Introduction
A rational term is a quotient of polynomials evaluated at the positive integers. Its numerator and denominator may each grow quickly, but for large \(n\), their leading powers determine the quotient’s scale. Finding that scale gives a direct route to classifying the associated positive-term series.
The key calculation is the difference between the polynomial degrees. If the numerator has degree \(p\) and the denominator has degree \(q\), the rational term behaves like a constant times \(n^{p-q}\), provided the leading-coefficient ratio is positive. The power \(p-q\), rather than either degree alone, determines the relevant comparison series.
Degrees and the leading power
Let \(P\) and \(Q\) be polynomials with positive leading coefficients. Write their leading terms as \(a_p n^p\) and \(b_q n^q\), where \(p\) and \(q\) are their degrees and \(a_p,b_q>0\). Consider the rational sequence \(u_n=P(n)/Q(n)\). Since the leading coefficients are positive, both polynomials are positive for all sufficiently large \(n\), and \(Q(n)\) is nonzero there. Thus \(u_n\) is positive from some index onward.
When \(p\ne q\), the leading power of \(u_n\) is \(n^{p-q}\), with leading factor \(a_p/b_q\). In particular, the case \(p<q\) gives a reciprocal power \(n^{-(q-p)}\). The degree difference identifies the exponent; the leading coefficients determine the positive constant multiplying that power.
Why the degree difference gives the equivalent
To justify the rule, divide the numerator by its leading term and the denominator by its leading term. Every lower-degree term then contains a factor \(1/n^k\) for some positive integer \(k\), which tends to zero as \(n\) grows. This gives the following derivation.
After dividing \(P(n)\) by \(a_p n^p\), its leading term becomes \(1\), and each lower-degree term tends to \(0\).
After dividing \(Q(n)\) by \(b_q n^q\), its leading term becomes \(1\), and each lower-degree term also tends to \(0\).
Compare \(P(n)/Q(n)\) with \((a_p/b_q)n^{p-q}\). The normalized numerator and denominator both tend to \(1\), so their quotient tends to \(1\).
In symbols, the normalized polynomials can be written as
Here \(a_j\) and \(b_j\) denote the coefficients of \(P\) and \(Q\), with a zero coefficient allowed for a missing power. Since \(j-p<0\) in each numerator correction and \(j-q<0\) in each denominator correction, every correction term tends to zero. Therefore,
This equivalent is useful for series because a positive constant factor does not change whether a series converges or diverges. More precisely, positive sequences with ratio tending to \(1\) have the same convergence behavior, by limit comparison. The power-series test states that \(\sum_{n=1}^{\infty}1/n^s\) converges when \(s>1\) and diverges when \(s\le 1\). These facts turn the degree difference into a classification rule.
Classifying the series from the degrees
If \(p<q\), set \(s=q-p\), a positive integer. The rational term is equivalent to \((a_p/b_q)/n^s\). Thus its positive-term series converges if \(q-p>1\), and diverges if \(q-p=1\). If \(p>q\), the term is equivalent to a positive constant times \(n^{p-q}\), so the terms do not tend to zero; the series diverges.
| Degree relationship | Leading scale | Series conclusion |
|---|---|---|
| \(p<q\), with \(q-p>1\) | A positive constant times \(1/n^{q-p}\) | Converges |
| \(p<q\), with \(q-p=1\) | A positive constant times \(1/n\) | Diverges |
| \(p>q\) | A positive constant times \(n^{p-q}\) | Diverges because the terms do not tend to zero |
The table assumes the rational terms are positive for all sufficiently large \(n\). If a finite number of initial terms are undefined or need separate checking, they do not affect convergence of the tail; the series must still be defined at any terms included in its stated index range. The equal-degree case is not part of this degree-difference rule.
Worked examples
Worked Example: A convergent series with a cubic reciprocal scale
Classify the series with terms
The numerator has degree \(2\) and leading coefficient \(3\); the denominator has degree \(5\) and leading coefficient \(1\). The degree difference is \(2-5=-3\), so the predicted equivalent is \(3/n^3\). Verify it by taking the ratio:
Thus \(u_n\sim 3/n^3\). Since \(3>1\), the series \(\sum_{n=1}^{\infty}1/n^3\) converges. Limit comparison gives that \(\sum_{n=1}^{\infty}u_n\) converges as well. The calculation checks the equivalent directly: the ratio of the original term to \(3/n^3\) tends to \(1\).
Worked Example: A divergent series whose terms grow
Classify the series with terms
The degrees are \(3\) and \(2\), so the degree difference is \(1\). The leading-coefficient ratio is \(2/5\), giving the predicted scale \((2/5)n\). Check by dividing each polynomial by its leading power:
Therefore \(v_n\sim (2/5)n\), so \(v_n\) does not tend to zero; in fact, it grows without bound. A necessary condition for convergence of a series is that its terms tend to zero. That condition fails, so \(\sum_{n=1}^{\infty}v_n\) diverges. The ratio calculation confirms that the proposed scale has the correct constant as well as the correct power.
Worked Example: The harmonic boundary
Classify the series with terms
The numerator has degree \(2\) and leading coefficient \(1\); the denominator has degree \(3\) and leading coefficient \(2\). Their degree difference is \(-1\), so the predicted equivalent is \(1/(2n)\). Compute the ratio:
Hence \(w_n\sim 1/(2n)\). The harmonic series \(\sum 1/n\) diverges, and multiplying its terms by the positive constant \(1/2\) does not change that conclusion. Limit comparison therefore shows that \(\sum_{n=1}^{\infty}w_n\) diverges. This is the boundary case: the denominator exceeds the numerator degree by exactly one, producing a harmonic rather than a convergent higher reciprocal power.
A common degree-counting mistake
A frequent error is to use the numerator’s degree alone as the exponent, or to subtract the degrees in the wrong order. For example, for \((n^2+1)/(n^5+2)\), one might incorrectly say that the term behaves like \(1/n^2\) because the numerator has degree \(2\). That overlooks the denominator’s degree.
The correct difference is \(2-5=-3\), so the candidate equivalent is \(1/n^3\). Checking the ratio makes the correction explicit:
By contrast, dividing the original term by the mistaken model \(1/n^2\) gives
not \(1\). So \(1/n^2\) is not an equivalent. The numerator degree is not the answer by itself: subtract the denominator degree from it.
Why the method matters
Rational terms can look complicated because several lower-order powers appear in both polynomials. The normalization argument shows why those lower-order terms do not determine the long-term scale: after dividing by the leading powers, each becomes a term tending to zero. The degree difference and leading-coefficient ratio capture exactly the information needed for this comparison.
For a positive-term series, the practical sequence is therefore short: identify both degrees, subtract denominator degree from numerator degree, form the leading-coefficient ratio, and compare with the corresponding power series. In particular, when the numerator degree is smaller, the size of the degree gap matters: a gap of two or more gives a convergent power-series model, while a gap of one gives the divergent harmonic model.
Check Your Understanding
For each question, identify the degree difference before deciding what comparison applies.
- What power of \(n\) is the leading scale of \((4n^3+n)/(2n^7+1)\)? Does the associated positive-term series converge?
- For \((3n^4+2)/(6n^3+n)\), what is the leading scale, and what necessary condition for series convergence is relevant?
- Why does a numerator degree two less than the denominator degree give a convergent power-series comparison?
- What limit should the ratio of a rational term to its proposed leading equivalent have?
- For a numerator degree one less than the denominator degree, which standard series determines the convergence behavior?