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Limit Comparison and Equivalents · Tutorial 626 of 998

Asymptotic Equivalence and Series Behavior

Learn to recognize when positive sequence terms are asymptotically equivalent and use that relationship to determine a series’ convergence behavior.

Intermediate 9 min read

What You'll Learn

  • Define asymptotic equivalence for positive sequences using a ratio limit.
  • Prove why equivalent positive terms give series with the same convergence behavior.
  • Replace complicated terms by simpler equivalents when testing positive series.
  • Distinguish asymptotic equivalence from equality and from a ratio limit other than one.
  • Identify why positive terms and the treatment of finitely many initial terms matter.

Why Compare Terms Asymptotically?

Some positive series have terms whose exact formulas are awkward, even though their long-term size resembles that of a familiar sequence. For instance, a term may contain a radical or a rational expression with several powers of \(n\). If its ratio to a simple comparison term approaches \(1\), the complicated expression can be replaced by that simpler term when deciding whether the series converges.

This replacement is about convergence behavior, not equality of terms or equality of sums. We will define the relationship precisely and prove why it preserves convergence for series with positive terms.

Definition of Asymptotic Equivalence

Let \((a_n)\) and \((b_n)\) be sequences that are positive for all sufficiently large \(n\). Their ratio is then defined for all sufficiently large \(n\).

Definition: The sequences \((a_n)\) and \((b_n)\) are asymptotically equivalent if \( \lim_{n\to\infty}\frac{a_n}{b_n}=1 \). We write \(a_n\sim b_n\). This notation describes the behavior as \(n\) tends to infinity; it does not assert that \(a_n=b_n\) for any particular \(n\).

The limit says that the ratio of the two terms becomes arbitrarily close to \(1\). Equivalently, for any chosen relative tolerance \(\varepsilon>0\), the terms eventually satisfy \(1-\varepsilon<a_n/b_n<1+\varepsilon\). In particular, choosing a fixed tolerance gives useful inequalities that compare the terms directly.

The Series Equivalence Theorem

Theorem: Suppose \(a_n>0\) and \(b_n>0\) for all sufficiently large \(n\), and \(a_n\sim b_n\). Then the positive-term series \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. Any finite number of initial terms does not affect this conclusion.

The theorem applies to positive terms because the ratio limit yields inequalities between the terms, and the comparison test transfers convergence through such inequalities. Positivity also ensures that the partial sums increase, so a divergent positive-term series diverges to \(+\infty\).

Why the Theorem Works

Here is a complete proof using the definition of the limit. Since \(a_n/b_n\to1\), take the tolerance \(\varepsilon=1/2\). There is an index \(N\) such that whenever \(n\ge N\),

$$ \left|\frac{a_n}{b_n}-1\right|<\frac12. $$

Removing the absolute value gives two inequalities:

$$ -\frac12<\frac{a_n}{b_n}-1<\frac12, \qquad\text{so}\qquad \frac12<\frac{a_n}{b_n}<\frac32. $$

Because \(b_n>0\), multiplying throughout by \(b_n\) preserves the inequality directions:

$$ \frac12 b_n<a_n<\frac32 b_n \qquad(n\ge N). $$

If \(\sum b_n\) converges, then the upper bound \(a_n<\frac32 b_n\) shows that the tail of \(\sum a_n\) converges by direct comparison. Adding the finite initial sum does not change convergence, so \(\sum a_n\) converges.

For the converse, rearrange the lower bound to obtain \(b_n<2a_n\) for \(n\ge N\). If \(\sum a_n\) converges, direct comparison shows that \(\sum b_n\) converges as well. Thus either both series converge or neither does. Since all terms are positive, if they do not converge, their partial sums are unbounded and tend to \(+\infty\). This proves the theorem.

The constants \(1/2\) and \(3/2\) are convenient choices, not special features of the result. Any two fixed positive bounds surrounding \(1\) would give the same comparison argument.

Applying the Theorem

Worked Example: A Rational Term Compared with the Harmonic Series

Determine the convergence behavior of

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

All terms are positive for \(n\ge1\). Choose the harmonic term \(b_n=1/n\). The ratio of the given term to this comparison term is

$$ \frac{\dfrac{2n^2+3n}{2n^3+n}}{\dfrac1n} =\frac{n(2n^2+3n)}{2n^3+n} =\frac{2n^3+3n^2}{2n^3+n} =\frac{2n^2+3n}{2n^2+1}. $$

Dividing the numerator and denominator of the last fraction by \(n^2\) gives

$$ \frac{2+3/n}{2+1/n^2}\longrightarrow\frac22=1. $$

Thus the given term is asymptotically equivalent to \(1/n\). The harmonic series \(\sum 1/n\) diverges, so the theorem shows that the given series diverges. As a check, the computed ratio is exactly the ratio of the original term to \(1/n\), and its limit is \(1\), as required by the definition.

Worked Example: A Radical Term Compared with a Constant

Consider the positive-term series

$$ \sum_{n=1}^{\infty}\bigl(\sqrt{n^2+n}-n\bigr). $$

The terms are positive because \(n^2+n>n^2\) for \(n\ge1\). Compare them with \(b_n=1/2\). Rationalizing the difference gives

$$ \sqrt{n^2+n}-n =\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)} {\sqrt{n^2+n}+n} =\frac{n}{\sqrt{n^2+n}+n}. $$

Therefore the ratio to \(1/2\) is

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

The terms are asymptotically equivalent to \(1/2\). The series \(\sum 1/2\) diverges because its partial sums are \(N/2\) after \(N\) terms and hence grow without bound. The theorem now shows that the radical series diverges too. The ratio calculation verifies the equivalence directly, and comparison with the constant series verifies the convergence conclusion.

Worked Example: A Rational Term Compared with a \(p\)-Series

Determine whether

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

converges. Its terms are positive for every \(n\ge1\). Compare them with \(b_n=1/n^2\). The ratio is

$$ \frac{\dfrac{n+2}{n^3+1}}{\dfrac1{n^2}} =\frac{n^2(n+2)}{n^3+1} =\frac{n^3+2n^2}{n^3+1} =\frac{1+2/n}{1+1/n^3} \longrightarrow 1. $$

Hence the given terms are asymptotically equivalent to \(1/n^2\). The \(p\)-series \(\sum 1/n^p\) converges when \(p>1\); here \(p=2\), so the comparison series converges. The theorem implies that the given series converges. Substituting the original terms into the ratio produces the displayed expression, whose numerator and denominator both tend to \(1\); this checks the equivalence used.

What Asymptotic Equivalence Does—and Does Not—Say

The theorem allows a simpler sequence to stand in for a complicated one when the question is whether a positive-term series converges. It does not say that the terms are equal, that their partial sums are equal, or that convergent series have the same sum. For example, a finite number of terms can be changed without altering convergence, while those changes can alter the value of a convergent sum.

It is also important that the ratio limit be \(1\) for the notation \(a_n\sim b_n\). A finite positive ratio limit other than \(1\) can still be useful in a comparison argument, but it is not asymptotic equivalence under this definition. The key conclusion here is specifically that a ratio tending to \(1\) supplies eventual upper and lower bounds by fixed positive multiples.

A Common Mistake: Confusing Equivalence with a Vanishing Difference

A frequent error is to replace the ratio condition with the claim that the difference \(a_n-b_n\) tends to zero. That is not the definition. For example, let \(a_n=n+\sqrt n\) and \(b_n=n\). Then

$$ \frac{a_n}{b_n} =\frac{n+\sqrt n}{n} =1+\frac1{\sqrt n} \longrightarrow 1, $$

so \(a_n\sim b_n\). But the difference is

$$ a_n-b_n=\sqrt n\longrightarrow+\infty, $$

not zero. The correction is to test the ratio \(a_n/b_n\), not the absolute difference. Asymptotic equivalence describes relative size: the difference can be large in absolute terms while remaining small compared with the terms themselves.

Another useful check is the direction of an inequality in the proof. From \(\frac12<a_n/b_n<\frac32\), positivity of \(b_n\) is what allows multiplication without reversing either inequality. If the terms are not eventually positive, this proof does not apply as written. The theorem is a result about positive-term series, not a general assertion about all series.

Practical Use

When a positive term has a complicated expression, a useful procedure is to identify a simpler positive comparison term, compute the ratio, and evaluate its limit. If the limit is \(1\), the two positive-term series have the same convergence behavior. The comparison series must still be one whose behavior is known; equivalence transfers the conclusion but does not determine it on its own.

1
Check positivity.
Confirm that both sequences are positive for all sufficiently large indices.
2
Compute the ratio.
Form \(a_n/b_n\), keeping the original terms visible until the algebra is verified.
3
Evaluate its limit.
If the limit is \(1\), the terms are asymptotically equivalent.
4
Use the comparison series.
Apply the equivalence theorem to transfer convergence or divergence, and remember that finitely many initial terms do not affect the result.

The central idea is concise: for positive terms, a ratio approaching \(1\) eventually traps each sequence between fixed positive multiples of the other. Those inequalities are precisely what make the convergence behavior agree.

Check Your Understanding

Use the definition and theorem to answer these questions.

  1. What limit must \(a_n/b_n\) have for \(a_n\sim b_n\)?
  2. Why does \(a_n\sim b_n\) give inequalities comparing the terms for all sufficiently large \(n\)?
  3. If \(a_n\sim b_n\) and \(\sum b_n\) converges, what can be concluded about \(\sum a_n\), assuming both sequences are eventually positive?
  4. For \(a_n=n+\sqrt n\) and \(b_n=n\), does the difference tend to zero? Does the ratio tend to one?
  5. Does asymptotic equivalence assert that two convergent series have the same sum? Explain briefly.

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