Limit Comparison for Boundary Coefficients

Why Compare Boundary Terms by Their Ratios?

At an endpoint of a power series, substituting the endpoint can leave a numerical series whose terms are more complicated than the familiar benchmarks \(1/n^p\) or \(1/[n(\log n)^q]\). Direct comparison can be awkward: the terms may not be visibly larger or smaller than a benchmark for every \(n\), even when their long-run sizes are very close.

Limit comparison focuses on that long-run size. If the ratio of two positive terms approaches a fixed positive number, then eventually each term is bounded above and below by a positive constant times the other. Those bounds force the two series to have the same convergence behavior.

Key idea: To classify a positive endpoint series, compare its terms with a known series whose terms capture their asymptotic size. A finite, strictly positive ratio limit transfers the benchmark’s convergence or divergence.

Definition and Test

Two positive sequences \(a_n\) and \(b_n\) are asymptotically equivalent, written \(a_n\sim b_n\), if their ratio tends to \(1\). More generally, limit comparison applies when the ratio tends to any finite positive constant.

Definition: Let \(a_n>0\) and \(b_n>0\) for all sufficiently large \(n\). If

\lim_{n\to\infty}\frac{a_n}{b_n}=L \qquad\text{with}\qquad 0<L<\infty,

then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. Changing or omitting finitely many terms does not affect convergence.

The positivity and the condition \(0<L<\infty\) matter. The test does not say that every pair of series with a ratio limit has the same behavior: a zero or infinite limit does not provide the two-sided bounds used in the result.

Why the Test Works

Here is the comparison argument behind the test. Suppose \(a_n,b_n>0\) eventually and \(a_n/b_n\to L\), where \(0<L<\infty\). By the definition of a limit, there is an index \(N\) such that for every \(n\ge N\),

\left|\frac{a_n}{b_n}-L\right|<\frac{L}{2}.

Rearranging this inequality gives

\frac{L}{2}<\frac{a_n}{b_n}<\frac{3L}{2}, \qquad\text{so}\qquad \frac{L}{2}b_n<a_n<\frac{3L}{2}b_n \quad(n\ge N).

If \(\sum b_n\) converges, the upper bound shows that the tail of \(\sum a_n\) is bounded by a constant multiple of a convergent positive series, so \(\sum a_n\) converges by direct comparison. If \(\sum b_n\) diverges, the lower bound shows that \(\sum a_n\) cannot converge: otherwise the inequality \(b_n<(2/L)a_n\) would force \(\sum b_n\) to converge by direct comparison. Thus the two series have the same verdict.

Choose a benchmark.
Match the main powers of \(n\), or of \(\log n\), in the boundary terms.

Compute the ratio.
Simplify \(a_n/b_n\) and find its limit as \(n\to\infty\).

Check the hypothesis.
The terms must be positive eventually, and the limit must be finite and strictly positive.

Transfer the verdict.
Use the known convergence behavior of the benchmark series.

Power-Series Benchmarks

The standard \(p\)-series \(\sum_{n=1}^{\infty}1/n^p\) converges when \(p>1\) and diverges when \(p\le 1\). Limit comparison therefore classifies any eventually positive terms with a positive finite ratio limit against \(1/n^p\). If \(a_n\sim C/n^p\) for a constant \(C>0\), then \(a_n/(1/n^p)\to C\), so \(\sum a_n\) has the same verdict as the \(p\)-series.

Worked Example: A Boundary Series with Harmonic Size

Consider the positive endpoint series

\sum_{n=1}^{\infty}\frac{4n+1}{n^2+3}.

Set \(a_n=(4n+1)/(n^2+3)\) and choose the harmonic benchmark \(b_n=1/n\). The ratio is

\frac{a_n}{b_n} = \frac{4n+1}{n^2+3}\cdot n = \frac{4n^2+n}{n^2+3}.

Divide the numerator and denominator by \(n^2\):

\frac{4n^2+n}{n^2+3} = \frac{4+1/n}{1+3/n^2} \longrightarrow 4.

The limit is finite and positive, and both sequences are positive. The benchmark \(\sum 1/n\) diverges, so limit comparison proves that the displayed endpoint series diverges as well. As a check, the terms have leading size \(4/n\), consistent with the ratio limit \(4\); the test gives the divergence verdict without requiring a term-by-term inequality for every \(n\).

Worked Example: A Boundary Series Comparable to a Convergent \(p\)-Series

Now consider

\sum_{n=1}^{\infty}\frac{2n^3+7n}{n^5+3n^2}.

The highest powers suggest comparison with \(1/n^2\). For \(a_n=(2n^3+7n)/(n^5+3n^2)\) and \(b_n=1/n^2\),

\frac{a_n}{b_n} = \frac{2n^3+7n}{n^5+3n^2}\cdot n^2 = \frac{2n^5+7n^3}{n^5+3n^2} = \frac{2+7/n^2}{1+3/n^3} \longrightarrow 2.

The ratio limit is positive and finite. Since the benchmark \(\sum 1/n^2\) converges, the given series converges. The calculation also checks the predicted scale: the dominant terms in the original numerator and denominator have degrees \(3\) and \(5\), giving a ratio of order \(1/n^2\).

Logarithmic Benchmarks

For logarithmic boundary terms, start the comparison at \(n=2\), so \(\log n>0\). The standard series \(\sum_{n=2}^{\infty}1/[n(\log n)^q]\) converges for \(q>1\) and diverges for \(q\le 1\). This classification is the benchmark used here; the integral-test treatment of logarithmic boundary series is a separate method.

In particular, if positive terms satisfy \(a_n\sim C/[n(\log n)^q]\) for \(C>0\), then their ratio to the benchmark tends to \(C\). The terms therefore have the same convergence behavior as that logarithmic series.

Worked Example: A Logarithmic Correction That Still Diverges

For \(n\ge2\), consider

\sum_{n=2}^{\infty} \frac{\log n+3}{n\big((\log n)^2+4\big)}.

The numerator has leading size \(\log n\), while the logarithmic factor in the denominator has leading size \((\log n)^2\). Compare with \(b_n=1/[n\log n]\). The ratio is

\frac{a_n}{b_n} = \frac{\log n+3}{n\big((\log n)^2+4\big)} \cdot n\log n = \frac{(\log n)^2+3\log n}{(\log n)^2+4}.

Dividing by \((\log n)^2\) gives

\frac{1+3/\log n}{1+4/(\log n)^2} \longrightarrow 1.

The benchmark has \(q=1\), so it diverges. The ratio limit is positive and finite, hence the given series diverges. This conclusion is consistent with the leading size \(1/[n\log n]\): the added constants do not change that asymptotic scale.

Worked Example: A Stronger Logarithmic Power Gives Convergence

For \(n\ge2\), let

\sum_{n=2}^{\infty} \frac{1+2/\log n}{n(\log n)^2}.

Compare with \(b_n=1/[n(\log n)^2]\). The ratio is exactly

\frac{a_n}{b_n} = \frac{1+2/\log n}{n(\log n)^2} \cdot n(\log n)^2 = 1+\frac{2}{\log n} \longrightarrow 1.

The comparison series has \(q=2>1\), so it converges. Since the ratio limit is \(1\) and the terms are positive, the given series converges too. The check is direct: the extra factor \(1+2/\log n\) tends to \(1\), so it does not change the benchmark’s asymptotic size.

Applying the Test at Power-Series Endpoints

At a boundary point of a power series, substitute the endpoint first. The resulting terms may include signs. Limit comparison as stated here is a test for positive terms, so for a signed endpoint series it is normally applied to the absolute values. If \(|a_n|\) has a finite positive ratio limit against a positive benchmark, that determines whether the endpoint series converges absolutely. It does not, by itself, decide whether a non-absolutely-convergent signed series converges.

For example, if substitution at an endpoint produces terms \((-1)^n c_n\), with \(c_n>0\), comparing \(c_n\) with \(1/n^p\) classifies absolute convergence. If the comparison shows \(\sum c_n\) diverges, this only rules out absolute convergence; a separate argument is needed to decide ordinary convergence. In particular, do not remove signs and then report the absolute-series verdict as the verdict for the signed series.

A Common Mistake: Using a Zero Ratio Limit

A common error is to assume that any ratio limit transfers convergence behavior. It does not. For example, take \(a_n=1/n^2\) and \(b_n=1/n\). Then

\frac{a_n}{b_n} = \frac{1/n^2}{1/n} = \frac{1}{n} \longrightarrow 0.

But \(\sum a_n=\sum 1/n^2\) converges, whereas \(\sum b_n=\sum 1/n\) diverges. Thus a zero ratio limit does not show that the series behave alike.

Correction: Use the limit comparison test only when the limit of the ratio is strictly positive and finite. If the limit is zero or infinite, this test gives no matching-convergence conclusion; another comparison or test is needed.

Practical Checklist

When classifying boundary coefficients by limit comparison, keep the argument focused:

Write the endpoint series after substituting the boundary value.
Check whether its terms are positive; if not, state clearly that you are comparing absolute values.
Choose \(1/n^p\) or \(1/[n(\log n)^q]\) to match the terms’ leading size.
Calculate the ratio and verify that its limit is strictly between zero and infinity.
Use the benchmark’s known verdict, and distinguish absolute convergence from convergence of a signed series.

The main benefit is that exact inequalities between complicated formulas are often unnecessary. A carefully chosen benchmark and a valid ratio limit reduce the endpoint question to a standard series.

Check Your Understanding

For each question, identify the appropriate benchmark and state what the limit comparison test allows you to conclude.

If \(a_n>0\) and \(a_n/(1/n^{3/2})\to 5\), does \(\sum a_n\) converge or diverge?
If \(a_n>0\) and \(a_n/[1/(n\log n)]\to 2\), what is the verdict for \(\sum_{n=2}^{\infty}a_n\)?
Why is a ratio limit of zero insufficient to conclude that two positive series have the same behavior?
If an endpoint series has terms \((-1)^n c_n\), where \(c_n>0\) is comparable to \(1/n^2\), what does limit comparison establish directly?
For terms asymptotic to \(1/[n(\log n)^{3/2}]\), which logarithmic benchmark verdict applies?