Worked Comparisons for Rational and Radical Terms

Comparing Rational and Radical Terms

A positive-term series may look complicated because each term contains several polynomial expressions, a radical, or both. Its convergence can nevertheless be determined when its terms have the same leading behavior as a familiar power \(1/n^p\). The practical task is to find that behavior carefully, especially when subtraction causes cancellation.

This tutorial develops that approach through worked classifications. In each case, we calculate a comparison ratio, identify its limit, and use the corresponding power-series benchmark to decide convergence or divergence.

Definition: For positive sequences \(a_n\) and \(b_n\), the notation \(a_n\sim b_n\) means \(\lim_{n\to\infty}a_n/b_n=1\). More generally, if \(a_n/b_n\) tends to a positive finite constant, then the two sequences have the same convergence behavior when used as terms of positive-term series.

The Comparison Principle

The benchmark used here is the \(p\)-series \(\sum_{n=1}^{\infty}1/n^p\). It converges when \(p>1\), and diverges when \(0<p\leq 1\). To use this benchmark, compare a given term \(a_n\) with \(b_n=1/n^p\) by computing \(a_n/b_n\). If that ratio tends to a positive finite number, the two series either both converge or both diverge.

Theorem: Suppose \(a_n>0\), \(b_n>0\) for all sufficiently large \(n\), and \(\lim_{n\to\infty}a_n/b_n=c\), where \(0<c<\infty\). Then \(\sum a_n\) and \(\sum b_n\) have the same convergence behavior.

Here is why the theorem works. Since \(a_n/b_n\to c\), eventually the ratio lies between \(c/2\) and \(3c/2\). Multiplying these inequalities by \(b_n>0\) gives

\frac{c}{2}b_n\leq a_n\leq\frac{3c}{2}b_n.

The terms are therefore bounded above and below by positive constant multiples of \(b_n\). The direct comparison test then gives the same convergence outcome for the two series. Changing or omitting finitely many initial terms does not change convergence, so the inequalities need only hold eventually.

A ratio limit of \(1\) says that the terms are equivalent in the precise sense \(a_n\sim b_n\). A ratio limit such as \(3\) is not equivalence to \(b_n\), but it still gives the same series classification. In particular, if a term behaves like \(C/n^p\) for a constant \(C>0\), compare it with \(1/n^p\).

Worked Examples

Worked Example: A Rational Term with a Convergent Benchmark

Classify the series \(\sum_{n=1}^{\infty}(3n^2+1)/(n^4+2n)\). The numerator has leading power \(3n^2\), and the denominator has leading power \(n^4\), suggesting a term of order \(1/n^2\). We verify that suggestion by taking the ratio with \(1/n^2\).

\frac{\dfrac{3n^2+1}{n^4+2n}}{\dfrac{1}{n^2}} = \frac{n^2(3n^2+1)}{n^4+2n} = \frac{3n^4+n^2}{n^4+2n} = \frac{3+\dfrac{1}{n^2}}{1+\dfrac{2}{n^3}}.

As \(n\to\infty\), \(1/n^2\to0\) and \(2/n^3\to0\). Thus the ratio tends to \(3\), a positive finite constant. The comparison theorem applies, and \(\sum 1/n^2\) converges because \(p=2>1\). Therefore the given series converges. As a check, the computed ratio has positive terms and tends to \(3\), not \(0\) or infinity, so it meets exactly the theorem’s required condition.

Worked Example: A Polynomial Radical with a Harmonic Benchmark

Classify \(\sum_{n=1}^{\infty}\sqrt{4n^6+n^2}/(2n^4+5n)\). For positive \(n\), factor the highest power inside the radical: \(\sqrt{4n^6+n^2}=n^3\sqrt{4+1/n^4}\). The denominator can be written \(n^4(2+5/n^3)\), so the term has the scale \(1/n\). Compute its ratio with \(1/n\):

\frac{\dfrac{\sqrt{4n^6+n^2}}{2n^4+5n}}{\dfrac{1}{n}} = \frac{n\sqrt{4n^6+n^2}}{2n^4+5n} = \frac{n^4\sqrt{4+\dfrac{1}{n^4}}}{n^4\left(2+\dfrac{5}{n^3}\right)} = \frac{\sqrt{4+\dfrac{1}{n^4}}}{2+\dfrac{5}{n^3}} \longrightarrow \frac{2}{2}=1.

The ratio is positive and tends to \(1\), so the term is equivalent to \(1/n\). The harmonic series \(\sum 1/n\) diverges, since its benchmark exponent is \(p=1\). The given series therefore diverges. The result is consistent with the leading terms: the radical contributes approximately \(2n^3\), while the denominator contributes approximately \(2n^4\), leaving approximately \(1/n\).

Worked Example: Rationalizing a Close Radical

Classify \(\sum_{n=1}^{\infty}(\sqrt{n^2+4n}-n)/n\). The numerator is a difference of two quantities that are both close to \(n\), so simply replacing the radical by its leading term would lose the information needed to classify the series. To compare with \(1/n\), divide the term by \(1/n\). The resulting ratio is the numerator \(\sqrt{n^2+4n}-n\). Rationalize it:

\frac{\dfrac{\sqrt{n^2+4n}-n}{n}}{\dfrac{1}{n}} = \sqrt{n^2+4n}-n = \frac{(\sqrt{n^2+4n}-n)(\sqrt{n^2+4n}+n)} {\sqrt{n^2+4n}+n} = \frac{4n}{\sqrt{n^2+4n}+n} = \frac{4}{\sqrt{1+\dfrac{4}{n}}+1}.

As \(n\to\infty\), the last expression tends to \(4/(1+1)=2\). The terms are positive, and this is a positive finite ratio limit. The comparison series is \(\sum 1/n\), which diverges. Hence the given series diverges. The ratio calculation also checks the size of the numerator: it tends to \(2\), so the full term has leading behavior \(2/n\).

Worked Example: A Radical with a Convergent Noninteger-Power Benchmark

Classify \(\sum_{n=1}^{\infty}\sqrt{4n^5+3n^2}/(n^4+2n)\). For positive \(n\), \(\sqrt{4n^5+3n^2}=n^{5/2}\sqrt{4+3/n^3}\), while the denominator is \(n^4(1+2/n^3)\). These factors suggest \(1/n^{3/2}\). Check by taking the ratio with that benchmark:

\frac{\dfrac{\sqrt{4n^5+3n^2}}{n^4+2n}}{\dfrac{1}{n^{3/2}}} = \frac{n^{3/2}\sqrt{4n^5+3n^2}}{n^4+2n} = \frac{n^4\sqrt{4+\dfrac{3}{n^3}}}{n^4\left(1+\dfrac{2}{n^3}\right)} = \frac{\sqrt{4+\dfrac{3}{n^3}}}{1+\dfrac{2}{n^3}} \longrightarrow 2.

The ratio limit is positive and finite, so the series has the same behavior as \(\sum 1/n^{3/2}\). This is a \(p\)-series with \(p=3/2>1\), so it converges. Checking the scale directly gives the same result: the radical contributes approximately \(2n^{5/2}\), and dividing by \(n^4\) gives approximately \(2/n^{3/2}\).

A Reliable Workflow

These examples follow a small number of steps. Making each step explicit helps prevent algebraic errors and makes clear which theorem justifies the final decision.

Check positivity.
Confirm that the terms and the proposed benchmark are positive for all sufficiently large \(n\). The limit-comparison theorem used here concerns positive terms.

Find the candidate power.
Compare the highest powers in polynomial numerators and denominators. For a radical, factor the highest power inside it before taking the square root.

Compute the exact ratio.
Divide the original term by \(1/n^p\), simplify, and only then take the limit. This confirms whether the guessed power is correct.

Apply the benchmark test.
If the ratio limit is positive and finite, classify the original series as the \(p\)-series is classified: convergence for \(p>1\), divergence for \(0<p\leq1\).

A Common Cancellation Pitfall

A leading-term approximation is useful for sums, products, and quotients when it preserves the relevant scale. It can fail when two leading terms are subtracted. For instance, \(\sqrt{n^2+4n}\sim n\) does not allow us to infer a useful approximation for \(\sqrt{n^2+4n}-n\) by subtracting \(n\) from both sides of that asymptotic statement.

A tempting but incorrect argument is: “The radical is approximately \(n\), so the difference is approximately \(n-n=0\); therefore the terms are much smaller than \(1/n\), and the series should converge.” This does not follow. Knowing that two quantities are each close to \(n\) does not determine the size of their difference. In this example, rationalization gives the exact identity

\sqrt{n^2+4n}-n = \frac{4}{\sqrt{1+4/n}+1} \longrightarrow 2.

Consequently, after division by \(n\), the term behaves like \(2/n\), and the series diverges. The correction is to simplify a close difference exactly—often by multiplying by the conjugate—before selecting a benchmark.

Why These Comparisons Matter

Rational and radical terms are common forms of series whose appearance can obscure a simple underlying power. Once the terms are expressed relative to a benchmark, the classification depends on the exponent, not on the original complexity of the formula. The ratio limit is also a useful check: it verifies the proposed scale and confirms that the comparison theorem applies.

The most important distinction is between a valid leading-order calculation and a guess based only on appearance. Factoring powers out of radicals is valid for identifying their scale; subtracting two leading approximations may not be. In every case, finish by checking the ratio and confirming that its limit is positive and finite before using the power-series classification.

Check Your Understanding

For each question, identify a suitable power-series benchmark before deciding whether the series converges.

Classify \(\sum_{n=1}^{\infty}(5n^3+2)/(n^5+7n)\). What is the limit of its ratio to \(1/n^2\)?
For \(\sum_{n=1}^{\infty}\sqrt{9n^4+n}/(3n^3+1)\), which \(p\)-series provides the benchmark, and does the series converge?
Why does \(\sqrt{n^2+6n}\sim n\) not by itself determine the leading behavior of \(\sqrt{n^2+6n}-n\)?
What conditions on the limit of \(a_n/b_n\) allow the limit-comparison theorem to transfer a convergence classification?
Classify \(\sum_{n=1}^{\infty}(\sqrt{n^2+2n}-n)/n\) by rationalizing the comparison ratio with \(1/n\).