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Comparison Tests
for Convergence

Learn to determine whether an infinite series converges or diverges by comparing it to a known series: the Direct Comparison Test and the Limit Comparison Test.

AP Calculus BC

Why Compare Series?

Many series don't fit neatly into the formulas for geometric or \(p\)-series. When you can't evaluate a series directly, you can bound it above or below by a simpler series whose behavior you already know.

Essential Benchmark Series

Geometric Series
\(\displaystyle\sum_{n=0}^{\infty} ar^n\) converges when \(|r| < 1\)
\(p\)-Series
\(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}\) converges when \(p > 1\)
Harmonic Series
\(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\) diverges (\(p = 1\))

Visual Intuition: Bounding Series

n \(a_n\) \(b_n\)
converges \(a_n \leq b_n\)
also converges

The Direct Comparison Test

Both series must have non-negative terms. The inequality only needs to hold eventually: from some index \(N\) onward.

DCT Logic at a Glance

\(0 \leq a_n \leq b_n\)
\(\sum b_n\) converges
\(\sum a_n\) converges ✓
\(\sum a_n\) diverges
\(\sum b_n\) diverges ✕

The Limit Comparison Test

This test is often easier than the DCT because you don't need to establish an inequality for every term: you just need the limit of the ratio.

DCT vs LCT: When to Use Which

Direct Comparison (DCT) Limit Comparison (LCT)
What you need An inequality \(a_n \leq b_n\) The limit \(\lim_{n\to\infty}\frac{a_n}{b_n} = L\)
Best when Inequality is obvious Inequality is hard to set up
Typical form Drop terms from denominator Dominant-term extraction
Limitation Wrong direction fails \(L = 0\) or \(L = \infty\) gives partial info

Step-by-Step Strategy

1
Identify the dominant behavior. Look at the highest-power terms in the numerator and denominator for large \(n\).
2
Choose a benchmark series. Simplify to a \(p\)-series or geometric series that behaves similarly.
3
Pick your test. Can you prove \(a_n \leq b_n\) easily? Use DCT. Otherwise, try LCT.
4
Execute and conclude. State the convergence/divergence of your benchmark, then apply the test to reach your conclusion.

Example 1: DCT for Convergence

Does \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 3n}\) converge or diverge?

1
Dominant term: For large \(n\), \(n^2 + 3n \approx n^2\). Compare to \(\frac{1}{n^2}\).
2
Establish inequality: Since \(n^2 + 3n > n^2\) for all \(n \geq 1\), we have \(\frac{1}{n^2 + 3n} < \frac{1}{n^2}\).
3
Benchmark: \(\sum \frac{1}{n^2}\) is a \(p\)-series with \(p = 2 > 1\), so it converges.
4
Conclusion: By DCT, \(\displaystyle\sum \frac{1}{n^2 + 3n}\) converges.

Example 2: LCT for Divergence

Does \(\displaystyle\sum_{n=1}^{\infty} \frac{3n + 1}{n^2 + 5}\) converge or diverge?

1
Dominant behavior: \(\frac{3n+1}{n^2+5} \approx \frac{3n}{n^2} = \frac{3}{n}\). Compare to \(b_n = \frac{1}{n}\).
2
Compute the limit: \(\displaystyle L = \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{(3n+1)/(n^2+5)}{1/n} = \lim_{n\to\infty} \frac{3n^2+n}{n^2+5} = 3\).
3
Result: Since \(L = 3\), a finite positive number, and \(\sum \frac{1}{n}\) diverges (harmonic series)...
4
Conclusion: By LCT, \(\displaystyle\sum \frac{3n+1}{n^2+5}\) diverges.

Example 3: LCT with Roots

Does \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 2}}\) converge or diverge?

1
Dominant behavior: \(\frac{1}{\sqrt{n^3+2}} \approx \frac{1}{n^{3/2}}\). Choose \(b_n = \frac{1}{n^{3/2}}\).
2
Limit: \(\displaystyle L = \lim_{n\to\infty}\frac{1/\sqrt{n^3+2}}{1/n^{3/2}} = \lim_{n\to\infty}\sqrt{\frac{n^3}{n^3+2}} = \lim_{n\to\infty}\sqrt{\frac{1}{1 + 2/n^3}} = 1\).
3
Benchmark: \(\sum \frac{1}{n^{3/2}}\) is a \(p\)-series with \(p = \frac{3}{2} > 1\), so it converges.
4
Conclusion: Since \(L=1\) (finite, positive), by LCT the original series converges.

When the Direction is Wrong

A common frustration: you set up an inequality but it points the wrong way. The DCT cannot conclude anything from these two scenarios:

No conclusion
\(a_n \leq b_n\) and \(\sum b_n\) diverges: says nothing about \(\sum a_n\).
No conclusion
\(a_n \geq b_n\) and \(\sum b_n\) converges: says nothing about \(\sum a_n\).

LCT Edge Cases: \(L = 0\) and \(L = \infty\)

\(L = 0\)
If \(\sum b_n\) converges, then \(\sum a_n\) converges. (But if \(\sum b_n\) diverges, no conclusion.)
\(L = \infty\)
If \(\sum b_n\) diverges, then \(\sum a_n\) diverges. (But if \(\sum b_n\) converges, no conclusion.)

Common Mistakes

✕ Wrong Direction
Showing \(a_n \leq b_n\) where \(\sum b_n\) diverges, then claiming \(\sum a_n\) diverges. Being smaller than a divergent series proves nothing.
✕ Forgetting Positivity
Both tests require positive terms. If \(a_n\) alternates in sign, comparison tests do not apply directly.
✕ Bad Benchmark Choice
Comparing \(\frac{1}{n^2+1}\) to \(\frac{1}{n}\) (divergent). This gives the wrong direction for DCT and a limit of \(L=0\) for LCT: inconclusive.

Summary of Comparison Tests

Direct (DCT) Limit (LCT)
Hypothesis \(0 \leq a_n \leq b_n\) \(a_n, b_n > 0\)
Convergence \(\sum b_n\) conv. \(\Rightarrow \sum a_n\) conv. \(0 < L < \infty\): same fate
Divergence \(\sum a_n\) div. \(\Rightarrow \sum b_n\) div. \(0 < L < \infty\): same fate
Key move Prove inequality algebraically Evaluate a single limit
Pitfall Wrong direction = no info \(L=0\) or \(L=\infty\) = partial info

Question 1 of 5

Which series would be the best benchmark to compare \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + n}\) against?

Question 2 of 5

Using the Limit Comparison Test with \(b_n = \frac{1}{n^2}\), what is \(\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}\) when \(a_n = \frac{5}{2n^2 + 3}\)?

Question 3 of 5

A student claims: "Since \(\frac{1}{n} \leq \frac{1}{\ln n}\) for large \(n\), and \(\sum\frac{1}{n}\) diverges, therefore \(\sum\frac{1}{\ln n}\) diverges by DCT." Is this reasoning valid?

Question 4 of 5

Does \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^3+4}\) converge or diverge?

Question 5 of 5

Which of the following is not a valid conclusion from the Direct Comparison Test?