Why Compare the Ratio of Terms?
Direct comparison asks whether one sequence of positive terms is eventually above or below another. Sometimes finding a useful inequality is difficult, even when the two sequences appear to have the same overall size. In that situation, the ratio of their terms can give a simpler route.
If the ratio approaches a positive finite number, then far enough along the sequences, each term is bounded above and below by a positive constant multiple of the other. Those bounds are precisely what is needed to transfer convergence or divergence between the two series.
Statement of the Test
Let \(a_n\) and \(b_n\) be positive terms for every \(n\) sufficiently large. The finite initial terms may be ignored: adding or removing finitely many finite terms does not affect whether a series converges.
The theorem does not say that the two series have the same sum. It says only that their convergence behavior agrees. The positivity of \(L\) matters: it allows the ratio to be bounded both above and away from zero.
For reference, the direct comparison rule used below is this: if \(0\leq u_n\leq v_n\) eventually and \(\sum v_n\) converges, then \(\sum u_n\) converges; if \(u_n\geq v_n\geq0\) eventually and \(\sum v_n\) diverges, then \(\sum u_n\) diverges.
Why the Test Works
Because the ratio tends to \(L>0\), its terms eventually lie in a fixed interval around \(L\). Choose the interval from \(L/2\) to \(3L/2\). The limit definition guarantees that this interval contains all sufficiently late ratios. Multiplying by the positive term \(b_n\) gives two comparison inequalities.
Since \(a_n/b_n\to L\), there is an index \(N\) such that for every \(n\geq N\), \(\left|a_n/b_n-L\right|<L/2\).
The inequality implies \(L/2<a_n/b_n<3L/2\). Since \(b_n>0\), multiplying preserves the inequalities.
Use the upper bound to transfer convergence from \(\sum b_n\) to \(\sum a_n\), and the lower bound to transfer divergence from \(\sum b_n\) to \(\sum a_n\).
Here is the complete argument. For \(n\geq N\), the limit definition gives
Therefore
Multiplication by \(b_n>0\) yields
If \(\sum b_n\) converges, then \(\sum a_n\) converges by the upper bound \(a_n<(3L/2)b_n\) and direct comparison. If \(\sum b_n\) diverges, then \(\sum a_n\) diverges by the lower bound \(a_n>(L/2)b_n\): a positive constant multiple of a divergent positive-term series also diverges, so direct comparison applies. The finitely many terms before \(N\) do not change either conclusion. This proves the test.
Applying Limit Comparison
A practical application has four parts: check that both term sequences are positive eventually, choose a comparison series whose behavior is known, compute \(a_n/b_n\), and verify that its limit is finite and strictly positive. Only after these checks should the known convergence result be transferred.
Worked Example: A Harmonic-Type Series
Determine the behavior of
Set \(a_n=(3n+1)/(n(n+1))\) and compare with the harmonic terms \(b_n=1/n\). For every positive integer \(n\), both terms are positive. Compute the ratio:
Since \(2/(n+1)\to0\),
The limit is finite and positive. The harmonic series \(\sum 1/n\) diverges, so the limit comparison test implies that the given series diverges as well.
Check: The terms are positive, the ratio was formed as \(a_n/b_n\), and its limit is \(3\), not \(0\) or infinity. Thus the hypotheses apply and the classification is divergence.
Worked Example: Comparison with a Convergent p-Series
Determine the behavior of
Let \(a_n=(2n^2+1)/(n^2(n^2+4))\) and choose \(b_n=1/n^2\). Both sequences are positive for every positive integer \(n\). Their ratio is
As \(n\to\infty\), the last fraction \(7/(n^2+4)\) tends to zero. Hence
This is a finite positive limit. The comparison series \(\sum 1/n^2\) is a convergent p-series, since its exponent is \(2>1\). Therefore the given series converges by the limit comparison test.
Check: The ratio simplifies exactly to \(2-7/(n^2+4)\), which tends to \(2\). The reference series converges, so the conclusion is convergence.
Worked Example: Comparison with a Geometric Series
Determine the behavior of
Let \(a_n=(2/5)^n(1+1/(n+1))\), and compare it with \(b_n=(2/5)^n\). Both are positive for every positive integer \(n\). The ratio is
Because \(1/(n+1)\to0\),
The reference series is geometric with common ratio \(2/5\). Since \(\lvert 2/5\rvert<1\), it converges. The limit comparison test therefore shows that the given series converges.
Check: The shared factor \((2/5)^n\) cancels in the ratio, leaving a quantity that tends to \(1\). The comparison series has ratio \(2/5\) and converges, so the classification is convergence.
Common Pitfalls
Mistake: treating the limit as a ratio of series sums. For example, compare \(a_n=2/n\) with \(b_n=1/n\). The term ratio is exactly \(2\), but this does not mean that the two series have equal sums. In fact, both series diverge. The test transfers convergence or divergence; it does not equate the values of the series.
Correction: State only what the theorem guarantees. A finite positive ratio limit establishes the same convergence behavior, not equality of terms, partial sums, or series sums.
Another error is to apply the result without checking its conditions. The comparison terms must be positive eventually, and the ratio limit must be a finite number strictly greater than zero. The theorem stated here does not cover a ratio limit of zero or an infinite ratio limit.
Choosing and Checking a Comparison
The most useful comparison is usually a familiar series whose behavior is already known, such as a p-series or a geometric series. The algebra should be arranged to make the ratio limit visible, but the conclusion depends on all the hypotheses, not just on a plausible-looking simplification.
| What to check | Why it matters |
|---|---|
| Are \(a_n\) and \(b_n\) positive eventually? | The comparison inequalities preserve their direction when multiplied by \(b_n\). |
| Is the ratio \(a_n/b_n\) computed correctly? | Reversing the ratio can change the limiting value and lead to an invalid conclusion. |
| Is the limit finite and strictly positive? | That condition produces both an eventual upper bound and an eventual positive lower bound. |
| Is the behavior of \(\sum b_n\) known? | The test transfers that convergence or divergence classification to \(\sum a_n\). |
Limit comparison is valuable when exact termwise inequalities are awkward, but the relative size of the terms is easy to evaluate. The proof explains why this works: a ratio tending to \(L>0\) traps the terms between fixed positive multiples of one another. Those bounds make the two series behave alike in the only sense the test claims: both converge, or both diverge.
Check Your Understanding
For each question, identify the hypotheses or comparison fact needed before drawing a conclusion.
- State the required condition on the limit of \(a_n/b_n\) in the finite-positive-limit test.
- If \(a_n/b_n\to 4\) and \(\sum b_n\) converges, what can be concluded about \(\sum a_n\)?
- If \(a_n/b_n\to 1/3\) and \(\sum b_n\) diverges, what can be concluded about \(\sum a_n\)?
- Why does the proof require \(b_n>0\) when multiplying the ratio bounds?
- If the term ratio tends to \(2\), does that imply the two series have equal sums? Explain briefly.