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Integral and Direct Comparison · Tutorial 622 of 998

Choosing Between Integration and Direct Comparison

Choose the test that makes the series’ behavior easiest to justify, and verify the conditions and comparison direction before drawing a conclusion.

Intermediate 10 min read

What You'll Learn

  • Identify when the integral test applies to a series with positive terms.
  • Select a direct upper or lower bound that proves convergence or divergence.
  • Use power-series benchmarks to assess rational and radical terms.
  • Recognize when a logarithmic factor makes the integral test convenient.
  • Check inequalities and method hypotheses before concluding.

Why the Choice of Test Matters

A positive series may invite more than one method. For example, a formula might have a continuous version that is straightforward to integrate, or its terms might be easy to bound by a familiar power of \(n\). In either case, the goal is not to perform the most elaborate calculation. It is to find a valid argument with hypotheses that can be checked clearly.

This tutorial focuses on choosing between the integral test and direct comparison. Both apply to series with nonnegative terms, but they use different information: the integral test studies the area under a function, while direct comparison uses an inequality between terms.

Working principle: For a positive series, use the integral test when the associated function is eventually continuous, positive, decreasing, and easy to integrate. Use direct comparison when a simple termwise bound by a known series is easier to establish.

The Two Methods and Their Conditions

Suppose \(a_n\ge 0\) for all sufficiently large \(n\). A finite number of terms does not affect whether a series converges, so the tests may be applied to a tail.

Integral Test: Suppose \(a_n=f(n)\) for all integers \(n\ge N\), where \(f\) is continuous, positive, and decreasing on \([N,\infty)\). Then \(\sum_{n=N}^{\infty}a_n\) and the improper integral \(\int_N^\infty f(x)\,dx\) either both converge or both diverge.
Direct Comparison: If \(0\le a_n\le b_n\) for all sufficiently large \(n\), and \(\sum b_n\) converges, then \(\sum a_n\) converges. If \(0\le b_n\le a_n\) for all sufficiently large \(n\), and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

The comparison series is often a \(p\)-series, \(\sum 1/n^p\), which converges for \(p>1\) and diverges for \(p\le 1\). The direction of the inequality matters: a smaller positive series cannot be forced to diverge merely because a larger one diverges.

Why the Integral Test Works

The integral test is based on the rectangles formed by a decreasing function. If \(f\) is positive and decreasing, then on each interval \([n,n+1]\), its values lie between \(f(n+1)\) and \(f(n)\). Integrating these bounds over an interval of width \(1\) gives

$$ f(n+1)\le \int_n^{n+1} f(x)\,dx\le f(n). $$

Adding these inequalities from \(n=N\) to \(n=M\) yields

$$ \sum_{n=N+1}^{M+1}f(n) \le \int_N^{M+1}f(x)\,dx \le \sum_{n=N}^{M}f(n). $$

Thus the partial sums and the accumulated areas bound one another, apart from a finite endpoint term. If the integral has finite total area, the series’ partial sums are bounded. If the series has bounded partial sums, the integral is bounded as well. Since both have nonnegative terms or area, boundedness of their increasing partial sums or accumulated areas is equivalent to convergence. This explains why the test requires a decreasing function: that condition makes the rectangle bounds valid.

Worked Example: A Logarithmic Factor Suited to Integration

Worked Example: A Logarithmic Factor Suited to Integration

Determine whether

$$ \sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2} $$

converges. The logarithm makes comparison with a plain \(p\)-series less direct, while the corresponding integral has a simple substitution. Set \(f(x)=1/[x(\ln x)^2]\) for \(x\ge2\). The function is continuous and positive on this interval. Its derivative is

$$ f'(x) =-\frac{1}{x^2(\ln x)^2} -\frac{2}{x^2(\ln x)^3} =-\frac{\ln x+2}{x^2(\ln x)^3}<0 \qquad (x\ge2). $$

Therefore \(f\) is decreasing, and the integral test applies. Evaluate the integral using \(u=\ln x\), so \(du=dx/x\):

$$ \int_2^\infty \frac{dx}{x(\ln x)^2} = \int_{\ln 2}^\infty \frac{du}{u^2} = \left[-\frac{1}{u}\right]_{\ln 2}^{\infty} = \frac{1}{\ln 2}. $$

This is finite, so the series converges. As a check on the antiderivative,

$$ \frac{d}{dx}\left(-\frac{1}{\ln x}\right) = \frac{1}{x(\ln x)^2}. $$

The function conditions and the finite integral agree with the convergence conclusion.

Worked Example: A Radical Term with a Harmonic Lower Bound

Worked Example: A Radical Term with a Harmonic Lower Bound

Determine whether

$$ \sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+n}} $$

converges. For \(n\ge1\), we have \(n\le n^2\), and hence

$$ n^2+n\le 2n^2\le \frac{9}{4}n^2. $$

Taking positive square roots gives \(\sqrt{n^2+n}\le 3n/2\). Taking reciprocals reverses the inequality because both sides are positive:

$$ \frac{1}{\sqrt{n^2+n}}\ge\frac{2}{3n}. $$

The harmonic series \(\sum 1/n\) diverges. Therefore, by direct comparison from below, the series in question diverges. The bound is usable for every \(n\ge1\); for instance, at \(n=1\), \(1/\sqrt2\ge2/3\), as the inequality predicts. The lower-bound direction is essential: it shows that the terms are at least as large as a divergent benchmark up to a positive constant.

Worked Example: A Rational Term with a Convergent Upper Bound

Worked Example: A Rational Term with a Convergent Upper Bound

Determine whether

$$ \sum_{n=1}^{\infty}\frac{3n+2}{n^3+4n+1} $$

converges. For \(n\ge1\), \(2\le2n\), so \(3n+2\le5n\). Also \(n^3+4n+1\ge n^3>0\). Consequently,

$$ 0<\frac{3n+2}{n^3+4n+1} \le \frac{5n}{n^3} =\frac{5}{n^2}. $$

The comparison series \(\sum 5/n^2\) converges because it is a constant multiple of a \(p\)-series with \(p=2>1\). The direct comparison test therefore proves convergence. The two estimates used in the bound hold for every \(n\ge1\), and dividing by the denominator preserves the stated upper bound because all denominators are positive.

Worked Example: A Radical Term with a Convergent Power Bound

Worked Example: A Radical Term with a Convergent Power Bound

Determine whether

$$ \sum_{n=1}^{\infty}\frac{1}{n\sqrt{n+1}} $$

converges. Since \(n+1\ge n>0\), taking square roots gives \(\sqrt{n+1}\ge\sqrt n\). Therefore,

$$ 0<\frac{1}{n\sqrt{n+1}} \le \frac{1}{n\sqrt n} = \frac{1}{n^{3/2}}. $$

The comparison series converges because \(3/2>1\). Direct comparison proves that the given series converges. The check is that the radical in the original denominator is at least \(\sqrt n\), making the original term no larger than \(n^{-3/2}\). A larger denominator produces a smaller positive fraction, which is the required direction for proving convergence.

A Practical Decision Process

For a concrete series, the following sequence of decisions helps prevent wasted work:

1
Check positivity on a tail.
Confirm that the terms are nonnegative for all sufficiently large \(n\). If only finitely many initial terms need attention, they do not change convergence.
2
Look for a simple comparison.
Try to bound a rational or radical term above by a convergent \(p\)-series, or below by a divergent one. Write down why the inequality holds and verify its direction.
3
Check whether integration is cleaner.
For a term involving logarithms or a familiar function, define \(f(x)\) by replacing \(n\) with \(x\). Verify continuity, positivity, and decrease on a suitable interval before evaluating its improper integral.
4
State the conclusion from the right hypothesis.
An upper bound by a convergent series proves convergence; a lower bound by a divergent series proves divergence. The integral test requires its function conditions, not just a finite or infinite-looking integral calculation.

These choices are not exclusive. A series may satisfy the hypotheses of both methods. Prefer the argument with the shortest transparent verification, rather than assuming that one test is always better for a particular kind of formula.

Common Pitfall: Using the Wrong Comparison Direction

Consider the false argument: “Since \(1/n^2\le1/n\), and \(\sum 1/n\) diverges, \(\sum 1/n^2\) diverges.” The inequality is true, but the conclusion does not follow. It gives an upper bound by a divergent series, which is not enough to establish divergence. Indeed, the series \(\sum1/n^2\) converges.

To use direct comparison to prove divergence, the divergent benchmark must be a lower bound. To use it to prove convergence, the convergent benchmark must be an upper bound. Always write the inequality in the form needed for the conclusion before invoking the comparison test.

Check Your Understanding

For each question, identify the needed hypothesis or comparison direction before deciding on a test.

  1. For \(\sum_{n=2}^{\infty}1/[n(\ln n)^3]\), what function would you use for the integral test, and what conditions must you verify?
  2. If \(0\le a_n\le 4/n^2\) for all sufficiently large \(n\), what conclusion follows, and why?
  3. To prove that a positive series diverges by comparison with \(\sum1/n\), should its terms be bounded above or below by a positive multiple of \(1/n\)?
  4. Why does \(1/[n\sqrt{n+1}]\le1/n^{3/2}\) establish convergence?
  5. What does the integral test require of the function \(f\), in addition to the existence of an improper integral?

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