Why Compare Ratios of Terms?
Some positive-term series are difficult to classify by direct comparison: it may be hard to find a useful inequality that holds for every term. But their terms can still have a simple relationship for large \(n\). If the ratio of two positive terms approaches a fixed positive number, then the terms eventually differ only by a bounded multiplicative factor. Their series must therefore share the same convergence behavior.
This is the idea behind the Limit Comparison Test. It is especially useful when the terms are rational expressions: the highest powers of \(n\) often reveal a suitable benchmark, such as a term of a \(p\)-series.
The Limit Comparison Test
The test concerns series with positive terms. It is enough that the terms be positive for all sufficiently large \(n\), because adding or removing finitely many terms does not change whether a series converges.
The constant \(c\) need not equal \(1\). The ratio may approach any finite positive value. The essential feature is that it is neither zero nor infinite.
When the limit is \(1\), the terms are called asymptotically equivalent. We write \(a_n\sim b_n\) to mean that both terms are positive eventually and
Asymptotically equivalent positive terms give series with the same convergence behavior. More generally, if \(a_n/b_n\to c\) for \(c>0\), then \(a_n\) is asymptotically proportional to \(b_n\), and the same conclusion holds.
Why the Test Works
A positive finite limit gives bounds above and below by constant multiples of the comparison terms. Those bounds let us use direct comparison in both directions.
Choose \(\varepsilon=c/2\), which is positive because \(c>0\). For all sufficiently large \(n\), the limit definition gives \(\left|a_n/b_n-c\right|<c/2\).
The inequality implies \(c/2<a_n/b_n<3c/2\). Since \(b_n>0\), multiplying by \(b_n\) preserves the inequalities.
For all sufficiently large \(n\), \((c/2)b_n<a_n<(3c/2)b_n\). If \(\sum b_n\) converges, the upper bound shows \(\sum a_n\) converges. If \(\sum b_n\) diverges, the lower bound shows \(\sum a_n\) diverges.
Thus each series has the same behavior as the other. The proof also explains why positivity and a positive finite limit matter: they give usable two-sided bounds.
Choosing a Benchmark for Rational Terms
For a rational expression involving polynomials in \(n\), the highest power in the numerator and denominator usually determines the order of the term. For example, if the numerator has degree \(r\) and the denominator has degree \(s\), with nonzero leading coefficients, then the expression is often comparable to a constant times \(n^{r-s}\). To select a benchmark, match this power with \(1/n^p\), then calculate the ratio limit to verify the choice.
Do not stop at noticing that the powers match. The leading coefficients determine the actual limit, and the test requires that the limit be positive and finite. Once the benchmark is established, use its known convergence behavior. In particular, the \(p\)-series \(\sum 1/n^p\) converges for \(p>1\) and diverges for \(p\leq 1\).
Worked Example: A Rational Series That Converges
Determine whether
converges or diverges. The terms are positive for every \(n\geq1\), since both \(3n^2+1\) and \(n^4+2n\) are positive. The highest powers suggest comparison with \(1/n^2\). Set \(a_n=(3n^2+1)/(n^4+2n)\) and \(b_n=1/n^2\). Then
As \(n\to\infty\), \(1/n^2\to0\) and \(2/n^3\to0\), so
This is a finite positive limit, so the Limit Comparison Test applies. The benchmark \(\sum 1/n^2\) is a \(p\)-series with \(p=2>1\), hence it converges. Therefore the given series converges. Check: the computed ratio tends to \(3\), not \(0\) or infinity, so the test has exactly the required ratio behavior.
Worked Example: A Rational Series That Diverges
Determine the behavior of
The numerator and denominator are positive for \(n\geq1\). Their highest powers suggest comparison with \(1/n\). Let \(a_n=(5n+2)/(n^2+3n+1)\) and \(b_n=1/n\). Calculate the ratio:
Taking the limit gives
The limit is positive and finite, so the two series have the same behavior. The benchmark \(\sum 1/n\) is the harmonic series, which diverges. Hence the given series diverges. Check: the ratio limit is \(5\), so the given terms are asymptotically proportional to the harmonic terms, consistent with divergence.
Asymptotic Equivalence in Practice
Sometimes the ratio limit is \(1\), so no constant factor remains in the comparison. This is asymptotic equivalence. For example, a square root or a lower-order term may make a formula look more complicated without changing its leading behavior. The ratio calculation confirms whether the simpler benchmark is valid.
Worked Example: Equivalent Terms with a Convergent Benchmark
Determine whether
converges or diverges. Both numerator and denominator are positive for \(n\geq1\). Since \(\sqrt{n^2+1}\) grows like \(n\) and the denominator grows like \(n^3\), use \(b_n=1/n^2\). With \(a_n=\sqrt{n^2+1}/(n^3+2)\), the ratio is
Here we divided the numerator inside the square root by \(n^2\), and divided the denominator by \(n^3\). Therefore,
Thus \(a_n\sim b_n\). The benchmark \(\sum 1/n^2\) converges because it is a \(p\)-series with \(p=2\). The Limit Comparison Test shows that the given series converges as well. Check: the ratio tends to \(1\), confirming that the square root and added constant do not change the benchmark's convergence behavior.
Common Pitfall: A Finite Ratio Does Not Mean Convergence
A finite positive ratio limit does not, by itself, say whether either series converges. It says only that the two series have the same behavior. You must still know whether the benchmark converges or diverges.
For a concrete example, let \(a_n=n/(n^2+1)\) and \(b_n=1/n\). Then
A mistaken conclusion would be: “The ratio tends to \(1\), so \(\sum a_n\) converges.” That does not follow. Since \(\sum b_n=\sum 1/n\) diverges, the Limit Comparison Test instead shows that \(\sum a_n\) also diverges. The ratio identifies a relationship; the known benchmark supplies the convergence or divergence conclusion.
A Reliable Procedure
When using the test, follow the reasoning in this order:
- Confirm that both sequences of terms are positive for all sufficiently large \(n\).
- Choose a simpler benchmark whose series has known behavior.
- Calculate \(a_n/b_n\) and evaluate its limit carefully.
- Check that the limit is finite and strictly positive.
- Use the benchmark's convergence or divergence to classify the original series.
If the ratio limit is zero or infinite, the theorem stated here does not apply; those cases require a different comparison result. Also, a ratio that fails to have a limit does not permit a conclusion from this test. The correct response is to choose another benchmark or another method, rather than assuming the test has succeeded.
Check Your Understanding
For each question, identify the benchmark and explain what the ratio limit lets you conclude.
- For \(a_n=(4n^2+3)/(2n^4+n)\), which \(p\)-series is a natural benchmark? What is the ratio limit?
- If \(a_n>0\), \(b_n>0\), and \(a_n/b_n\to 7\), while \(\sum b_n\) converges, what can be concluded about \(\sum a_n\)?
- If \(a_n/b_n\to 1\) and \(\sum b_n\) diverges, what is the behavior of \(\sum a_n\)?
- Why does knowing only that \(a_n/b_n\to 2\) not determine whether either series converges?
- What condition on the ratio limit is required for the Limit Comparison Test stated here?
