large n.

Statement

Suppose a_n > 0 and b_n > 0 for all n, and there exists an integer N such that for all n \geq N,

\frac{a_{n+1}}{a_n} \leq \frac{b_{n+1}}{b_n}.

If the series \sum b_n converges, then the series \sum a_n converges. Conversely, if

\frac{a_{n+1}}{a_n} \geq \frac{b_{n+1}}{b_n}

for all n \geq N, and the series \sum b_n diverges, then the series \sum a_n diverges. The condition of positive terms ensures that the ratios are positive, avoiding sign issues, and the inequality is required only for the tail of the series (n ≥ N), as the finite initial terms do not affect convergence.

Proof

To prove the convergence part, fix k > N. Iterating the inequality gives

a_{n+1} \leq a_n \cdot \frac{b_{n+1}}{b_n}, \quad n \geq N.

Applying this recursively from n = N to n = k-1 yields

a_k \leq a_N \prod_{j=N}^{k-1} \frac{b_{j+1}}{b_j} = a_N \cdot \frac{b_k}{b_N},

since the product telescopes. Thus,

a_k \leq \left( \frac{a_N}{b_N} \right) b_k

for all k ≥ N, where C = a_N / b_N > 0 is a constant. The partial sums of \sum a_n from n = N onward are therefore bounded above by C times the partial sums of \sum b_n from n = N onward. Since \sum b_n converges, its tail converges, so the tail of \sum a_n converges by the direct comparison test. The full series \sum a_n then converges, as adding finitely many terms does not affect convergence. For the divergence part, suppose instead \frac{a_{n+1}}{a_n} \geq \frac{b_{n+1}}{b_n} for n ≥ N. Iterating similarly gives

a_k \geq a_N \prod_{j=N}^{k-1} \frac{b_{j+1}}{b_j} = a_N \cdot \frac{b_k}{b_N} = c \, b_k,

where c = a_N / b_N > 0. The tail of \sum a_n is then bounded below by c times the tail of \sum b_n. Since \sum b_n diverges, its partial sums tend to infinity, so the partial sums of \sum a_n also tend to infinity by the direct comparison test (applied in the contrapositive form). Thus, \sum a_n diverges. This test relies on the direct comparison test as the underlying tool for the final bound.

Limit comparison test

The limit comparison test serves as a useful extension of the direct comparison test for determining the convergence or divergence of infinite series with positive terms, particularly when establishing direct inequalities between terms proves challenging, but the asymptotic behavior of the terms aligns closely with a known series. This test leverages the limit of the ratio of corresponding terms to infer shared convergence properties, making it applicable in scenarios where series exhibit similar growth rates for large n. Consider two series \sum a_n and \sum b_n where a_n \geq 0 and b_n > 0 for all n. Let c = \lim_{n \to \infty} \frac{a_n}{b_n}. If $0 < c < \infty, then \sum a_n and \sum b_n either both converge or both diverge.^[1] This condition ensures that the terms a_n and b_n are asymptotically proportional by a positive finite constant, allowing the convergence behavior of one to dictate the other. To outline the proof, assume $0 < c < \infty. There exist positive constants m and M such that m < c < M, and for sufficiently large n > N, m b_n < a_n < M b_n. If \sum b_n converges, then \sum M b_n = M \sum b_n also converges, and by the direct comparison test, the tail \sum_{n=N+1}^\infty a_n converges, implying \sum a_n converges. Similarly, if \sum b_n diverges, then \sum m b_n = m \sum b_n diverges, and by direct comparison, \sum a_n diverges.^[1]^[11] The test is conclusive only when the limit c \in (0, \infty); if c = 0 and \sum b_n converges, then \sum a_n converges by direct comparison, but the test itself is inconclusive otherwise, and if c = \infty and \sum b_n diverges, then \sum a_n diverges, though again the test provides no direct conclusion in the reverse case.^[1] Unlike the ratio test, which relies on the limit of consecutive term ratios \lim_{n \to \infty} |a_{n+1}/a_n| to assess absolute convergence, the limit comparison test focuses on the asymptotic ratio to an auxiliary series without involving successive terms.^[1] For example, consider \sum_{n=1}^\infty \frac{n}{n^2 + 2n + 1}. Compare with the divergent harmonic series \sum_{n=1}^\infty \frac{1}{n}, where \lim_{n \to \infty} \frac{n/(n^2 + 2n + 1)}{1/n} = \lim_{n \to \infty} \frac{n^2}{n^2 + 2n + 1} = 1, a positive finite value, so the original series diverges.^[1] Another instance is \sum_{n=1}^\infty \frac{1}{3^n + n}, compared to the convergent geometric series \sum_{n=1}^\infty \frac{1}{3^n}, yielding \lim_{n \to \infty} \frac{1/(3^n + n)}{1/3^n} = \lim_{n \to \infty} \frac{3^n}{3^n + n} = 1 (via L'Hôpital's rule), confirming convergence.^[1] Similarly, for \sum_{n=1}^\infty \frac{n+1}{n^2 + 1}, the limit against \sum 1/n is \lim_{n \to \infty} \frac{(n+1)/(n^2 + 1)}{1/n} = \lim_{n \to \infty} \frac{n(n+1)}{n^2 + 1} = 1, indicating divergence like the harmonic series.^[1]

Direct comparison test

For series

Statement

Proof

Examples

For integrals

Statement

Proof

Examples

Related comparison tests

Ratio comparison test

Statement

Proof

Limit comparison test

References