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Convergence Tests · Tutorial 840 of 999

Dominant Powers in Rational Series

You will learn how polynomial degrees predict the behavior of a rational term and how to use p-series comparison to classify its series.

Advanced 10 min read

What You'll Learn

  • Find the dominant power of \(n\) in a rational term from its numerator and denominator degrees.
  • Convert the degree difference into the p-series exponent used for comparison.
  • Use limit comparison to a p-series when the terms are eventually positive.
  • Use direct comparison when simple polynomial bounds are available.
  • Distinguish a term tending to zero from a series that converges.
  • Check the hypotheses before drawing a convergence or divergence conclusion.

Why Leading Powers Matter

Rational terms often contain several powers of \(n\), but their behavior for large \(n\) is governed by the highest powers in the numerator and denominator. That makes the degrees of the polynomials a useful first guide when deciding whether a series might converge.

The key is to connect the dominant power of the term to a familiar benchmark, the p-series \(\sum_{n=1}^{\infty}1/n^p\). The degree difference tells us which p-series to compare with; the comparison test then gives a justified conclusion. The degree calculation is a guide, not a substitute for checking positivity and applying a valid test.

Key idea: If a rational term has numerator degree \(m\) and denominator degree \(d\), and their leading-coefficient ratio is positive, then the term behaves like a positive constant times \(n^{m-d}=1/n^{d-m}\). The comparison exponent is \(p=d-m\), not \(m-d\).

The Degree Difference and the p-Series Exponent

Let \(P(n)\) and \(Q(n)\) be polynomials with degrees \(m\) and \(d\), respectively. Write their leading terms as \(a_m n^m\) and \(b_d n^d\), where \(a_m\ne 0\) and \(b_d\ne 0\). Dividing numerator and denominator by their leading powers shows that

$$ \frac{P(n)}{Q(n)} \sim \frac{a_m}{b_d}n^{m-d} = \frac{a_m}{b_d}\frac{1}{n^{d-m}}. $$

Here, “behaves like” means that the ratio of the two expressions approaches \(1\) as \(n\to\infty\). When \(a_m/b_d>0\), the rational term is positive for all sufficiently large \(n\), provided the denominator has no zeros there. Define \(p=d-m\). Then the p-series benchmark has term \(1/n^p\).

Definition: For a rational term \(a_n=P(n)/Q(n)\), its dominant-power comparison exponent is \(p=\deg(Q)-\deg(P)\), provided the leading coefficients are nonzero. This exponent identifies the p-series term \(1/n^p\) to use for comparison.

The p-series test states that \(\sum 1/n^p\) converges when \(p>1\) and diverges when \(p\le 1\). Thus, for an eventually positive rational term with a positive leading-coefficient ratio, the degree calculation predicts convergence when the denominator degree exceeds the numerator degree by more than \(1\), and divergence when the difference is at most \(1\). We will justify this prediction using limit comparison below.

Worked Example: A Convergent Rational Series

Worked Example: Comparing a Rational Series with a p-Series

Classify

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

Step 1: Check positivity and identify the degrees. For every integer \(n\ge 1\), both \(2n^2+3\) and \(n^5+4n^2\) are positive. The numerator degree is \(2\), and the denominator degree is \(5\). Therefore \(p=5-2=3\).

Step 2: Use direct comparison. Since \(n\ge 1\), we have \(3\le 3n^2\), so

$$ 2n^2+3\le 2n^2+3n^2=5n^2. $$

Also \(n^5+4n^2\ge n^5>0\). Therefore

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

The comparison series \(\sum 5/n^3\) converges because it is \(5\) times a p-series with \(p=3>1\). By direct comparison, the given series converges.

Check: The limit comparison ratio confirms the predicted exponent:

$$ \lim_{n\to\infty} \frac{\dfrac{2n^2+3}{n^5+4n^2}}{1/n^3} = \lim_{n\to\infty} \frac{2n^5+3n^3}{n^5+4n^2} = \lim_{n\to\infty} \frac{2+3/n^2}{1+4/n^3} =2. $$

The limit is finite and positive, consistent with comparison to \(\sum 1/n^3\).

Limit Comparison: Why the Degree Rule Works

The limit comparison test applies to series with positive terms. If \(a_n>0\) and \(b_n>0\) for all sufficiently large \(n\), and

$$ \lim_{n\to\infty}\frac{a_n}{b_n}=L \qquad\text{with}\qquad 0<L<\infty, $$

then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. A finite number of initial terms does not change whether a series converges, so eventual positivity is sufficient.

For a rational term \(a_n=P(n)/Q(n)\), choose \(b_n=1/n^p\), where \(p=d-m\). Here is the complete calculation that produces the comparison limit.

1
Write the leading terms.
Suppose \(P(n)=a_m n^m+\) lower-degree terms and \(Q(n)=b_d n^d+\) lower-degree terms, with \(a_m,b_d\ne 0\).
2
Form the ratio to the benchmark.
For \(p=d-m\), divide \(P(n)/Q(n)\) by \(1/n^p\), which is the same as multiplying by \(n^p\).
3
Divide by the highest powers.
The numerator and denominator have equal resulting degree \(d\), so their leading coefficients determine the limit.
$$ \frac{a_n}{1/n^{d-m}} = \frac{n^{d-m}P(n)}{Q(n)} \longrightarrow \frac{a_m}{b_d}. $$

For clarity, the limit follows by writing

$$ P(n)=n^m\left(a_m+\frac{a_{m-1}}{n}+\cdots+\frac{a_0}{n^m}\right), $$

and

$$ Q(n)=n^d\left(b_d+\frac{b_{d-1}}{n}+\cdots+\frac{b_0}{n^d}\right). $$

Since \(p=d-m\), substituting these expressions gives

$$ \frac{n^{d-m}P(n)}{Q(n)} = \frac{a_m+a_{m-1}/n+\cdots+a_0/n^m} {b_d+b_{d-1}/n+\cdots+b_0/n^d}. $$

As \(n\to\infty\), every term involving \(1/n\) approaches \(0\). The denominator approaches the nonzero number \(b_d\), so the ratio approaches \(a_m/b_d\). If this ratio is positive, the limit comparison test applies. Because the p-series benchmark converges for \(p>1\) and diverges for \(p\le 1\), the degree rule follows.

Worked Example: The Harmonic Boundary

Worked Example: A Term That Behaves Like \(1/n\)

Classify

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

Step 1: Check positivity. For \(n\ge1\), the numerator \(5n^4+2\) and denominator \(2n^5+n^2\) are both positive. The term is therefore positive.

Step 2: Find the comparison exponent. The degrees are \(4\) and \(5\), so \(p=5-4=1\). The relevant benchmark is the harmonic series \(\sum 1/n\).

Step 3: Compute the limit comparison ratio.

$$ \lim_{n\to\infty} \frac{\dfrac{5n^4+2}{2n^5+n^2}}{1/n} = \lim_{n\to\infty} \frac{5n^5+2n}{2n^5+n^2} = \lim_{n\to\infty} \frac{5+2/n^4}{2+1/n^3} = \frac{5}{2}. $$

The limit \(5/2\) is finite and positive. The harmonic series diverges, so the given series diverges by limit comparison.

Check: The term behaves like \((5/2)n^{-1}=(5/2)/n\). It tends to zero, but comparison with the harmonic series shows that the sum still diverges.

Worked Example: A Denominator Degree Two Higher

Worked Example: Limit Comparison with \(1/n^2\)

Classify

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

Step 1: Check positivity and degrees. For \(n\ge1\), both polynomials are positive. Their degrees are \(2\) and \(4\), so \(p=4-2=2\). Compare with \(\sum 1/n^2\).

Step 2: Calculate the ratio.

$$ \lim_{n\to\infty} \frac{\dfrac{n^2+1}{3n^4+2n}}{1/n^2} = \lim_{n\to\infty} \frac{n^4+n^2}{3n^4+2n} = \lim_{n\to\infty} \frac{1+1/n^2}{3+2/n^3} = \frac{1}{3}. $$

The limit is finite and positive, and the p-series with \(p=2\) converges. Therefore the rational series converges by limit comparison.

Check: The ratio calculation gives the leading-term behavior \((1/3)n^{-2}\), matching the convergent benchmark \(1/n^2\).

When the Degree Difference Is Zero or Negative

The exponent \(p=d-m\) need not be positive. If \(p=0\), the rational term approaches the nonzero leading-coefficient ratio \(a_m/b_d\). If \(p<0\), its magnitude grows like a positive constant times \(n^{-p}\). In either case, the terms do not approach zero. The divergence test says that if the terms of a series do not tend to zero, the series diverges.

This is different from the case \(p=1\). When \(p=1\), the terms do approach zero, but the harmonic comparison still proves divergence. A term tending to zero is necessary for convergence of a series, but it is not sufficient.

A Common Mistake: Reversing the Exponent

A frequent error is to calculate the power \(m-d\) and then treat it as the p-series exponent. These are related, but they have opposite signs: the term has a factor \(n^{m-d}\), while the benchmark is written \(1/n^p=n^{-p}\). Therefore \(p=d-m\).

Common mistake: For \((5n^4+2)/(2n^5+n^2)\), the power difference \(4-5=-1\) does not mean the p-series exponent is \(-1\). It means the term behaves like a constant times \(n^{-1}=1/n\). Equivalently, \(p=5-4=1\), so the series diverges by limit comparison with the harmonic series.

Another mistake is to infer convergence just because the terms tend to zero. The harmonic series is a counterexample to that reasoning: its terms approach zero, but its partial sums do not approach a finite limit.

Conditions to Check Before Concluding

The degree method is useful only when its hypotheses are respected. Before applying a comparison test, check the following:

  • Defined terms: The denominator must be nonzero at the indices in the series. A rational expression with a zero denominator is not a valid term at that index.
  • Positivity: Direct and limit comparison in the forms used here require positive terms, at least eventually. Check the signs of the leading coefficients and the polynomials, rather than assuming positivity from the degrees alone.
  • Correct benchmark: Use \(p=d-m\), because \(n^{m-d}=1/n^{d-m}\).
  • Valid conclusion: For \(p>1\), comparison with a convergent p-series establishes convergence. For \(p\le1\), comparison with a divergent p-series establishes divergence, provided the limit comparison ratio is positive and finite.
  • Term test: When \(p\le0\), the terms fail to approach zero, so use the divergence test directly.

If a rational term is eventually negative instead of positive, do not apply a positive-term comparison test to it as written. One can often analyze its magnitude or multiply the entire tail by \(-1\); a finite initial segment never changes convergence. The important point is to establish the relevant sign before using comparison.

Why This Classification Is Useful

Dominant powers turn a complicated-looking rational expression into a familiar benchmark without requiring every lower-degree term to be manipulated separately. The method is especially efficient when the numerator and denominator are polynomials with clear leading terms. It also provides a reliable first decision: a degree gap greater than one leads to a convergent p-series comparison, while a gap of one leads to the harmonic boundary and divergence.

The final conclusion still comes from a theorem, not from visual resemblance alone. Direct comparison requires an inequality; limit comparison requires eventual positivity and a finite positive ratio. Keeping those conditions visible makes the degree calculation both quick and rigorous.

Check Your Understanding

For each question, identify the comparison exponent and state which condition or test justifies the conclusion.

  1. For \(\sum (4n+1)/(n^4+3)\), what is the dominant-power comparison exponent, and does the series converge?
  2. For \(\sum (3n^3+2)/(6n^4+n)\), which p-series is the appropriate benchmark, and what is the limit comparison ratio?
  3. Why does a rational series with equal numerator and denominator degrees fail the term test when its leading-coefficient ratio is nonzero?
  4. For \(\sum (n^2+5)/(2n^3+1)\), explain why the degree calculation indicates divergence even though the terms approach zero.
  5. What must be checked about the signs of the rational terms before using direct or limit comparison?

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