The nth Term Test for Divergence: Study Guide (Calculus BC 10.3)

BC-only material. The nth term test is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You need four earlier ideas. If any is shaky, revisit it first. The Calculus BC roadmap shows where each one sits.

Prerequisite	Topic	What you use it for here
Limits at infinity and dominant terms	1.15	Finding the limit of a rational term such as (5n² − 2n)/(8n² + 3)
L’Hospital’s Rule	4.7	Limits of the form ∞ · 0 or 1^∞, such as (1 + 2/n)ⁿ
Partial sums and convergence	10.1	The test is proved from partial sums
Geometric series	10.2	A quick source of examples; see the Topic 10.2 study guide

Notation on this page: ∑ from n = 1 to ∞ of aₙ is the infinite series a₁ + a₂ + a₃ + …. Its nth partial sum is Sₙ = a₁ + a₂ + … + aₙ. The series converges to S when Sₙ → S as n → ∞. If the partial sums have no finite limit, the series diverges.

Two sequences, not one

Every series comes with two sequences. Keep them apart.

• The sequence of terms: a₁, a₂, a₃, …. These are the numbers you add.
• The sequence of partial sums: S₁, S₂, S₃, …. These are the running totals.

The series converges only if the partial sums settle down to a finite number. The terms settling down is a different question.

Take aₙ = 3n/(4n + 1). The terms are 0.6, 0.667, 0.692, 0.706, … and they approach 3/4. So the sequence of terms converges. But each new term adds between 0.6 and 0.75 to the running total, and the amounts added get closer to 0.75. The partial sums are 0.6, 1.267, 1.959, 2.665, …, and by n = 10 the total is already about 7.01. The totals keep climbing by about 0.7 each step, so they cannot settle down. The series diverges.

Why the terms of a convergent series must go to 0

Each term is the difference of two neighbouring partial sums:

aₙ = Sₙ − Sₙ₋₁ (for n ≥ 2)

Suppose the series converges to a finite number S. Then Sₙ → S, and also Sₙ₋₁ → S, because it is the same sequence shifted by one place. So

lim aₙ = lim Sₙ − lim Sₙ₋₁ = S − S = 0.

That proves one fact: if a series converges, its terms approach 0.

Now turn the fact around (this is its contrapositive, which is always equally true): if the terms do not approach 0, the series cannot converge. That is the nth term test.

The nth term test for divergence. If lim (n → ∞) of aₙ ≠ 0, or the limit does not exist, then ∑ aₙ diverges. If lim (n → ∞) of aₙ = 0, the test gives no conclusion.

Read the second line carefully. The test has only two possible outputs: “diverges” or “no conclusion”. There is no “converges” output. The name says it: it is a test for divergence.

How to use the test, step by step

1. Identify aₙ, the general term being added. Check where the sum starts; it does not affect the test, but you need the right formula.
2. Find lim (n → ∞) of aₙ. Use dominant terms for rational expressions, L’Hospital’s Rule (treating n as a continuous variable x) for indeterminate forms, or known limits.
3. Compare with 0.
   • Limit is a non-zero number, or ±∞, or does not exist → the series diverges by the nth term test.
   • Limit is 0 → no conclusion from this test. Say so, then use another method (partial sums now; later tests in Topics 10.4 to 10.9).
4. Write the justification. Name the limit, say that it is not 0, name the test, and conclude.

A complete justification looks like this: “lim (n → ∞) of 3n/(4n + 1) = 3/4 ≠ 0, so by the nth term test, ∑ 3n/(4n + 1) diverges.”

Quick links with geometric series. For ∑ a rⁿ with a ≠ 0 and |r| ≥ 1, the terms a rⁿ do not approach 0, so the series diverges by the nth term test. This explains half of the geometric series rule from Topic 10.2. The other half (convergence when |r| < 1) needs the partial-sum formula: the nth term test cannot supply it.

Worked example 1: four series, four limits

Question. For each series, find the limit of the terms and state what the nth term test shows.

(a) ∑ from n = 1 to ∞ of (5n² − 2n)/(8n² + 3) (b) ∑ from n = 1 to ∞ of (−1)ⁿ n/(3n + 2) (c) ∑ from n = 1 to ∞ of 7/(2n + 5) (d) ∑ from n = 1 to ∞ of (1.02)ⁿ

(a) The highest power of n is n² on the top and the bottom. Divide both by n²:

(5 − 2/n)/(8 + 3/n²) → 5/8 as n → ∞.

5/8 ≠ 0, so the series diverges by the nth term test.

(b) Look at the size first: n/(3n + 2) → 1/3. The sign factor (−1)ⁿ flips the sign every step. The terms are −1/5, 1/4, −3/11, 2/7, −5/17, 3/10, …. The odd terms approach −1/3 and the even terms approach +1/3. The terms do not approach a single number, so lim aₙ does not exist. In particular it is not 0, so the series diverges by the nth term test.

(c) 7/(2n + 5) → 0, because the denominator grows without bound. The limit is 0, so the nth term test gives no conclusion. Do not write “converges”. (This series does in fact diverge, but you need a different test to show it, such as the integral test in Topic 10.4.)

(d) This is geometric with r = 1.02 > 1. The terms (1.02)ⁿ grow without bound, so lim aₙ = ∞ ≠ 0. The series diverges by the nth term test. (The geometric rule from Topic 10.2 gives the same answer, since |r| ≥ 1.)

Check. In (a), (b) and (d) the terms stay a fixed distance away from 0 for large n: about 0.625, about 0.333 in size, and more than 1. Adding infinitely many such numbers cannot give a finite total. In (c) the terms shrink, so the picture is unclear, which is why the test stays silent.

Worked example 2: a limit that needs L’Hospital’s Rule

Question. Does the series ∑ from n = 1 to ∞ of (1 + 2/n)ⁿ converge or diverge? Justify your answer.

1. Identify the term. aₙ = (1 + 2/n)ⁿ. As n → ∞, the base approaches 1 and the power grows: this is the indeterminate form 1^∞. You cannot just say “1 to any power is 1”.
2. Take logarithms. Replace n by a continuous variable x and let y = (1 + 2/x)ˣ. Then ln y = x ln(1 + 2/x) = ln(1 + 2/x) / (1/x). As x → ∞, the top → ln 1 = 0 and the bottom → 0. This is 0/0, so L’Hospital’s Rule applies.
3. Differentiate top and bottom.
   • d/dx [ln(1 + 2/x)] = (−2/x²) / (1 + 2/x).
   • d/dx [1/x] = −1/x².
   • Ratio: [(−2/x²)/(1 + 2/x)] / (−1/x²) = 2/(1 + 2/x) → 2.
4. Undo the logarithm. ln y → 2, so y → e². Therefore lim aₙ = e² ≈ 7.389.
5. Conclude. e² ≠ 0, so by the nth term test the series diverges.

Check with numbers. a₁ = 3, a₁₀ ≈ 6.192, a₁₀₀ ≈ 7.245, a₁₀₀₀ ≈ 7.374. The terms are creeping up towards 7.389. Adding infinitely many numbers that are bigger than 3 cannot give a finite total. ✓

Note. Using a continuous variable x is allowed: if f(x) → L as x → ∞ and aₙ = f(n), then aₙ → L too. (The reverse is not always true, but you only need this direction.)

Worked example 3: terms go to 0, but the series still diverges

Question. A series is built in blocks. Block 1 is one copy of 1. Block 2 is two copies of 1/2. Block 3 is three copies of 1/3, and so on: block m is m copies of 1/m.

1 + (1/2 + 1/2) + (1/3 + 1/3 + 1/3) + (1/4 + 1/4 + 1/4 + 1/4) + …

(a) Do the terms approach 0? What does the nth term test show? (b) Does the series converge?

(a) Every term in block m equals 1/m. As you go further along the series (n → ∞), you move into later and later blocks, so m → ∞ and the terms 1/m → 0. The nth term test gives no conclusion.

(b) Look at the partial sums at the end of each block.

• Each block adds m copies of 1/m, which is exactly 1.
• Blocks 1 to m use 1 + 2 + … + m = m(m + 1)/2 terms.
• So the partial sum after m(m + 1)/2 terms is exactly m.

For example, after 3 terms the total is 2, after 55 terms it is 10, and after 5050 terms it is 100. All the terms are positive, so the partial sums only ever go up. They pass every whole number m, so they grow without bound. The partial sums have no finite limit, and the series diverges.

Interpretation. The terms do shrink to 0, but slowly: each value 1/m appears m times, so together they always add up to 1. Shrinking terms are not enough. The total depends on how fast the terms shrink.

The lesson. “Terms approach 0” is necessary for convergence but not sufficient. This is exactly why the nth term test can never be used to prove that a series converges. You will meet the most famous series with this behaviour, the harmonic series ∑ 1/n, in Topic 10.5.

Using the fact forwards

The statement behind the test, “if ∑ aₙ converges, then aₙ → 0”, can also be used directly.

Example. You are told that ∑ aₙ converges and that aₙ ≠ −1 for all n. Find lim (n → ∞) of (5 − 2aₙ)/(1 + aₙ).

Since the series converges, aₙ → 0. The expression is continuous at 0, so the limit is (5 − 0)/(1 + 0) = 5. You do not need to know the sum of the series, only that its terms approach 0.

Common misconceptions

• “The terms go to 0, so the series converges.” False. The block series in Worked example 3 has terms that approach 0 and still diverges. When aₙ → 0, write “the nth term test is inconclusive”.
• “The terms approach 3/4, so the series converges to 3/4.” This mixes up the two sequences. The terms approaching 3/4 means the series diverges. The sum of a series is the limit of the partial sums, not of the terms.
• “The sequence converges, so the series converges.” The sequence n/(n + 1) converges (to 1), but the series ∑ n/(n + 1) diverges by the nth term test.
• Treating “limit does not exist” as “no conclusion”. If the terms oscillate, as in Worked example 1(b), the limit is not 0, so the series diverges.
• Getting the limit wrong in a 1^∞ form. (1 + 2/n)ⁿ does not approach 1. It approaches e². Use logarithms and L’Hospital’s Rule.
• A justification with no limit stated. “The terms are big, so it diverges” earns little. State the value of lim aₙ, say it is not 0, and name the test.
• Thinking the starting index matters. Changing where the sum starts, or removing finitely many terms, does not change lim aₙ, so it never changes the outcome of the nth term test.

Where this leads

The nth term test is a quick first check: if the terms do not approach 0, you are done. When they do approach 0, you need a sharper tool. Topic 10.4 introduces the first one, which compares a series with an improper integral. Continue with Topic 10.4, Integral Test for Convergence, or see the order on the Calculus BC roadmap.

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