Comparison Tests for Convergence: Study Guide (Calculus BC 10.6)

BC-only material. The comparison tests are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You compare a new series with one you already understand, so you need a good stock of known series. If any row is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Limits at infinity	1.15	Finding lim aₙ/bₙ by dividing by the highest power
Geometric series	10.2	Benchmarks such as ∑ (3/4)ⁿ
nth term test	10.3	Checking first whether the terms even go to 0
p-series	10.5	Benchmarks such as ∑ 1/n² and ∑ 1/√n

Notation on this page: aₙ is the term of the series you are testing; bₙ is the term of the benchmark series you compare it with. ∑ means the sum from n = 1 to ∞ unless stated.

The idea: compare with a series you know

Consider ∑ 1/(n² + 3). It is not a p-series, and the integral test would need an arctangent. But its terms are close to 1/n², and in fact a little smaller:

1/(n² + 3) < 1/n² for every n ≥ 1.

∑ 1/n² converges (p = 2). A series of positive terms that are each smaller than the terms of a convergent series cannot add up to more than that series does. So ∑ 1/(n² + 3) converges too.

That is the whole idea of this topic. You need two things: a benchmark series whose behaviour you know (usually a p-series or a geometric series), and a link between the two series. The link is either an inequality (direct comparison) or a limit (limit comparison).

Why positivity matters. When every term is positive, the partial sums only go up. Increasing partial sums do one of two things: they stay below some ceiling and settle to a limit, or they grow without bound. A comparison supplies the ceiling, or proves there is none. With negative terms this reasoning breaks down, so both tests require positive terms.

The direct comparison test

Test. Suppose 0 ≤ aₙ ≤ bₙ for every n from some N onwards.

• If ∑ bₙ converges, then ∑ aₙ converges. (Smaller than convergent.)
• If ∑ aₙ diverges, then ∑ bₙ diverges. (Bigger than divergent.)

The inequality only needs to hold “eventually”, because the first few terms never decide convergence (Topic 10.5).

The other two directions are useless. A series that is smaller than a divergent series might converge or diverge. A series that is bigger than a convergent series might converge or diverge. Figure 1 sums this up.

Useful inequality tricks.

• Bounded pieces: −1 ≤ sin n ≤ 1, 0 ≤ cos² n ≤ 1, 0 < arctan n < π/2.
• Making a fraction bigger: make the numerator bigger or the denominator smaller (for example, drop a positive term from the denominator).
• Making a fraction smaller: make the numerator smaller or the denominator bigger.
• ln n > 1 when n ≥ 3, because e ≈ 2.718 < 3.

Worked example 1: direct comparison

Question. Decide whether each series converges. (a) ∑ (2 + sin n)/n² (b) ∑ from n = 2 to ∞ of (ln n)/√n

(a)

1. Check positivity. −1 ≤ sin n ≤ 1, so 1 ≤ 2 + sin n ≤ 3. Every term is positive.
2. Guess. The top stays between 1 and 3; the bottom is n². The terms behave like a constant over n², so expect convergence. Aim to show the terms are smaller than a convergent series.
3. Inequality. (2 + sin n)/n² ≤ 3/n² for every n ≥ 1.
4. Benchmark. ∑ 3/n² is 3 times a p-series with p = 2 > 1, so it converges.
5. Conclusion. 0 < aₙ ≤ 3/n² and ∑ 3/n² converges, so by the direct comparison test, ∑ (2 + sin n)/n² converges.

(b)

1. Guess. ln n grows, but only slowly; the bottom is n^(1/2). The terms are at least as big as 1/√n (from some point), which suggests divergence. Aim to show the terms are bigger than a divergent series.
2. Inequality. For n ≥ 3, ln n ≥ ln 3 > 1, so (ln n)/√n > 1/√n > 0.
3. Benchmark. ∑ 1/√n is a p-series with p = 1/2 ≤ 1, so it diverges.
4. Conclusion. The terms are bigger than those of a divergent series for every n ≥ 3. Leaving out the n = 2 term does not affect divergence. By the direct comparison test, the series diverges.

A wrong turn to avoid in (a). Comparing with (2 + sin n)/n² ≤ 3/n is true, but ∑ 3/n diverges. “Smaller than divergent” gives no conclusion. Choose the comparison that matches your guess.

The limit comparison test

Direct comparison needs an inequality in the right direction, and sometimes the natural one points the wrong way. Take ∑ 1/(n² − n + 1). Since n² − n + 1 ≤ n² for n ≥ 1, we get 1/(n² − n + 1) ≥ 1/n². That is “bigger than convergent”: no conclusion. Yet for large n the terms are almost exactly 1/n². The limit comparison test captures “almost exactly”.

Test. Suppose aₙ > 0 and bₙ > 0 for every n from some point on, and

lim as n → ∞ of aₙ/bₙ = L, where L is a finite positive number (0 < L < ∞).

Then ∑ aₙ and ∑ bₙ both converge or both diverge.

Why it works. If aₙ/bₙ → L > 0, then eventually aₙ/bₙ is between L/2 and 3L/2. So (L/2)·bₙ < aₙ < (3L/2)·bₙ. Each inequality gives a direct comparison: the right one shows ∑ aₙ converges if ∑ bₙ does, and the left one shows ∑ aₙ diverges if ∑ bₙ does.

Choosing bₙ. Keep only the term that grows fastest in the numerator and the one that grows fastest in the denominator, and drop constant multiples.

aₙ	Dominant terms	Benchmark bₙ
(n + 4)/(2n² − n + 3)	n/(2n²)	1/n
(√n + 5)/(n² + 1)	√n/n²	1/n^(3/2)
3ⁿ/(4ⁿ + 7)	3ⁿ/4ⁿ	(3/4)ⁿ

You can keep the constant (for example bₙ = 1/(2n)); it only changes L, not the conclusion.

Worked example 2: limit comparison

Question. Decide whether each series converges. (a) ∑ 1/(n² − n + 1) (b) ∑ (n + 4)/(2n² − n + 3)

(a)

1. Positivity. n² − n + 1 = 1, 3, 7, … for n = 1, 2, 3, and it is positive for every n ≥ 1. So aₙ > 0.
2. Benchmark. Dominant term n², so bₙ = 1/n² (positive; ∑ bₙ converges, p = 2).
3. Limit. aₙ/bₙ = n²/(n² − n + 1). Divide top and bottom by n²: 1/(1 − 1/n + 1/n²) → 1/(1 − 0 + 0) = 1.
4. Conclusion. L = 1 is finite and positive, and ∑ 1/n² converges, so by the limit comparison test ∑ 1/(n² − n + 1) converges.

(b)

1. Positivity. For n ≥ 1, n + 4 > 0, and 2n² − n + 3 = n(2n − 1) + 3 > 0. So aₙ > 0.
2. Benchmark. Dominant terms n/(2n²) = 1/(2n). Take bₙ = 1/n.
3. Limit. aₙ/bₙ = n(n + 4)/(2n² − n + 3) = (n² + 4n)/(2n² − n + 3). Divide by n²: (1 + 4/n)/(2 − 1/n + 3/n²) → 1/2.
4. Conclusion. 0 < 1/2 < ∞ and ∑ 1/n diverges (harmonic), so ∑ (n + 4)/(2n² − n + 3) diverges.

Check the guess. In (b) the terms behave like 1/(2n), half the harmonic terms. Half of an infinite sum is still infinite, which matches L = 1/2 and divergence.

Worked example 3: geometric benchmarks and choosing a method

Question. (a) Show that ∑ 3ⁿ/(4ⁿ + 7) converges, and find an upper bound for its sum. (b) Decide whether ∑ (√n + 5)/(n² + 1) converges.

(a)

1. Inequality. Adding 7 makes the denominator bigger, so 0 < 3ⁿ/(4ⁿ + 7) < 3ⁿ/4ⁿ = (3/4)ⁿ for every n ≥ 1.
2. Benchmark. ∑ (3/4)ⁿ is geometric with ratio 3/4, and |3/4| < 1, so it converges. Starting at n = 1, its sum is (3/4)/(1 − 3/4) = 3.
3. Conclusion. By direct comparison, the series converges, and its sum is less than 3. (Background: adding terms numerically gives about 2.30.)

Here direct comparison is the natural choice: the inequality points the right way and also gives a bound.

(b)

1. Benchmark. Dominant terms √n/n² = 1/n^(3/2), so bₙ = 1/n^(3/2), a convergent p-series (p = 3/2).
2. Why not direct comparison? The +5 on top makes aₙ bigger than √n/n², and the +1 below makes it smaller. The two effects pull in opposite directions, so a quick inequality is awkward.
3. Limit. aₙ/bₙ = (√n + 5) n^(3/2)/(n² + 1) = (n² + 5n^(3/2))/(n² + 1). Divide by n²: (1 + 5/√n)/(1 + 1/n²) → 1.
4. Conclusion. All terms are positive and L = 1, so by the limit comparison test the series converges.

Writing a complete justification

In this unit a decision without the conditions of the test checked is not a full justification. For a comparison test, write down:

1. That the terms are positive (for every n, or from some N on).
2. The benchmark series and why it converges or diverges (p-series with the value of p, or geometric with the ratio).
3. Either the inequality and the n for which it holds, or the limit L with working, and that 0 < L < ∞.
4. The name of the test and the conclusion.

Background: the cases L = 0 and L = ∞. If aₙ/bₙ → 0, then eventually aₙ < bₙ, so convergence of ∑ bₙ gives convergence of ∑ aₙ (nothing follows if ∑ bₙ diverges). If aₙ/bₙ → ∞, then eventually aₙ > bₙ, so divergence of ∑ bₙ gives divergence of ∑ aₙ. These are direct comparisons in disguise. The safest habit is to choose bₙ from the dominant terms, so that L is a positive number.

Common misconceptions

• Comparing in the useless direction. “1/(n + 2) < 1/n and ∑ 1/n diverges, so ∑ 1/(n + 2) diverges” is invalid reasoning (even though the conclusion happens to be true). Smaller than divergent tells you nothing.
• Forgetting to check positivity. Both tests assume positive terms. Do not use them on ∑ (−1)ⁿ/n² directly.
• Taking the limit of aₙ instead of aₙ/bₙ. lim aₙ = 0 is the nth term test, which cannot prove convergence.
• Accepting L = 0 or L = ∞ as “both behave the same”. The basic test needs 0 < L < ∞. With bₙ = 1/n for aₙ = 1/n², L = 0, and the two series behave differently.
• Choosing a benchmark you have not classified. The comparison is only as good as your knowledge of ∑ bₙ. State why it converges or diverges.
• An inequality that fails for small n but is claimed “for all n”. Say “for n ≥ 3” (or whatever is true). That is enough.
• Thinking the comparison gives the sum. It decides convergence, and at most gives a bound.

Where this leads

Next, Topic 10.7, Alternating Series Test for Convergence, deals with series whose signs alternate, where comparison tests cannot be used directly. The ratio test (Topic 10.8) compares a series with a geometric series automatically. In Topic 10.9 you will apply comparison tests to |aₙ| to show absolute convergence. See the order on the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.