Test for Divergence
Textbook Reference
| Primary source | Stewart, Calculus, 9th edition, Section 11.2: “Series,” pages 783 to 784 (Theorem 6, the Test for Divergence, and Example 9) |
| Open companion | OpenStax Calculus Volume 2, Section 5.3: “The Divergence and Integral Tests” |
| Companion link | https://openstax.org/books/calculus-volume-2/pages/5-3-the-divergence-and-integral-tests |
The divergence test rests on two earlier ideas: the partial-sum definition of a series and the limit of a sequence.
Key idea
The first thing to check on any series is whether its terms even shrink to zero. If they do not, the series cannot possibly add up to a finite number.
The reasoning is short. If a series $\sum a_n$ converges to a sum $S$, then the running totals $s_n$ approach $S$, and so do the totals one step earlier, $s_{n-1}$. But the $n$th term is the jump between consecutive totals, $a_n = s_n - s_{n-1}$. If both totals close in on the same $S$, that jump must shrink to $0$. So convergence forces $a_n \to 0$.
Flip that around and you get a fast rejection tool. If the terms do not go to $0$, the series cannot converge. That is the Test for Divergence, and it is the cheapest check you own: glance at the terms, take their limit, and if that limit is anything other than $0$ (or fails to exist), you are done. The series diverges.
There is one trap, and it is the most important caveat in the chapter. The test only ever says “diverges.” It can never say “converges.” Terms going to $0$ is necessary but not sufficient, and the harmonic series is the standing proof: its terms $\dfrac{1}{n}$ go to $0$, yet it diverges. So a limit of $0$ tells you nothing; you must reach for a different test.
Prerequisite Check
If these are solid, you are ready.
Quick Reference
Theorem (necessary condition). If $\displaystyle\sum_{n=1}^{\infty} a_n$ converges, then $\displaystyle\lim_{n\to\infty} a_n = 0$.
Test for Divergence (the usable contrapositive). If $\displaystyle\lim_{n\to\infty} a_n$ does not exist, or if $\displaystyle\lim_{n\to\infty} a_n \neq 0$, then $\displaystyle\sum_{n=1}^{\infty} a_n$ diverges.
What the test can and cannot do.
- It can prove a series diverges (when the term limit is nonzero or missing).
- It can never prove a series converges.
- If $\displaystyle\lim_{n\to\infty} a_n = 0$, the test is inconclusive; use another test.
Procedure.
- Compute $\displaystyle\lim_{n\to\infty} a_n$.
- If the limit is nonzero or does not exist, conclude the series diverges. Stop.
- If the limit is $0$, the test says nothing; move to a different test.
Key Concepts
1. Why Convergence Forces the Terms to Zero
The theorem behind the test is a one-paragraph argument worth understanding, because it explains exactly why the test works and why its converse fails.
Suppose $\sum a_n$ converges with sum $S$. Let $s_n = a_1 + a_2 + \cdots + a_n$ be the $n$th partial sum. Convergence means $\displaystyle\lim_{n\to\infty} s_n = S$. The partial sums one step back also approach $S$, since shifting the index by one does not change a limit: \[ \lim_{n\to\infty} s_{n-1} = S. \] The $n$th term is the difference of consecutive partial sums, $a_n = s_n - s_{n-1}$. Therefore \[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} (s_n - s_{n-1}) = S - S = 0. \] So any convergent series has terms tending to $0$. The Test for Divergence is just the contrapositive: if the terms do not tend to $0$, the series is not convergent.
2. Using the Test
Example 1 (terms with a nonzero limit). Show that $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{5n^2 + 4}$ diverges.
Goal. Take the limit of the terms. If it is nonzero, the Test for Divergence applies.
Work. \[ \lim_{n\to\infty} \frac{n^2}{5n^2 + 4} = \lim_{n\to\infty} \frac{1}{5 + \frac{4}{n^2}} = \frac{1}{5} \neq 0. \] Since the terms approach $\dfrac{1}{5}$, not $0$, the series diverges by the Test for Divergence.
Answer: the series diverges.
Recap. The terms settle toward $\dfrac{1}{5}$, so the running total keeps gaining about $\dfrac{1}{5}$ per term forever and cannot converge. One limit finishes the problem.
(Common error: trying to “add up” the series or hunting for a sum. Once the term limit is nonzero, there is nothing more to do; the series diverges and no sum exists.)
Example 2 (terms with no limit). Determine whether $\displaystyle\sum_{n=1}^{\infty} (-1)^n$ converges.
Goal. Check the limit of the terms.
Work. The terms are $-1, 1, -1, 1, \dots$, which oscillate and have no limit. Because $\displaystyle\lim_{n\to\infty} (-1)^n$ does not exist, the Test for Divergence applies.
Answer: the series diverges.
Recap. The test fires in two situations: a nonzero limit, or no limit at all. Oscillating terms are the second situation.
3. The Caveat That Matters Most: a Limit of Zero Proves Nothing
A term limit of $0$ is the one case the test cannot rule on, and mistaking it for convergence is the most common error in the chapter.
Example 3 (the test is inconclusive). What does the Test for Divergence say about $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$?
Goal. Apply the test honestly and report what it does and does not establish.
Work. The terms satisfy $\displaystyle\lim_{n\to\infty} \frac{1}{n} = 0$. The Test for Divergence requires a nonzero or missing limit to conclude divergence, so with a limit of $0$ the test is silent. It gives no information.
Answer: the test is inconclusive. (Separately, the harmonic series is known to diverge, but the Test for Divergence does not show that.)
Recap. A limit of $0$ is exactly the case the test cannot rule on. The harmonic series proves the converse of the theorem is false: $a_n \to 0$ does not force convergence. When you get a limit of $0$, switch to another test.
(Common error: writing “since $a_n \to 0$, the series converges by the Test for Divergence.” The test never concludes convergence. A zero limit means try a different test, not that the series converges.)
Inline Self-Check
Question. A student writes: “The terms of $\displaystyle\sum_{n=1}^{\infty}\frac{2n+1}{3n+1}$ approach $\dfrac{2}{3}$, so by the Test for Divergence the series diverges.” Is the conclusion correct?
Show answer
Yes, the conclusion is correct. The term limit is $\displaystyle\lim_{n\to\infty}\frac{2n+1}{3n+1} = \frac{2}{3} \neq 0$, so the Test for Divergence applies and the series diverges. (This is the test used correctly: a nonzero term limit does prove divergence.)
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Concluding convergence from the test | “$a_n \to 0$, so it converges” | The test never proves convergence |
| Calling a zero limit inconclusive-then-converged | Treating $\sum 1/n$ as convergent | A zero limit is inconclusive; $\sum 1/n$ in fact diverges |
| Looking for a sum after the test fires | Summing $\sum \frac{n^2}{5n^2+4}$ | A divergent series has no finite sum |
| Missing the no-limit case | Only checking for a nonzero limit | A term limit that does not exist also forces divergence |
| Confusing terms with partial sums | Taking $\lim s_n$ for the test | The test uses $\lim a_n$, the limit of the terms |
Leveled Practice
Attempt each problem before opening the answer.
Level 1: Direct Application
Problem 1. Use the Test for Divergence on $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2n+1}$.
Show answer
\[ \lim_{n\to\infty}\frac{n}{2n+1} = \frac{1}{2} \neq 0, \] so the series diverges by the Test for Divergence.
Problem 2. Use the Test for Divergence on $\displaystyle\sum_{n=1}^{\infty} \cos n$.
Show answer
The terms $\cos n$ oscillate and do not approach any single value, so $\displaystyle\lim_{n\to\infty}\cos n$ does not exist. The series diverges by the Test for Divergence.
Problem 3. Apply the Test for Divergence to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$. What can you conclude?
Show answer
The terms satisfy $\displaystyle\lim_{n\to\infty}\frac{1}{n^2} = 0$, so the Test for Divergence is inconclusive: it tells you nothing about whether this series converges or diverges. (A different test, the integral test, shows it converges, but not this one.)
Level 2: Decide Which Conclusion Is Available
Problem 4. For each series, state whether the Test for Divergence proves divergence or is inconclusive: (a) $\displaystyle\sum \frac{3n^2+1}{n^2+5}$, (b) $\displaystyle\sum \frac{1}{\sqrt{n}}$.
Show answer
(a) Term limit $= 3 \neq 0$, so the test proves the series diverges.
(b) Term limit $= 0$, so the test is inconclusive. (This series in fact diverges, as a $p$-series with $p = \tfrac{1}{2}$, but the Test for Divergence does not show that.)
Problem 5. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{2n! + 1}$ diverges.
Show answer
\[ \lim_{n\to\infty}\frac{n!}{2\,n! + 1} = \lim_{n\to\infty}\frac{1}{2 + \frac{1}{n!}} = \frac{1}{2} \neq 0, \] so the series diverges by the Test for Divergence.
Level 3: Reasoning
Problem 6. Explain why the Test for Divergence can never be used to prove a series converges, referencing the harmonic series.
Show answer
The theorem behind the test says only that convergence implies $a_n \to 0$. The test is the contrapositive: $a_n \not\to 0$ implies divergence. The converse, “$a_n \to 0$ implies convergence,” is false, and the harmonic series $\sum 1/n$ is the counterexample: its terms go to $0$ but the series diverges. So a term limit of $0$ is consistent with both convergence and divergence, which is exactly why the test cannot conclude convergence.
Problem 7. A geometric series $\sum ar^{n-1}$ has $|r| \ge 1$. Show that the Test for Divergence applies.
Show answer
The terms are $ar^{n-1}$ with $a \neq 0$. If $|r| > 1$, then $|ar^{n-1}| \to \infty$, so the terms do not go to $0$. If $r = 1$, the terms are the constant $a \neq 0$. If $r = -1$, the terms are $a, -a, a, -a, \dots$, which oscillate with no limit. In every case with $|r| \ge 1$ the term limit is nonzero or missing, so the Test for Divergence proves the series diverges. (This matches the geometric-series rule that convergence requires $|r| < 1$.)
Mastery Checklist
You have mastered this skill when you can do all of the following without notes:
Mental Model
Think of a series as a tally you keep adding to, one mark of size $a_n$ at a time. For the running tally to settle at a final number, the marks you add must fade to nothing. If the marks keep arriving at full size, or even at a steady fraction like $\dfrac{1}{5}$, the tally never stops growing.
So the Test for Divergence is a doorway check: do the incoming marks fade to zero? If they clearly do not, you can turn the series away at the door, it diverges, no further work. But passing the doorway check (marks fading to zero) does not get you inside. The harmonic series fades to zero at the door yet still drifts off to infinity once admitted. The doorway only rejects; it never admits. Admission requires one of the later tests.
Connections
Built From
- Limits of sequences: the test is the computation of a single sequence limit, the limit of the terms.
- Infinite series and partial sums: the theorem behind the test is proved from $a_n = s_n - s_{n-1}$ and the convergence of the partial sums.
Leads To
- The integral test and the comparison tests: the divergence test is the mandatory first check; when it comes back with a zero limit (inconclusive), these tests take over to decide convergence.
- The ratio and root tests, alternating series: every later convergence test assumes you have already run the divergence test, because if the terms do not go to $0$ there is no point applying anything subtler.
Why It Matters
The Test for Divergence is the triage step of series analysis: it is the fastest possible filter, rejecting any series whose terms do not vanish before you spend effort on it. In numerical computing the same idea is a sanity check on an iterative sum: if the terms being added are not shrinking, the running total will never stabilize, so the computation cannot converge. Knowing that the converse fails is equally practical: a shrinking term is not a guarantee of a finite total, which is why a careful algorithm cannot simply stop when the next term looks small.
Audience Notes
For students who find math intimidating: this is the easiest test in the chapter. Take one limit. If it is not zero, the series diverges and you are finished.
For students who want depth: the one-line proof from $a_n = s_n - s_{n-1}$ is worth knowing, because it makes transparent both why the test works and why a zero limit cannot imply convergence. The gap it leaves is the reason the rest of the chapter exists.
For students aimed at a career in computing or engineering: the test is a cheap precondition check. Before trusting any iterative summation to converge, confirm the terms are actually shrinking, and remember that shrinking alone is not enough to stop early.
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