Tutorials › Calculus II › Direct-Comparison Workshop: Rational Sequences

Home  /  Tutorials  /  Integral and Direct Comparison

Integral and Direct Comparison · Tutorial 620 of 998

Direct-Comparison Workshop: Rational Sequences

Use direct comparison with p-series to classify rational-term series by proving explicit inequalities.

Intermediate 9 min read

What You'll Learn

  • State the direct comparison rules for convergence and divergence.
  • Use p-series criteria to interpret a rational-term comparison.
  • Bound rational expressions with positive polynomial coefficients using powers of the index.
  • Prove convergence with an explicit upper bound by a convergent p-series.
  • Prove divergence with an explicit lower bound by a divergent p-series.
  • Identify why an upper bound by a divergent series does not classify the original series.

Why Compare Rational Terms Directly?

A series whose terms are rational expressions in \(n\) can be difficult to classify by looking at its formula alone. Direct comparison replaces the expression with a simpler power of \(n\), then uses a known result about p-series. The essential work is not guessing the likely behavior; it is proving an inequality in the correct direction.

For example, an upper bound by a convergent p-series proves convergence. A lower bound by a divergent p-series proves divergence. The reverse pairings do not, by themselves, establish either conclusion. In this workshop, each classification will rest on an explicit inequality valid for every index in the series.

Definition: A p-series is a series of the form \(\sum_{n=1}^{\infty}1/n^p\), where \(p>0\). It converges when \(p>1\) and diverges when \(0<p\leq 1\).

The Direct Comparison Rules

The terms being compared must be nonnegative, at least from some index onward. A finite number of initial terms does not affect whether a series converges or diverges, so an inequality need only hold for all \(n\geq N\) for some fixed \(N\).

Theorem: Suppose \(a_n\geq 0\) and \(b_n\geq 0\) for all \(n\geq N\). If \(a_n\leq b_n\) for all \(n\geq N\), and \(\sum b_n\) converges, then \(\sum a_n\) converges. If \(a_n\geq b_n\) for all \(n\geq N\), and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

The direction of each inequality matches the conclusion. In the first case, the partial sums of \(a_n\) are bounded above by the partial sums of a convergent series. In the second case, the partial sums of \(a_n\) are at least as large as those of a divergent series, so they cannot remain bounded.

$$ 0\leq a_n\leq b_n,\quad \sum b_n\text{ converges} \quad\Longrightarrow\quad \sum a_n\text{ converges}, $$
$$ a_n\geq b_n\geq 0,\quad \sum b_n\text{ diverges} \quad\Longrightarrow\quad \sum a_n\text{ diverges}. $$

For rational terms, the useful comparison is often \(C/n^p\), where \(C>0\) is a constant. Multiplication by a positive constant does not change whether a p-series converges or diverges. Thus, an explicit upper bound by \(C/n^p\) proves convergence when \(p>1\), and an explicit lower bound by \(C/n^p\) proves divergence when \(0<p\leq1\).

A Coefficient Bound for Rational Expressions

A convenient method applies when the numerator and denominator polynomials have nonnegative coefficients. Write

$$ P(n)=\sum_{j=0}^{r}p_jn^j,\qquad Q(n)=\sum_{j=0}^{s}q_jn^j, $$

where \(p_j,q_j\geq0\), and the leading coefficients \(p_r\) and \(q_s\) are positive. For \(n\geq1\), every power \(n^j\) with \(j\leq r\) satisfies \(n^j\leq n^r\). Consequently,

$$ P(n)\leq \left(\sum_{j=0}^{r}p_j\right)n^r. $$

Also, the leading term is one of the nonnegative terms in \(Q(n)\), so

$$ Q(n)\geq q_sn^s. $$

Because the denominator is positive, combining these bounds gives an upper bound for the quotient:

$$ 0<\frac{P(n)}{Q(n)} \leq \frac{\sum_{j=0}^{r}p_j}{q_s}\frac{1}{n^{s-r}} \qquad(n\geq1). $$

This proves convergence by direct comparison when \(s-r>1\). For a lower bound, use the leading term \(P(n)\geq p_rn^r\) and the coefficient-sum bound \(Q(n)\leq(\sum_{j=0}^{s}q_j)n^s\), also valid for \(n\geq1\). Then

$$ \frac{P(n)}{Q(n)} \geq \frac{p_r}{\sum_{j=0}^{s}q_j}\frac{1}{n^{s-r}}. $$

When \(s-r=1\), this is a positive constant times \(1/n\), so it proves divergence by comparison with the harmonic p-series. If \(s-r\leq0\), the same lower bound is at least a positive constant for \(n\geq1\), which also proves divergence. These bounds rely on nonnegative polynomial coefficients; with negative coefficients, the inequalities must be established separately rather than assumed.

Worked Examples

Worked Example: An Upper Bound Proves Convergence

Classify the series

$$ \sum_{n=1}^{\infty}\frac{4n+3}{n^3+2n+5}. $$

For \(n\geq1\), \(3\leq3n\), so \(4n+3\leq4n+3n=7n\). The denominator satisfies \(n^3+2n+5\geq n^3\). Both expressions are positive. Dividing an upper bound for the numerator by a positive lower bound for the denominator gives

$$ 0< \frac{4n+3}{n^3+2n+5} \leq \frac{7n}{n^3} = \frac{7}{n^2}. $$

The comparison series \(\sum 7/n^2\) converges because it is a constant multiple of a p-series with \(p=2>1\). The direct comparison convergence rule therefore proves that the original series converges.

Check the inequality at the first index: when \(n=1\), the original term is \(7/8\), while the bound is \(7\), and \(0<7/8\leq7\). The general proof above establishes the required inequality for every \(n\geq1\).

Worked Example: A Lower Bound Proves Divergence

Classify the series

$$ \sum_{n=1}^{\infty}\frac{2n^2+3n+1}{n^3+5n^2+4}. $$

The numerator is at least \(2n^2\). For \(n\geq1\), \(5n^2\leq5n^3\) and \(4\leq4n^3\), so the denominator satisfies

$$ n^3+5n^2+4 \leq n^3+5n^3+4n^3 =10n^3. $$

The denominator is positive. A lower bound for the numerator and an upper bound for the denominator therefore give a lower bound for the quotient:

$$ \frac{2n^2+3n+1}{n^3+5n^2+4} \geq \frac{2n^2}{10n^3} = \frac{1}{5n} >0. $$

The series \(\sum 1/(5n)\) diverges because it is one-fifth of the p-series with \(p=1\). Since the original terms are at least as large as these divergent comparison terms, the original series diverges.

At \(n=1\), the original term is \(6/10=3/5\), and the comparison term is \(1/5\), so the lower-bound direction is consistent: \(3/5\geq1/5\). The inequalities used above prove it for all \(n\geq1\).

Worked Example: A Larger Degree Gap Gives Convergence

Classify the series

$$ \sum_{n=1}^{\infty}\frac{6n^2+2n+1}{3n^4+7n^2+2n+8}. $$

For \(n\geq1\), \(2n\leq2n^2\) and \(1\leq n^2\). Hence

$$ 6n^2+2n+1 \leq6n^2+2n^2+n^2 =9n^2. $$

The denominator contains the positive term \(3n^4\) and other positive terms, so it is at least \(3n^4\). Therefore,

$$ 0< \frac{6n^2+2n+1}{3n^4+7n^2+2n+8} \leq \frac{9n^2}{3n^4} = \frac{3}{n^2}. $$

The comparison series \(\sum 3/n^2\) converges because \(p=2>1\), and thus the given series converges by direct comparison. At \(n=1\), the original term is \(9/20\), and the bound is \(3\), confirming \(0<9/20\leq3\). The inequalities establish this direction for every positive integer \(n\).

Worked Example: Equal Degrees Give a Constant Lower Bound

Classify the series

$$ \sum_{n=1}^{\infty}\frac{3n^2+n+2}{2n^2+5n+1}. $$

For \(n\geq1\), the numerator is at least \(3n^2\). Also, \(5n\leq5n^2\) and \(1\leq n^2\), so the denominator is at most

$$ 2n^2+5n+1 \leq2n^2+5n^2+n^2 =8n^2. $$

It follows that

$$ \frac{3n^2+n+2}{2n^2+5n+1} \geq \frac{3n^2}{8n^2} = \frac{3}{8}. $$

Thus the partial sum through \(N\) terms is at least \(3N/8\), which grows without bound as \(N\) increases. The series diverges. This verifies divergence directly, without needing to express the constant lower bound as a p-series. At \(n=1\), the original term is \(6/8=3/4\), which is indeed at least \(3/8\).

Why the Direction Matters

A comparison is useful only when its inequality points toward the conclusion it supports. For convergence, the original terms need to be no larger than terms of a known convergent series. For divergence, the original terms need to be no smaller than terms of a known divergent series.

Common mistake: The fact that a series is bounded above by a divergent series does not prove that it diverges.

For a concrete example, \(1/n^2\leq1/n\) for every \(n\geq1\), and \(\sum 1/n\) diverges. It would be incorrect to conclude that \(\sum 1/n^2\) diverges from this upper bound. The inequality has the wrong direction for proving divergence. In fact, \(\sum 1/n^2\) converges, as the p-series criterion shows. To prove convergence, compare \(1/n^2\) from above with a convergent series; to prove divergence, compare it from below with a divergent series.

A second error is reversing the effect of a denominator bound. If \(P(n)>0\) and \(Q(n)>0\), then \(P(n)\geq L(n)\) and \(Q(n)\leq U(n)\) imply \(P(n)/Q(n)\geq L(n)/U(n)\). A larger numerator and smaller positive denominator make the quotient larger, not smaller. Checking this direction before concluding can prevent an otherwise plausible-looking but invalid comparison.

A Practical Decision Process

1
Check positivity.
Confirm that the rational terms and the comparison terms are nonnegative for the indices under consideration.
2
Choose a power.
Use the polynomial degrees to suggest a comparison of the form \(C/n^p\), then prove the needed inequality explicitly.
3
Check the inequality direction.
An upper bound is the useful direction for convergence; a lower bound is the useful direction for divergence.
4
Apply the p-series criterion.
For \(p>1\), the comparison p-series converges; for \(0<p\leq1\), it diverges.
5
State the conclusion and its basis.
Name the comparison series and say whether the direct comparison rule proves convergence or divergence.

The coefficient bounds give a systematic route when all polynomial coefficients are nonnegative: a denominator degree at least two greater than the numerator degree yields an upper comparison by a convergent p-series. A denominator degree exactly one greater yields a lower comparison by a constant multiple of the harmonic series. Equal or smaller denominator degree can be handled with a positive constant lower bound under the same coefficient assumptions. In every case, the explicit inequality—not the degree observation alone—is what justifies the classification.

Check Your Understanding

For each question, identify the comparison direction and explain what conclusion it permits.

  1. If \(0\leq a_n\leq 6/n^3\) for all \(n\geq1\), what can be concluded about \(\sum a_n\), and why?
  2. If \(a_n\geq 2/(7n)\) for all \(n\geq1\), what can be concluded about \(\sum a_n\), and why?
  3. Why does an upper bound by \(4/n\) fail to prove that a positive-term series diverges?
  4. For \(P(n)=2n^2+3n+1\) and \(Q(n)=n^4+2n^2+5\), use \(n\geq1\) to produce an explicit upper bound by a constant multiple of \(1/n^2\).
  5. State the coefficient-sign condition needed for the degree-based bounds in this tutorial, and explain why it matters.

Previous

Next

Back to Tutorials