Tutorials › Calculus II › Radicals of Polynomial Expressions

Limit Comparison and Equivalents · Tutorial 631 of 998

Radicals of Polynomial Expressions

Extract the dominant power from a polynomial radical and compare the resulting series terms with a suitable power series.

Intermediate 9 min read

What You'll Learn

  • Factor the highest power of the index from each polynomial inside a radical.
  • Determine the exponent and leading constant in the resulting power-law model.
  • Check when a radical expression is positive and suitable for limit comparison.
  • Use a finite positive limit to transfer convergence or divergence from a power series.
  • Recognize and correct degree-counting errors when simplifying radicals.

Why Radicals Have Power-Law Models

A radical can make the growth of a sequence look less obvious than it is. For example, a term involving the square root of a quotient of polynomials may appear complicated, even though its size for large \(n\) is controlled by only two pieces of information: the degrees and leading coefficients of those polynomials.

Finding that simpler model is useful when testing a series. Once a positive term is shown to behave like a constant multiple of \(n^{-p}\), the familiar power-series threshold tells us whether the series converges or diverges.

Definition: Two positive sequences \(a_n\) and \(b_n\) are asymptotically equivalent, written \(a_n\sim b_n\), if \(\lim_{n\to\infty}a_n/b_n=1\). In this tutorial, a power-law model for a radical sequence is a sequence \(c n^\alpha\), where \(c>0\), such that the radical sequence is asymptotically equivalent to that model.

Extracting the Leading Powers

Let \(P\) and \(Q\) be polynomials of degrees \(m\) and \(k\), with positive leading coefficients \(a\) and \(b\). Their leading terms determine their large-\(n\) behavior: \(P(n)/n^m\) tends to \(a\), and \(Q(n)/n^k\) tends to \(b\). If both polynomials are positive for all sufficiently large \(n\), then their quotient is also eventually positive.

For an integer \(r\geq 2\), consider the positive \(r\)th root of that quotient. Factoring the leading powers gives a power-law model. The exponent is the difference of the polynomial degrees divided by \(r\); the constant is the \(r\)th root of the ratio of leading coefficients.

$$ \sqrt[r]{\frac{P(n)}{Q(n)}} \sim \left(\frac{a}{b}\right)^{1/r} n^{(m-k)/r}. $$

The formula applies when \(P(n)/Q(n)\) is positive for all sufficiently large \(n\), and the radical denotes the positive real \(r\)th root. In particular, it applies to square roots of polynomial ratios that are eventually positive.

Why the Formula Works

Write each polynomial as its leading power times a factor that tends to its leading coefficient. Taking the quotient separates the power of \(n\) from a remaining factor whose limit is the ratio of the leading coefficients. Taking the \(r\)th root then gives the stated constant and divides the power exponent by \(r\).

$$ P(n)=n^m\left(a+\frac{a_{m-1}}{n}+\cdots+\frac{a_0}{n^m}\right), \qquad Q(n)=n^k\left(b+\frac{b_{k-1}}{n}+\cdots+\frac{b_0}{n^k}\right). $$

Here the coefficients \(a_j\) and \(b_j\) are those of the polynomials, with \(a\) and \(b\) their leading coefficients. As \(n\) tends to infinity, every term involving a negative power of \(n\) tends to zero. Thus the parenthetical factors tend to \(a\) and \(b\), respectively.

$$ \sqrt[r]{\frac{P(n)}{Q(n)}} = n^{(m-k)/r} \sqrt[r]{\frac{a+\frac{a_{m-1}}{n}+\cdots+\frac{a_0}{n^m}} {b+\frac{b_{k-1}}{n}+\cdots+\frac{b_0}{n^k}}} \sim \left(\frac{a}{b}\right)^{1/r}n^{(m-k)/r}. $$

The final equivalence follows because the quotient inside the root tends to \(a/b>0\), and the positive \(r\)th-root function is continuous at that value. This also shows why the positivity condition matters: the real radical must be defined for the terms under consideration.

From a Power Model to a Series Test

For positive terms \(u_n\) and \(v_n\), the limit-comparison test says that if \(\lim_{n\to\infty}u_n/v_n=L\) with \(0<L<\infty\), then \(\sum u_n\) and \(\sum v_n\) either both converge or both diverge. The reason is that, eventually, the ratio is bounded above and below by positive constants, so the terms are bounded by constant multiples of one another.

The power series \(\sum 1/n^p\) converges when \(p>1\) and diverges when \(p\leq 1\). Therefore, if a positive radical term is equivalent to \(c n^{-p}\) for some \(c>0\), its series has the same behavior as \(\sum 1/n^p\). The constant \(c\) does not change the convergence result.

Practical rule: For a positive polynomial quotient inside an \(r\)th root, subtract the denominator degree from the numerator degree, divide by \(r\), and take the \(r\)th root of the leading-coefficient ratio. Then compare the resulting power with \(n^{-p}\).

Worked Examples

Worked Example: A Square Root with a Convergent Power Model

Determine whether the series with terms \(u_n=\sqrt{(3n^4+6n^2+3)/(5n^7+5n^4+10n^2)}\) converges.

The numerator has degree \(4\) and leading coefficient \(3\); the denominator has degree \(7\) and leading coefficient \(5\). The expected power is \(n^{(4-7)/2}=n^{-3/2}\), and the leading constant is \(\sqrt{3/5}\). To verify the model directly, divide the numerator and denominator inside the radical by their leading powers:

$$ u_n = \sqrt{\frac{3n^4\left(1+\frac{2}{n^2}+\frac{1}{n^4}\right)} {5n^7\left(1+\frac{1}{n^3}+\frac{2}{n^5}\right)}} = \sqrt{\frac{3}{5}}\,n^{-3/2} \sqrt{\frac{1+\frac{2}{n^2}+\frac{1}{n^4}} {1+\frac{1}{n^3}+\frac{2}{n^5}}}. $$

The last radical tends to \(\sqrt{1/1}=1\), since each negative power of \(n\) tends to zero. Consequently,

$$ \lim_{n\to\infty}\frac{u_n}{n^{-3/2}} = \sqrt{\frac{3}{5}}. $$

This limit is finite and positive. Limit comparison with \(\sum 1/n^{3/2}\), which converges because \(3/2>1\), proves that \(\sum u_n\) converges. As a check, the extracted constant and exponent give exactly the ratio limit: \((\sqrt{3/5}\,n^{-3/2})/n^{-3/2}=\sqrt{3/5}\).

Worked Example: A Square Root with the Harmonic Power

Determine whether the series with terms \(v_n=\sqrt{(2n^4+2n^2)/(8n^6+4n^3)}\) converges.

The numerator has degree \(4\), and the denominator has degree \(6\). The degree difference divided by \(2\) is \((4-6)/2=-1\), while the leading-coefficient ratio inside the root is \(2/8=1/4\). Thus the model is \(\tfrac12 n^{-1}\). Factoring those leading terms gives

$$ v_n = \sqrt{\frac{2n^4\left(1+\frac{1}{n^2}\right)} {8n^6\left(1+\frac{1}{2n^3}\right)}} = \frac{1}{2n} \sqrt{\frac{1+\frac{1}{n^2}}{1+\frac{1}{2n^3}}}. $$

The remaining radical tends to \(1\), so the comparison ratio is

$$ \lim_{n\to\infty}\frac{v_n}{1/n} = \lim_{n\to\infty} \frac12\sqrt{\frac{1+\frac{1}{n^2}}{1+\frac{1}{2n^3}}} = \frac12. $$

The terms are positive, and the limit is finite and positive. The harmonic series \(\sum 1/n\) diverges, so limit comparison proves that \(\sum v_n\) diverges. As a check on the constant, the ratio of the leading coefficients is \(2/8=1/4\), whose positive square root is \(1/2\), matching the limit.

Worked Example: A Cube Root with a Convergent Power Model

Determine whether the series with terms \(w_n=\sqrt[3]{(8n^2+4n)/(n^6+3n^4+1)}\) converges.

The polynomial degrees are \(2\) and \(6\), so the power is \(n^{(2-6)/3}=n^{-4/3}\). The leading-coefficient ratio is \(8/1\), whose cube root is \(2\). Factoring the leading powers makes the comparison explicit:

$$ w_n = \sqrt[3]{\frac{8n^2\left(1+\frac{1}{2n}\right)} {n^6\left(1+\frac{3}{n^2}+\frac{1}{n^6}\right)}} = 2n^{-4/3} \sqrt[3]{\frac{1+\frac{1}{2n}} {1+\frac{3}{n^2}+\frac{1}{n^6}}}. $$

The remaining cube root tends to \(1\), hence

$$ \lim_{n\to\infty}\frac{w_n}{n^{-4/3}} = 2\lim_{n\to\infty} \sqrt[3]{\frac{1+\frac{1}{2n}} {1+\frac{3}{n^2}+\frac{1}{n^6}}} = 2. $$

The limit is finite and positive, so \(\sum w_n\) has the same behavior as \(\sum 1/n^{4/3}\). Since \(4/3>1\), that power series converges, and therefore \(\sum w_n\) converges. The check on the leading constant is \(\sqrt[3]{8/1}=2\), as obtained by factoring.

A Common Degree-Counting Mistake

A frequent error is to subtract the polynomial degrees correctly but forget that the exponent is then divided by the root index. In the first example, the quotient has a leading power \(n^{4-7}=n^{-3}\). That is the power inside the square root, not the power of the full term.

$$ \text{Incorrect: }\sqrt{n^{4-7}}=n^{-3}. \qquad \text{Correct: }\sqrt{n^{4-7}}=n^{-3/2}. $$

The correction matters: \(n^{-3}\) and \(n^{-3/2}\) are different comparison models. A reliable check is to write the factored expression explicitly and apply the radical to both the leading constant and the power of \(n\).

When the Method Applies

Before using this method, check that the radical terms are real and positive for all sufficiently large \(n\). For an even root, the expression inside the radical must be nonnegative; for positive-term limit comparison, the terms must in fact be positive eventually. Also check that the denominator polynomial is nonzero for all sufficiently large \(n\). A denominator with a nonzero leading coefficient has no zeros for sufficiently large positive \(n\), because its leading term determines its eventual sign.

Once these conditions hold, lower-degree terms do not determine the power model: after division by the leading power, their contributions tend to zero. The degrees determine the exponent, and the leading coefficients determine the positive constant. The resulting power model then provides a direct route to a convergence or divergence conclusion.

Check Your Understanding

For each question, identify the degree-based power before deciding what the comparison shows.

  1. For \(\sqrt{(5n^6+n)/(2n^{10}+1)}\), what power of \(n\) is the model, and what is its leading constant?
  2. If a positive radical term is equivalent to \(7n^{-5/4}\), does its series converge or diverge? State the comparison series.
  3. For \(\sqrt[3]{(27n^2+1)/(8n^5+3)}\), what are the exponent and leading constant in its power model?
  4. Why must the limit of the ratio to the comparison term be finite and positive to use the ordinary limit-comparison conclusion?
  5. In a square root of a polynomial quotient, what is the error in using the degree difference itself as the final exponent?