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Convergence Comparisons · Tutorial 329 of 998

Direct Comparisons for Infinite Tails

Use upper and lower power bounds to decide whether an integral over an infinite interval converges or diverges.

Intermediate 10 min read

What You'll Learn

  • State the conditions for direct comparison on an infinite interval.
  • Use the power-integral threshold to interpret a comparison bound.
  • Prove convergence by bounding above with an integrable power.
  • Prove divergence by bounding below with a nonintegrable power.
  • Classify rational tails using inequalities valid beyond a chosen cutoff.
  • Bound finite sums of nonnegative terms term by term.

Why Compare an Integrand With a Power?

Improper integrals over infinite intervals are defined by taking a limit. For a function \(f\) continuous on \([A,\infty)\), the integral \(\int_A^\infty f(x)\,dx\) converges if the limit of \(\int_A^R f(x)\,dx\) as \(R\) tends to infinity exists and is finite. Otherwise, it diverges.

Often, finding an antiderivative is difficult or unnecessary. If the integrand is nonnegative, its size can instead be compared with a function whose tail behavior is known. Powers are especially useful because the integral of \(x^{-p}\) has a sharp convergence threshold.

Definition: Let \(f\) be nonnegative and continuous for all sufficiently large \(x\). A direct comparison classifies \(\int_A^\infty f(x)\,dx\) by bounding \(f(x)\), for every \(x\) beyond some fixed cutoff, above or below by a function with a known improper integral.

The Power Benchmark

For \(A\geq 1\), the benchmark integral \(\int_A^\infty x^{-p}\,dx\) converges exactly when \(p>1\). When \(p\neq 1\), integration up to \(R\) gives

$$ \int_A^R x^{-p}\,dx = \frac{R^{1-p}-A^{1-p}}{1-p}. $$

If \(p>1\), then \(R^{1-p}\) tends to zero, so the improper integral equals \(A^{1-p}/(p-1)\), a finite value. If \(p<1\), then \(R^{1-p}\) grows without bound, and the integral diverges. At the boundary \(p=1\), direct integration gives

$$ \int_A^R \frac{1}{x}\,dx = \ln R-\ln A, $$

which also grows without bound. Thus a power with exponent \(p>1\) provides a convergent upper bound; a power with exponent \(p\leq 1\) can provide a divergent lower bound.

Power benchmark: For \(A\geq1\), \(\int_A^\infty x^{-p}\,dx\) converges if \(p>1\) and diverges if \(p\leq1\). The comparison only needs to hold for all \(x\) beyond a cutoff; changing a finite initial part does not affect convergence.

Direct Comparison on an Infinite Interval

The direct comparison theorem translates pointwise inequalities into inequalities between accumulated areas. In the convergence direction, the useful inequality puts the integrand below a function with finite total area. In the divergence direction, it puts the integrand above a function whose accumulated area grows without bound.

Theorem: Suppose \(f\) and \(g\) are nonnegative and continuous for \(x\geq A\).
  • If \(f(x)\leq g(x)\) for all \(x\geq A\), and \(\int_A^\infty g(x)\,dx\) converges, then \(\int_A^\infty f(x)\,dx\) converges.
  • If \(f(x)\geq g(x)\) for all \(x\geq A\), and \(\int_A^\infty g(x)\,dx\) diverges, then \(\int_A^\infty f(x)\,dx\) diverges.

The reason is that the inequalities hold on every finite interval \([A,R]\). In the first case, \(0\leq\int_A^R f(x)\,dx\leq\int_A^R g(x)\,dx\), so the increasing accumulated areas for \(f\) remain bounded by a quantity with a finite limit. In the second case, \(\int_A^R f(x)\,dx\geq\int_A^R g(x)\,dx\); if the latter grows without bound, so must the former. This is why the direction of the inequality matters.

For comparison with powers, a useful pattern is \(f(x)\leq Cx^{-p}\) with \(C>0\) and \(p>1\), which proves convergence. Conversely, \(f(x)\geq cx^{-q}\) with \(c>0\) and \(q\leq1\) proves divergence. These inequalities must be valid throughout a tail, not merely at a few sample values.

Worked Example: A Rational Tail That Converges

Worked Example: A Rational Tail That Converges

Classify

$$ \int_1^\infty \frac{3x+2}{x^3+1}\,dx. $$

For \(x\geq1\), the numerator satisfies \(3x+2\leq 5x\), because \(2\leq2x\). The denominator satisfies \(x^3+1\geq x^3\). Both numerator and denominator are positive, so these bounds give

$$ 0\leq \frac{3x+2}{x^3+1} \leq \frac{5x}{x^3} =\frac{5}{x^2}. $$

The comparison integral is finite:

$$ \int_1^\infty \frac{5}{x^2}\,dx = 5\left[-\frac{1}{x}\right]_1^\infty = 5(0-(-1)) = 5. $$

The integrand is nonnegative and bounded above by an integrable function on the entire interval \([1,\infty)\). By direct comparison, the given improper integral converges. As a check, the inequalities used above hold for every \(x\geq1\), and the comparison integral has the finite value \(5\).

Worked Example: A Rational Tail That Diverges

Worked Example: A Rational Tail That Diverges

Classify

$$ \int_1^\infty \frac{x^2+1}{x^3+4x+1}\,dx. $$

For \(x\geq1\), the numerator is at least \(x^2\). Also, \(x\leq x^3\) and \(1\leq x^3\), so

$$ x^3+4x+1 \leq x^3+4x^3+x^3 =6x^3. $$

The denominator is positive. Using the lower bound on the numerator and upper bound on the denominator therefore gives

$$ \frac{x^2+1}{x^3+4x+1} \geq \frac{x^2}{6x^3} = \frac{1}{6x}. $$

The lower comparison has divergent integral:

$$ \int_1^R \frac{1}{6x}\,dx = \frac{1}{6}\ln R, \qquad \lim_{R\to\infty}\frac{1}{6}\ln R = \infty. $$

Thus the given integral diverges by direct comparison. The check is the direction of the bound: the integrand is at least \(1/(6x)\) for every \(x\geq1\), so its accumulated area cannot remain finite while the lower bound's area grows without bound.

Finite Sums of Positive Terms

A finite sum of nonnegative functions can be bounded by bounding its terms. If each term has a convergent power upper bound, adding those bounds gives a convergent upper bound for the sum. For divergence, it is enough to find one nonnegative term that is already bounded below by a divergent comparison function: the full sum is at least that term.

These arguments use nonnegativity. With terms of different signs, cancellation can change the behavior, and the same inequalities cannot be applied without further care. The examples here concern finite sums of nonnegative terms.

Worked Example: A Sum of Positive Rational Terms

Classify

$$ \int_1^\infty \left( \frac{1}{x^2+1} + \frac{1}{x^3+1} \right)\,dx. $$

Both terms are positive. Since \(x^2+1\geq x^2\) and \(x^3+1\geq x^3\) for \(x\geq1\), taking reciprocals gives

$$ 0\leq \frac{1}{x^2+1}+\frac{1}{x^3+1} \leq \frac{1}{x^2}+\frac{1}{x^3}. $$

Integrate the upper bound:

$$ \int_1^\infty\left(\frac{1}{x^2}+\frac{1}{x^3}\right)\,dx = \left[-\frac{1}{x}\right]_1^\infty + \left[-\frac{1}{2x^2}\right]_1^\infty = 1+\frac{1}{2} = \frac{3}{2}. $$

The upper bound has finite integral, so the original integral converges. The check is that both termwise bounds hold for every \(x\geq1\), and their sum has finite integral \(3/2\).

Choosing a Useful Bound

For rational functions, inequalities can often be obtained directly from the dominant powers in the numerator and denominator, without finding an exact antiderivative. The goal here is not to identify a precise asymptotic equivalent; it is to find a simple inequality that works for every \(x\) beyond a chosen cutoff.

For example, when \(x\geq1\), a lower-degree term such as \(4x\) is no larger than \(4x^3\). This lets us bound a denominator above by a multiple of its highest power, which is useful when seeking a lower bound on a positive fraction. Conversely, a denominator can be bounded below by one of its positive terms to obtain an upper bound. Keep track of which way each operation goes: dividing by a larger positive denominator makes a fraction smaller.

1
Identify the needed conclusion.
To prove convergence, seek an integrable upper bound. To prove divergence, seek a nonintegrable lower bound.
2
Choose a power benchmark.
For convergence, choose \(x^{-p}\) with \(p>1\). For divergence, choose \(x^{-p}\) with \(p\leq1\).
3
Establish the inequality on a tail.
Use positive-term bounds valid for every \(x\geq A\), and check that reciprocals or divisions preserve the intended direction.
4
Apply the comparison theorem.
Evaluate or classify the power integral, then state the conclusion for the original integrand.

A Common Comparison Mistake

A frequent error is to use an upper bound by a divergent function to claim convergence. For instance, the inequality \(0\leq f(x)\leq 1/x\) does not prove that \(\int_1^\infty f(x)\,dx\) converges: the upper comparison integral diverges, so it supplies no finite bound on the accumulated area of \(f\).

To see why this is not enough, both \(f(x)=1/x^2\) and \(f(x)=1/x\) satisfy \(0\leq f(x)\leq1/x\) for \(x\geq1\), but their improper integrals have different outcomes. The correction is to find an upper bound whose integral converges, such as \(C/x^p\) with \(p>1\), or to use a suitable lower bound with a divergent integral when proving divergence.

Takeaway: A comparison is useful only when both the inequality's direction and the comparison integral's behavior support the conclusion. A convergent upper bound proves convergence; a divergent lower bound proves divergence.

Check Your Understanding

For each question, identify the comparison direction before deciding what conclusion follows.

  1. If \(0\leq f(x)\leq 4/x^3\) for all \(x\geq2\), what can you conclude about \(\int_2^\infty f(x)\,dx\)?
  2. If \(f(x)\geq 2/x\) for all \(x\geq1\), what can you conclude about \(\int_1^\infty f(x)\,dx\)?
  3. Why does \(0\leq f(x)\leq1/x\) not by itself settle whether the integral of \(f\) converges?
  4. For a finite sum of nonnegative functions, what bound is sufficient to prove convergence of its improper integral?
  5. When comparing a positive rational function with a power, why must a denominator's upper bound be handled carefully if the goal is a lower bound for the fraction?

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