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Limit Comparison and Equivalents · Tutorial 637 of 998

Cancellation Pitfalls in Asymptotic Comparison

Learn to detect cancellation, retain the terms that determine a difference, and use comparison tests only when their positivity and limit conditions are satisfied.

Intermediate 9 min read

What You'll Learn

  • Explain why equivalent sequences can have a difference much smaller than either sequence.
  • Distinguish positive comparison terms from signed expressions formed by subtraction.
  • Use exact subtraction and common denominators to expose the surviving terms.
  • Check whether a difference is eventually positive before applying a positive-series test.
  • Identify when asymptotic error terms are small enough to justify an estimate for a difference.

Why Subtraction Needs Special Care

When positive sequences are added, their largest terms usually determine the scale of the sum: two positive quantities cannot cancel each other. Subtraction is different. Two sequences can each be close to the same large leading term, while their difference is controlled by much smaller terms.

This matters in series problems. A comparison based on leading terms may correctly describe each of two positive sequences separately, yet give no useful information about the difference between them. Before comparing a difference with a familiar positive sequence, we must expose the terms that actually survive the subtraction and check that the resulting series terms have the required sign.

Key takeaway: Asymptotic equivalence is generally preserved by addition of positive quantities, but not by subtraction. If leading terms cancel, the smaller terms and the errors in the approximations become decisive.

Equivalence Does Not Pass Automatically Through Subtraction

For sequences \(u_n\) and \(v_n\), with \(v_n\ne 0\) for all sufficiently large \(n\), the notation \(u_n\sim v_n\) means that \(u_n/v_n\to 1\) as \(n\to\infty\). This says that \(u_n\) and \(v_n\) have the same leading scale and relative size. It does not say that their difference is small in any particular way.

For example, if \(a_n\sim n^2\) and \(b_n\sim n^2\), both sequences have leading term \(n^2\). Subtracting those leading terms gives zero, but the actual difference \(a_n-b_n\) may grow, approach a nonzero constant, tend to zero, or change sign. The separate equivalences do not decide which outcome occurs.

The issue can be stated in terms of errors. Write two approximations as \(a_n=A_n+e_n\) and \(b_n=B_n+f_n\). Then their exact difference is

$$ a_n-b_n=(A_n-B_n)+(e_n-f_n). $$

If \(A_n\) and \(B_n\) are large and close, \(A_n-B_n\) may be much smaller than either one. Knowing only that \(e_n\) is small relative to \(A_n\) and \(f_n\) is small relative to \(B_n\) may not control the difference: those errors could be larger than the surviving term.

Worked Example: Leading Terms Cancel but the Difference Grows

Worked Example: Leading Terms Cancel but the Difference Grows

Let \(A_n=n^2+3n+1\) and \(C_n=n^2+n\). Find the exact difference, then compare it with what the leading terms alone suggest.

Both sequences are asymptotic to \(n^2\), since

$$ \frac{A_n}{n^2}=1+\frac{3}{n}+\frac{1}{n^2}\longrightarrow 1, \qquad \frac{C_n}{n^2}=1+\frac{1}{n}\longrightarrow 1. $$

Subtract the full expressions, not just their leading terms:

$$ A_n-C_n =(n^2+3n+1)-(n^2+n) =2n+1. $$

The leading \(n^2\) terms cancel, but the difference is \(2n+1\), which grows without bound. Check at \(n=1\): \(A_1=1+3+1=5\) and \(C_1=1+1=2\), so \(A_1-C_1=3\), agreeing with \(2(1)+1=3\). Substituting \(2n+1\) into the subtraction also recovers \(A_n-C_n\) exactly. Thus the two sequences are both equivalent to \(n^2\), while their difference is equivalent to \(2n\), not to zero.

A Precise Condition for Estimating a Difference

The subtraction becomes manageable when the errors are small compared with the surviving scale, not merely compared with the original large terms. For sequences \(e_n\) and \(r_n\), write \(e_n=o(r_n)\) when \(r_n\ne0\) eventually and \(e_n/r_n\to0\).

Theorem: Suppose \(a_n=A_n+e_n\), \(b_n=B_n+f_n\), and \((A_n-B_n)/r_n\to 1\), where \(r_n\ne0\) eventually. If \(e_n=o(r_n)\) and \(f_n=o(r_n)\), then \(a_n-b_n\sim r_n\).

Proof. Divide the exact subtraction identity by \(r_n\):

$$ \frac{a_n-b_n}{r_n} = \frac{A_n-B_n}{r_n} +\frac{e_n}{r_n} -\frac{f_n}{r_n}. $$

The first term on the right tends to \(1\), and each of the last two terms tends to \(0\). Therefore \((a_n-b_n)/r_n\to1\), which is precisely \(a_n-b_n\sim r_n\). The condition is stronger than requiring \(e_n=o(A_n)\) and \(f_n=o(B_n)\): when \(r_n\) is much smaller than \(A_n\) and \(B_n\), the errors must be controlled on the smaller scale \(r_n\).

Worked Example: A Positive Difference on a Much Smaller Scale

Worked Example: A Positive Difference on a Much Smaller Scale

For \(n\in\mathbb N\), let \(a_n=n^2+n+1/n\) and \(b_n=n^2+n\). Determine the difference and decide whether the series of differences converges.

First, each sequence is equivalent to \(n^2\):

$$ \frac{a_n}{n^2} =1+\frac{1}{n}+\frac{1}{n^3}\longrightarrow1, \qquad \frac{b_n}{n^2} =1+\frac{1}{n}\longrightarrow1. $$

But exact subtraction gives

$$ a_n-b_n =(n^2+n+1/n)-(n^2+n) =\frac{1}{n}. $$

The difference is positive for every \(n\), and its series is the harmonic series, which diverges. The substitution check is immediate: \(b_n+1/n=n^2+n+1/n=a_n\). At \(n=1\), the terms are \(a_1=3\) and \(b_1=2\), and their difference is \(1=1/1\).

If we replace both sequences by their leading equivalent \(n^2\) before subtracting, we get \(n^2-n^2=0\). That replacement loses the entire actual term \(1/n\), and therefore cannot support a conclusion about convergence.

Positive Terms and Signed Expressions Are Not Interchangeable

Limit comparison for series is used with terms that are positive eventually. In its standard finite-positive-limit form, if \(p_n>0\) and \(q_n>0\) eventually, and \(p_n/q_n\to L\) for some \(0<L<\infty\), then \(\sum p_n\) and \(\sum q_n\) either both converge or both diverge. Positivity ensures that the terms can be compared in the required direction.

A difference \(a_n-b_n\) is not automatically positive just because \(a_n\) and \(b_n\) are each positive. It may be positive, negative, or change sign. Even when it is positive, its scale must be determined from the subtraction itself. First establish the sign eventually; then, if the terms are positive, use a comparison result whose hypotheses hold. A signed expression cannot simply be treated as a positive-series term.

Worked Example: Use a Common Denominator Before Comparing

Worked Example: Use a Common Denominator Before Comparing

For \(n\in\mathbb N\), consider \(x_n=(n+1)/n\) and \(y_n=n/(n+1)\). Find an equivalent for the positive difference \(x_n-y_n\), and determine the behavior of its series.

Each sequence tends to \(1\), so subtracting their leading constants would give \(1-1=0\), which does not estimate the difference. Instead, use a common denominator:

$$ x_n-y_n =\frac{(n+1)^2-n^2}{n(n+1)} =\frac{n^2+2n+1-n^2}{n(n+1)} =\frac{2n+1}{n(n+1)}. $$

The numerator and denominator are positive for every \(n\), so the difference is positive. To identify its scale, compare it with \(2/n\):

$$ \frac{x_n-y_n}{2/n} = \frac{2n+1}{n(n+1)}\cdot\frac{n}{2} = \frac{2n+1}{2(n+1)} \longrightarrow 1. $$

Thus \(x_n-y_n\sim 2/n\), and the series of differences diverges by limit comparison with the harmonic series. Check at \(n=1\): \(x_1-y_1=2-1/2=3/2\), while the exact expression gives \((2+1)/(1\cdot2)=3/2\). The common-denominator rewrite preserves the small residual that the separate limits of \(x_n\) and \(y_n\) conceal.

A Reliable Procedure

When a term is built by subtracting two quantities that may have the same leading behavior, use a deliberate sequence of checks.

1
Identify the cancellation.
Find which leading terms are equal or nearly equal. Do not assume their difference has the scale of either original quantity.
2
Rewrite exactly.
Subtract polynomial terms directly, or combine fractions over a common denominator. Keep the lower-order terms that remain.
3
Determine sign and scale.
Check whether the resulting expression is positive eventually, then compare it with a candidate scale using a ratio limit.
4
Apply a series test only after checking its conditions.
For positive-term comparison, verify eventual positivity and a finite, strictly positive comparison limit where required.

Common Mistake: Subtracting Equivalent Replacements

A tempting but invalid step is to replace each sequence by its leading equivalent and then subtract those replacements. For the sequences in the first example, that calculation would be

$$ A_n\sim n^2,\qquad C_n\sim n^2 \quad\Longrightarrow\quad A_n-C_n\text{ is treated as }n^2-n^2=0. $$

The conclusion does not follow: the exact subtraction is \(A_n-C_n=2n+1\). The mistake is treating two separate approximations as though their errors remain negligible after cancellation. They need not. The correction is to subtract the original expressions, or to prove that the approximation errors are negligible relative to the surviving difference.

Another useful warning is that a limit of zero for a ratio does not by itself identify a difference after cancellation. The numerator must first be the actual difference, simplified exactly enough to reveal its scale. A ratio test or comparison limit applied to an expression that has already discarded the residual cannot repair the lost information.

Why the Distinction Matters

Asymptotic estimates are powerful because they replace complicated terms with simpler models. Their validity depends on what operation is being performed. For positive sums, a large positive contribution cannot be erased by another positive contribution. In a difference, however, shared leading behavior disappears, and terms that seemed secondary may become the whole answer.

The safe habit is therefore not to avoid asymptotic comparison, but to use it at the right stage: first simplify the difference without losing its residual; then establish its sign and leading scale; finally compare the resulting terms under the hypotheses of the series test. That order prevents a plausible leading-term calculation from producing a false convergence conclusion.

Check Your Understanding

For each question, pay attention to the exact subtraction and to the sign of the resulting terms.

  1. If \(u_n\sim n^3\) and \(v_n\sim n^3\), what information about \(u_n-v_n\) follows from these equivalences alone?
  2. For \(u_n=n^2+4n\) and \(v_n=n^2+n\), find the exact difference and its leading scale.
  3. Why is positivity of both \(a_n\) and \(b_n\) insufficient to guarantee positivity of \(a_n-b_n\)?
  4. In the difference \(n^2+n+1/n-(n^2+n)\), what scale is lost by replacing each parenthesized sequence with its leading equivalent?
  5. State the positivity and limit conditions needed to use finite-positive-limit comparison for two series.

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