Equal-Degree Polynomial Ratios and Limit Constants

Why the Limit Constant Matters

When two polynomials have the same degree, their values may grow without bound, but their ratio can approach a finite number. That number is determined by the leading coefficients. In series problems, the same calculation can reveal whether a complicated positive term behaves like a familiar comparison term.

For example, a ratio of two quadratic polynomials can be paired with a power such as $1/n^2$. If the polynomial ratio approaches a positive constant, the resulting sequence can be compared with $1/n^2$ using limit comparison. The constant need not be $1$: any finite, strictly positive limit gives the same convergence behavior.

Degree and Leading Coefficient

A nonzero polynomial $P(n)=a_dn^d+a_{d-1}n^{d-1}+\cdots+a_1n+a_0$, with $a_d\ne0$, has degree $d$ and leading coefficient $a_d$. The leading term is $a_dn^d$. It determines the polynomial’s highest-power growth as $n$ becomes large.

Theorem: Let $P$ and $Q$ be polynomials of the same degree $d$, with leading coefficients $a_d\ne0$ and $b_d\ne0$, respectively. Then $Q(n)\ne0$ for all sufficiently large $n$, and $$ \lim_{n\to\infty}\frac{P(n)}{Q(n)}=\frac{a_d}{b_d}. $$ If $a_d/b_d>0$, this limit is finite and strictly positive.

The theorem concerns the ratio as $n$ tends to infinity. It does not say that the ratio equals the quotient of the leading coefficients at each particular $n$. The lower-degree terms still affect the values at finite $n$; their relative effect tends to zero.

Why the Leading Coefficients Determine the Limit

Write both polynomials with all their terms shown:

P(n)=a_dn^d+a_{d-1}n^{d-1}+\cdots+a_0,\qquad Q(n)=b_dn^d+b_{d-1}n^{d-1}+\cdots+b_0.

For $n\ne0$, divide the numerator and denominator by $n^d$. The ratio becomes a quotient in which every lower-degree term has a factor $1/n$, $1/n^2$, or a higher reciprocal power. Each such factor tends to zero. The denominator therefore tends to $b_d$, which is nonzero, so the quotient tends to $a_d/b_d$.

Write the leading terms.
Record the common degree $d$ and the leading coefficients $a_d$ and $b_d$.

Divide by the common highest power.
Divide the numerator and denominator by $n^d$.

Take the limit.
All lower-degree contributions tend to zero, leaving the quotient of the leading coefficients.

Worked Examples

Worked Example: A Ratio of Quartic Polynomials

Find the limit of the following ratio:

\lim_{n\to\infty}\frac{6n^4-3n^2+1}{2n^4+7n-5}.

Both polynomials have degree $4$. Their leading coefficients are $6$ and $2$. Divide the numerator and denominator by $n^4$:

\frac{6n^4-3n^2+1}{2n^4+7n-5} = \frac{6-\frac{3}{n^2}+\frac{1}{n^4}}{2+\frac{7}{n^3}-\frac{5}{n^4}}.

As $n\to\infty$, the terms $3/n^2$, $1/n^4$, $7/n^3$, and $5/n^4$ all tend to zero. The denominator tends to $2\ne0$, so

\lim_{n\to\infty}\frac{6n^4-3n^2+1}{2n^4+7n-5} = \frac{6-0+0}{2+0-0} = \frac{6}{2} = 3.

Check: the numerator’s leading term divided by the denominator’s leading term is $6n^4/(2n^4)=3$. This agrees with the limit obtained by dividing every term by $n^4$.

Worked Example: Comparing a Series with a Convergent Power Series

Determine the convergence of the series whose terms are

u_n=\frac{6n^2+2n+1}{3n^2+4}\cdot\frac{1}{n^{3/2}},\qquad n\ge1.

Use $v_n=1/n^{3/2}$ as the comparison sequence. Both $u_n$ and $v_n$ are positive for $n\ge1$, since each polynomial in the ratio is positive there. Their ratio is

\frac{u_n}{v_n} = \frac{\left(\frac{6n^2+2n+1}{3n^2+4}\right)\frac{1}{n^{3/2}}}{\frac{1}{n^{3/2}}} = \frac{6n^2+2n+1}{3n^2+4}.

The two polynomials have degree $2$, with leading coefficients $6$ and $3$. Dividing through by $n^2$ gives

\frac{u_n}{v_n} = \frac{6+\frac{2}{n}+\frac{1}{n^2}}{3+\frac{4}{n^2}} \longrightarrow \frac{6}{3} = 2.

The limit $2$ is finite and strictly positive. The limit comparison test therefore gives the same behavior for $\sum u_n$ and $\sum v_n$. The series $\sum 1/n^{3/2}$ converges by the $p$-series test, since $3/2>1$. Hence $\sum u_n$ converges.

Check: substituting the proposed comparison term into the ratio cancels the common factor $1/n^{3/2}$, leaving precisely the equal-degree polynomial ratio. Its leading-coefficient quotient is $6/3=2$, as computed.

Worked Example: Comparing a Series with the Harmonic Series

Determine the convergence of the series with positive terms

w_n=\frac{7n^3+2n}{2n^3+5n^2+1}\cdot\frac{1}{n},\qquad n\ge1.

Compare with $z_n=1/n$. The numerator and denominator polynomials are positive for $n\ge1$, so both sequences are positive. Their ratio is

\frac{w_n}{z_n} = \frac{\left(\frac{7n^3+2n}{2n^3+5n^2+1}\right)\frac{1}{n}}{\frac{1}{n}} = \frac{7n^3+2n}{2n^3+5n^2+1}.

Both polynomials have degree $3$, and their leading coefficients are $7$ and $2$. Divide by $n^3$:

\frac{w_n}{z_n} = \frac{7+\frac{2}{n^2}}{2+\frac{5}{n}+\frac{1}{n^3}} \longrightarrow \frac{7}{2}.

Because $7/2$ is finite and strictly positive, limit comparison says that $\sum w_n$ and $\sum 1/n$ have the same behavior. The harmonic series $\sum 1/n$ diverges, so $\sum w_n$ diverges.

Check: the ratio used for comparison simplifies to the displayed cubic-over-cubic expression. Its leading-term quotient is $7n^3/(2n^3)=7/2$, confirming the limit.

Limit Comparison and Its Conditions

For positive sequences $u_n$ and $v_n$, the limit comparison test applies when

\lim_{n\to\infty}\frac{u_n}{v_n}=c \qquad\text{with}\qquad 0<c<\infty.

Under these conditions, the series $\sum u_n$ and $\sum v_n$ either both converge or both diverge. Positivity is important: it allows inequalities between terms to give the corresponding comparison between series. It is sufficient that both sequences be positive for all sufficiently large $n$, because changing finitely many terms does not change convergence.

The reason the positive finite limit gives this conclusion can be seen directly. Since $u_n/v_n\to c$, take an error tolerance of $c/2$. For all sufficiently large $n$, the definition of the limit gives

\left|\frac{u_n}{v_n}-c\right|<\frac{c}{2}.

This implies $c/2<u_n/v_n<3c/2$. Multiplying by $v_n>0$ preserves the inequality:

\frac{c}{2}v_n<u_n<\frac{3c}{2}v_n.

Thus, for all sufficiently large $n$, each sequence is bounded above and below by positive constant multiples of the other. The direct comparison test then ensures that they have the same convergence behavior. This argument also explains why the comparison limit may be $2$, $7/2$, or any other positive finite constant: multiplying terms by such a constant does not change whether a positive-term series converges.

Practical rule: When the ratio of two positive sequences reduces to a quotient of equal-degree polynomials, its limit is the numerator’s leading coefficient divided by the denominator’s leading coefficient. If that quotient is strictly positive, limit comparison is available.

Common Mistake: Using the Wrong Coefficients

A frequent error is to compare constant terms or coefficients of $n$, rather than the coefficients of the highest powers. For instance, consider

\frac{5n^2+2n+1}{2n^2+7n+4}.

A wrong calculation might use the coefficients of $n$ and claim that the limit is $2/7$. That ignores the fact that the $n^2$ terms grow faster than the $n$ terms. Dividing by $n^2$ makes the relevant contributions clear:

\frac{5n^2+2n+1}{2n^2+7n+4} = \frac{5+\frac{2}{n}+\frac{1}{n^2}}{2+\frac{7}{n}+\frac{4}{n^2}} \longrightarrow \frac{5}{2}.

The correction is to identify the common degree first, then use the leading coefficients of that degree. Lower-degree coefficients affect finite values, but after division by the highest power their contributions tend to zero.

Why This Calculation Is Useful

Many series terms are products of a simple power of $n$ and a rational factor involving polynomials. The rational factor may be complicated, but if its numerator and denominator have the same degree, its limit is a constant found from just two coefficients. That constant can identify a useful comparison with a familiar power sequence.

The order of reasoning matters: first establish that the comparison sequences are positive eventually; then compute the ratio limit; finally check that the limit is strictly positive and finite before applying limit comparison. A zero or infinite ratio is outside this version of the test. Such cases require different comparison reasoning, so the leading-coefficient calculation alone should not be treated as a conclusion about series behavior.

Check Your Understanding

Use the equal-degree ratio rule and the stated conditions for limit comparison to answer these questions.

What is the limit of $(8n^3-4n+1)/(2n^3+n^2+6)$ as $n\to\infty$? Show the normalization by $n^3$.
For $u_n=((4n^2+3)/(2n^2+5))\,n^{-2}$, what comparison sequence gives a positive finite ratio limit, and what is that limit?
Why does a ratio limit of $0$ not satisfy the positive finite limit comparison condition stated here?
Find and correct the error in claiming that $(3n^2+8n)/(6n^2+n+4)$ has limit $8$.
If $u_n/v_n\to 5/3$ and both sequences are eventually positive, what can be concluded about the behavior of $\sum u_n$ and $\sum v_n$?