Why Use Limit Comparison?
A series can be difficult to classify by direct comparison when its terms do not fit neatly into a simple inequality. The limit comparison test offers another route: compare the terms’ relative sizes by taking a ratio. If that ratio approaches a finite positive number, the two terms eventually differ only by a constant factor, so their series have the same convergence behavior.
The test is useful only when its conditions are met. In particular, the terms being compared must be positive, at least eventually, and the ratio must have a finite positive limit. A ratio that approaches zero or grows without bound does not allow the same conclusion.
“Positive for all sufficiently large \(n\)” means that there is some index after which every term is positive. A finite number of initial terms does not affect whether a series converges or diverges, so positivity of every term from the first index is not essential. In many AP problems, however, it is straightforward to verify positivity for every index in the series.
Choose a Benchmark First
The benchmark series should have a convergence classification that is already known. For AP Calculus problems, common choices are geometric series and p-series. A p-series has the form \(\sum 1/n^p\); it converges when \(p>1\) and diverges when \(p\leq 1\). A geometric series with ratio \(r\) converges when \(|r|<1\) and diverges when \(|r|\geq 1\).
A practical way to choose a benchmark is to look at the dominant powers of \(n\). For a rational expression, the highest powers in the numerator and denominator often indicate whether the terms behave like \(1/n^p\). For other expressions, algebra such as rationalizing a radical can reveal the size of a term. The comparison is not complete until the ratio limit and the benchmark’s classification are both stated.
Confirm that the terms in both series are positive for all sufficiently large indices.
Choose a geometric series or p-series whose terms have a similar size.
Evaluate \(\lim_{n\to\infty}a_n/b_n\) and check that its value is finite and strictly positive.
State whether the benchmark converges or diverges, then apply the limit comparison test to reach the same conclusion for the given series.
Why a Finite Positive Limit Works
The reason for the test is that a ratio close to a positive number \(L\) bounds one term above and below by constant multiples of the other. Those bounds are enough to transfer convergence or divergence.
To justify the result, choose the positive number \(\varepsilon=L/2\). By the definition of the limit, there is an index \(N\) such that for every \(n\geq N\),
This inequality implies
Since \(b_n>0\), multiplying through by \(b_n\) preserves the inequalities:
If \(\sum b_n\) converges, then \(\sum (3L/2)b_n\) converges. The upper inequality and the direct comparison test imply that \(\sum a_n\) converges. If \(\sum b_n\) diverges, then \(\sum (L/2)b_n\) diverges. The lower inequality implies that \(\sum a_n\) diverges. These two implications establish that the series have the same classification. The direct comparison facts used here follow from partial sums: sums of nonnegative terms are increasing, and an increasing sequence of partial sums converges exactly when it is bounded above.
Worked Examples
Worked Example: A Rational Series Compared with a p-Series
Determine whether \[ \sum_{n=1}^{\infty}\frac{3n+2}{n^3+1} \] converges or diverges.
The numerator and denominator are positive for \(n\geq 1\), so the terms are positive. Their leading powers suggest comparison with the p-series \(\sum 1/n^2\). Let \[ a_n=\frac{3n+2}{n^3+1}, \qquad b_n=\frac{1}{n^2}. \]
Compute the ratio:
As \(n\to\infty\), \(2/n\to 0\) and \(1/n^3\to 0\), so
The limit \(3\) is finite and positive. The benchmark \(\sum 1/n^2\) is a p-series with \(p=2>1\), so it converges. By the limit comparison test, the given series converges. As a check, the computed ratio tends to \(3\), not zero or infinity, and both compared sequences are positive.
Worked Example: A Rational Series Compared with the Harmonic Series
Determine whether \[ \sum_{n=1}^{\infty}\frac{2n^2+1}{n^3-4n+7} \] converges or diverges.
The numerator is positive for \(n\geq 1\). For the denominator, its values at \(n=1\) and \(n=2\) are \(4\) and \(7\). For \(n\geq 2\), \(n^2-4\geq 0\), so \[ n^3-4n+7=n(n^2-4)+7\geq 7. \] Thus both terms are positive. The leading powers suggest the harmonic benchmark \(b_n=1/n\). Set \(a_n=(2n^2+1)/(n^3-4n+7)\).
The ratio is
Taking the limit gives
The limit is finite and positive. The benchmark is the harmonic series \(\sum 1/n\), which is the p-series with \(p=1\) and diverges. The limit comparison test therefore shows that the given series diverges. The denominator positivity check and the ratio limit verify the required conditions.
Worked Example: Rationalizing Before Comparing
Determine whether \[ \sum_{n=1}^{\infty}\frac{\sqrt{n^2+4n}-n}{n^2} \] converges or diverges.
Since \(n^2+4n>n^2\) for \(n\geq 1\), the square root is greater than \(n\), so the given terms are positive. Compare with \(b_n=1/n^2\), and let \[ a_n=\frac{\sqrt{n^2+4n}-n}{n^2}. \] Then
Rationalize this difference by multiplying by its conjugate:
Divide numerator and denominator by \(n\):
Therefore,
The ratio limit is finite and positive, and \(\sum 1/n^2\) converges because it is a p-series with \(p=2\). The limit comparison test proves that the given series converges. Rationalizing was useful because it exposed the ratio’s limit directly.
What the Test Does Not Say
The condition \(0<L<\infty\) matters. If the ratio limit is zero or infinity, the limit comparison test in this form gives no conclusion about whether both series share a classification. The series might still be classified by another argument, but the finite-positive-limit conclusion cannot be used.
Worked Example: A Zero Ratio Limit Is Not Enough
Consider \(a_n=1/n^2\) and \(b_n=1/n\). Both sequences have positive terms, and
The series \(\sum a_n=\sum 1/n^2\) converges, while \(\sum b_n=\sum 1/n\) diverges. Thus these series do not have the same classification. This is not a contradiction: the ratio limit is zero, not finite and positive. The calculation verifies why it would be incorrect to apply the test’s conclusion here.
Writing a Complete Justification
A strong AP-style explanation identifies the terms and benchmark, verifies positivity, evaluates the ratio limit, classifies the benchmark, and names the conclusion. Merely saying that two expressions “look similar” does not establish the needed condition.
For a finite positive limit, the core of a concise justification can be written as: “The terms are positive, and the ratio of the given term to the benchmark term approaches \(L\), where \(0<L<\infty\). The benchmark series [converges/diverges] because [state its classification]. Therefore, by the limit comparison test, the given series [converges/diverges].”
The benchmark must actually be a series with a known classification, and the limit must be computed for the ratio of corresponding terms. Reversing the ratio is acceptable when the resulting limit is still finite and positive, but it is clearest to state explicitly which term is in the numerator and which is in the denominator.
Check Your Understanding
For each question, include the conditions and reasoning that justify the conclusion.
- What conditions on \(a_n\), \(b_n\), and \(\lim a_n/b_n\) are required for the limit comparison test?
- For \(\sum (5n+1)/(n^3+2)\), which p-series is a natural benchmark, and what is the ratio limit?
- If the ratio limit is \(4\) and the benchmark series diverges, what follows about the given series?
- Why does a ratio limit of zero not establish that both series converge or both diverge?
- For a series with terms \((\sqrt{n^2+6n}-n)/n^2\), what benchmark could be used, and what does rationalization show about the ratio limit?
