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.

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.

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 \(n\ge N\),

Rearranging this inequality gives

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.

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.

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.

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

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.

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.