Tutorials › Calculus II › Rational Terms Compared with p-Series

Home  /  Tutorials  /  Integral and Direct Comparison

Integral and Direct Comparison · Tutorial 613 of 998

Rational Terms Compared with p-Series

Use polynomial degree and elementary bounds to compare positive rational terms with p-series and decide whether their series converge.

Intermediate 9 min read

What You'll Learn

  • Identify the effective degrees and leading coefficients in a rational term.
  • Bound a rational term above and below by constant multiples of a power of n.
  • Translate the difference in polynomial degrees into a p-series comparison.
  • Apply direct comparison in the correct direction for convergence and divergence.
  • Recognize why a bound by a divergent p-series does not establish convergence.

Why Polynomial Degrees Help

A rational term is a quotient of polynomials evaluated at an integer \(n\). For large \(n\), the highest-degree terms determine useful bounds on that quotient. Those bounds can connect the series to a familiar \(p\)-series, whose convergence behavior is already known.

The aim is not merely to notice which powers appear. We need inequalities in the right direction: an upper bound by a convergent series can prove convergence, while a lower bound by a divergent series can prove divergence. A degree calculation suggests the comparison; explicit bounds justify it.

The Degree Rule for Rational Terms

Let \(P\) and \(Q\) be polynomials with degrees \(m\) and \(d\), respectively, and leading coefficients \(a\) and \(b\). Suppose \(a/b>0\), so \(P(n)/Q(n)\) is positive for all sufficiently large \(n\). The degrees here are the actual degrees after any cancellation within each polynomial.

Degree rule: For all sufficiently large \(n\), the rational term \(P(n)/Q(n)\) is bounded above and below by positive constant multiples of \(n^{m-d}\). Equivalently, it is comparable by such bounds to \(1/n^{d-m}\).

In particular, the candidate \(p\)-series exponent is the denominator degree minus the numerator degree: \(p=d-m\). The word “candidate” matters. To use direct comparison, we still need to establish an appropriate inequality.

The \(p\)-series \(\sum_{n=1}^{\infty}1/n^p\) converges when \(p>1\) and diverges when \(p\leq 1\). The direct comparison rules say that for eventually nonnegative terms, an upper bound by a convergent series proves convergence, and a lower bound by a divergent series proves divergence. Removing or adding finitely many initial terms does not change convergence.

Why the Degree Rule Works

Write the leading terms and all lower-degree terms of \(P\) as

$$ P(n)=a n^m+\sum_{j=0}^{m-1}a_j n^j. $$

After dividing by \(n^m\), each lower-degree term becomes \(a_j/n^{m-j}\), which tends to zero as \(n\) increases. Thus, for sufficiently large \(n\), the sum of the absolute values of those terms is at most \(|a|/2\). The triangle inequality then gives

$$ \frac{|a|}{2}n^m \leq |P(n)| \leq \frac{3|a|}{2}n^m. $$

Applying the same argument to \(Q\), for all sufficiently large \(n\) we have

$$ \frac{|b|}{2}n^d \leq |Q(n)| \leq \frac{3|b|}{2}n^d. $$

Because \(a/b>0\), the two polynomials have the same sign for all sufficiently large \(n\), so their quotient is positive. Combining the lower bound for \(|P(n)|\) with the upper bound for \(|Q(n)|\), and then the upper bound for \(|P(n)|\) with the lower bound for \(|Q(n)|\), yields

$$ \frac{|a|}{3|b|}n^{m-d} \leq \frac{P(n)}{Q(n)} \leq \frac{3|a|}{|b|}n^{m-d}. $$

These constants are positive and do not depend on \(n\). This proves the degree rule. It also explains why the degree difference alone is not a complete direct-comparison argument: the rule guarantees eventual constant-multiple bounds, but a particular proof should exhibit the bound it needs.

Using the Degree Difference to Classify a Series

Set \(p=d-m\). If \(p>1\), then \(n^{m-d}=1/n^p\), and the degree rule gives an upper bound by a constant multiple of a convergent \(p\)-series. Direct comparison proves convergence. If \(p\leq1\), the rule gives a lower bound by a positive constant multiple of a divergent \(p\)-series. Direct comparison proves divergence. When \(p\leq0\), the lower bound does not even tend to zero; when \(0<p\leq1\), the \(p\)-series itself diverges.

This classification applies to rational terms that are positive from some index onward and have the stated leading-coefficient condition. The finite number of earlier terms, if any, cannot affect the result. Here are explicit comparisons for several degree patterns.

Worked Example: A Convergent Rational Series

Classify

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

The numerator has degree \(1\), and the denominator has degree \(3\). The degree difference suggests comparison with \(1/n^2\). For every integer \(n\geq1\),

$$ 2n+3\leq 5n \qquad\text{and}\qquad n^3+4n+1\geq n^3. $$

The terms are positive, so these inequalities give the required upper bound:

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

The series \(\sum 5/n^2\) converges because it is a constant multiple of a \(p\)-series with \(p=2>1\). By direct comparison, the given series converges. The inequality checks the degree prediction: the rational term is indeed bounded above by a constant multiple of \(1/n^2\).

Worked Example: A Divergent Series with a Harmonic Lower Bound

Classify

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

The numerator has degree \(2\) and the denominator degree \(3\), suggesting \(p=1\). For \(n\geq1\), the numerator is at least \(3n^2\). Also, \(n\leq n^3\), so

$$ 2n^3+5n\leq 2n^3+5n^3=7n^3. $$

Since the denominator is positive, using the numerator’s lower bound and the denominator’s upper bound gives

$$ \frac{3n^2+1}{2n^3+5n} \geq \frac{3n^2}{7n^3} = \frac{3}{7n}. $$

The harmonic series \(\sum 1/n\) diverges. Therefore \(\sum 3/(7n)\) diverges, and the lower bound proves by direct comparison that the given series diverges.

Check the inequality direction: to make a lower bound for the quotient, we used a lower bound for its positive numerator and an upper bound for its positive denominator. The result is a positive constant multiple of the harmonic term.

Worked Example: Equal Degrees Give Terms That Do Not Vanish

Classify

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

Both polynomials have degree \(2\), so the degree difference gives \(p=0\): the terms are bounded below by a positive constant, rather than by a decaying \(p\)-series. For every \(n\geq1\),

$$ n^2+4n+1\geq n^2 \qquad\text{and}\qquad n^2+2\leq 3n^2. $$

Consequently,

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

The terms do not tend to zero: each is at least \(1/3\). Therefore the series diverges. Equivalently, the partial sum of the first \(N\) terms is at least \(N/3\), which grows without bound.

This example illustrates what happens when the numerator and denominator have the same degree. The term does not behave like a reciprocal power that decays to zero; its leading-degree comparison is with a positive constant.

Choose the Comparison Direction Carefully

A bound is useful only if its direction supports the conclusion. For convergence, we need to show that the terms are no larger than terms of a known convergent series. For divergence, we need to show that the terms are no smaller than terms of a known divergent series.

Goal Useful inequality Comparison series
Prove convergence \(0\leq a_n\leq b_n\) \(\sum b_n\) converges
Prove divergence \(a_n\geq b_n\geq0\) \(\sum b_n\) diverges
Not enough for convergence \(a_n\leq b_n\) \(\sum b_n\) diverges
Not enough for divergence \(a_n\geq b_n\) \(\sum b_n\) converges

In particular, being bounded above by a divergent \(p\)-series does not establish convergence. A divergent comparison series can still have terms larger than the series being studied.

Common Pitfall: An Upper Bound by the Harmonic Series

Consider \(\sum (n+1)/(n^2+1)\). A tempting calculation is

$$ \frac{n+1}{n^2+1} \leq \frac{2n}{n^2} = \frac{2}{n} \qquad(n\geq1), $$

followed by the incorrect claim that the original series converges. The inequality itself is correct, but its conclusion does not follow: \(\sum 2/n\) diverges, so this is not an upper bound by a known convergent series.

For the correct conclusion, use a lower bound. Since \(n+1\geq n\) and \(n^2+1\leq2n^2\) for \(n\geq1\),

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

The harmonic comparison series \(\sum 1/(2n)\) diverges, so the given series diverges. The degree difference predicted the borderline \(p=1\); the explicit lower bound supplies the direct-comparison proof.

A Practical Procedure

For a rational-term series, a dependable analysis can be organized into a few steps:

1
Identify the actual degrees.
Determine the numerator and denominator degrees after simplifying each polynomial, and check that the terms are positive eventually.
2
Find the candidate exponent.
Subtract the numerator degree from the denominator degree to obtain \(p=d-m\).
3
Choose the needed bound.
For convergence, seek an upper bound by \(C/n^p\) with \(p>1\). For divergence, seek a lower bound by \(c/n^p\) with \(p\leq1\), or show that the terms stay above a positive constant.
4
State the comparison conclusion.
Verify the inequality for all sufficiently large \(n\), cite the \(p\)-series behavior, and apply the appropriate direct comparison rule.

The degrees often make the likely answer apparent, but the written inequality is what makes the direct comparison valid. Keeping those two roles separate prevents a degree estimate from becoming an unsupported conclusion.

Check Your Understanding

For each question, consider both the degree prediction and the inequality needed to justify a direct comparison.

  1. For \(\sum (4n+1)/(n^3+2)\), what \(p\)-series exponent is suggested by the degrees, and what comparison direction would prove convergence?
  2. For \(\sum (n^2+2)/(n^3+1)\), is a lower bound by a constant multiple of \(1/n\) useful for proving convergence or divergence?
  3. What conclusion is suggested when the numerator and denominator have the same degree and their leading coefficients have the same sign?
  4. Why does an upper bound by \(3/n\) not prove that a positive-term series converges?
  5. In establishing a lower bound for a positive rational quotient, why is an upper bound on the denominator useful?

Previous

Next

Back to Tutorials