What Is the Divergence Test?
The divergence test, also called the nth-term test, is the first and simplest check to run on any infinite series:
if limn→∞ aₙ ≠ 0 (or the limit does not exist), then Σ aₙ diverges
The reasoning is unavoidable: to add up infinitely many terms and get a finite total, the terms must eventually shrink toward zero. If they instead settle at 1/2, or oscillate, or grow, then you are perpetually adding non-vanishing chunks, and the running total cannot converge. This test is fast and decisive when it applies, one limit computation. But it carries a crucial asymmetry: it can only prove divergence, never convergence. If the terms do go to zero, the test is silent, and a sharper tool must decide. This calculator evaluates lim aₙ with a table of large-n values, delivers the verdict, and, when the limit is zero, points you toward the right follow-up test.
How to Use the Divergence Test Calculator
Enter the general term aₙ as a function of n. The steps display the test, a table of aₙ at n = 10, 100, up to 10 million, and the conclusion: divergence if the limit is a nonzero number or fails to exist, or "inconclusive" if the limit is zero (with a nudge toward the integral, ratio, or comparison tests). The graph plots the terms against the zero line they must approach. This is the gateway to the convergence-test toolkit: when it returns inconclusive, continue to the integral test, ratio test, or limit comparison test.
Worked Example
Test Σ n/(2n + 1). The limit of the terms is
limn→∞ n/(2n + 1) = 1/2 ≠ 0
Since the terms approach 1/2 rather than 0, the series diverges by the nth-term test: you are essentially adding 1/2 over and over, so the partial sums grow without bound. The verdict takes one line. Now contrast the harmonic series Σ 1/n, the most important cautionary tale in all of series theory. Its terms 1/n do go to zero, so the divergence test says nothing, yet the harmonic series famously diverges (its partial sums crawl to infinity like ln n). This is the whole point of the test's asymmetry: terms going to zero is necessary for convergence but nowhere near sufficient. The oscillating chip (−1)ⁿ has no limit at all, terms bounce between +1 and −1, which also forces divergence.
Why Terms Must Vanish, but Vanishing Is Not Enough
The forward direction is airtight. If a series converges to sum S, then its partial sums Sn approach S, so the individual terms aₙ = Sn − Sn−1 approach S − S = 0. Contrapositive: if the terms do not approach zero, the series cannot converge. That is the divergence test, and it is a genuine theorem, not a heuristic. The reverse, however, is false, and understanding why is a rite of passage in analysis. The harmonic series shows that terms can shrink to zero while their sum still diverges, because they shrink too slowly, ln n growth beats every finite bound even though each added term is tiny. The dividing line is subtle: Σ 1/n diverges but Σ 1/n² converges, and both have terms going to zero. This is precisely why an entire arsenal of convergence tests exists, the divergence test clears the easy cases (terms not vanishing) instantly, leaving the genuinely hard cases (terms vanishing, but how fast?) to the integral, comparison, ratio, and root tests. Running the divergence test first is always worthwhile: it is cheap, and when it fires, it settles the question immediately.
Common Mistakes to Avoid
- Using it to prove convergence. The single most common error. Terms going to zero does not imply the series converges; the harmonic series is the eternal counterexample. The test only ever proves divergence.
- Stopping at "inconclusive" thinking it means convergent. When lim aₙ = 0, the test gives no information whatsoever. You must apply a different test; the question is genuinely open until you do.
- Confusing the limit of terms with the sum. lim aₙ is about individual terms shrinking, not about the total. A series can have terms approaching zero and still sum to infinity, or to a finite value; the term-limit alone cannot tell which.
- Mishandling terms with no limit. If aₙ oscillates without settling (like (−1)ⁿ), the limit does not exist, which counts as "not zero" and forces divergence. This is a valid divergence conclusion, not an inconclusive one.
- Evaluating at too small an n. A term like n/(2n+1) is 0.476 at n = 10 but 0.4995 at n = 1000; the limit is 1/2. Judging from small n can hide the true limiting value, which is why the tool tabulates out to 10 million.
Real-World Applications
The divergence test is the mandatory first gate in any convergence analysis, and its lesson, that convergence requires vanishing terms but demands more, shapes how series are used throughout applied mathematics. In numerical computing, before summing a series to approximate a function or a physical quantity, one checks that the terms decay; a series whose terms do not vanish signals a modeling error or a divergent expansion that no amount of computation can rescue. Perturbation series in physics are notorious here: many are asymptotic and their terms eventually grow, so the divergence test warns that summing more terms makes things worse, not better, a crucial practical caution in quantum field theory.
Frequently Asked Questions
What does the divergence test conclude?
Only one thing: if the terms aₙ do not approach zero (nonzero limit or no limit), the series diverges. It never proves convergence. When the terms do go to zero, the test is inconclusive and a different test is required.
Why is it called the nth-term test?
Because it examines only the limit of the general nth term, lim aₙ, not the partial sums or any deeper structure. It is the most basic thing you can check about a series, looking at where the individual terms are headed.
If the terms go to zero, does the series converge?
Not necessarily. This is the test's fundamental limitation. The harmonic series Σ 1/n has terms going to zero yet diverges. Vanishing terms are necessary for convergence but not sufficient; how fast they vanish determines the outcome, which other tests measure.
Why does the harmonic series diverge despite vanishing terms?
Its terms 1/n shrink too slowly. Grouping them (1/2, then 1/3+1/4 > 1/2, then the next four > 1/2, and so on) shows infinitely many blocks each exceeding 1/2, so the partial sums grow without bound like ln n. Slow decay defeats convergence.
What if the limit of the terms does not exist?
Then the terms fail to approach zero (a limit of zero would have to exist), so the series diverges. Oscillating terms like (−1)ⁿ that bounce between values without settling are a clear divergence by the test, since no single limit, let alone zero, is reached.
Should I always run the divergence test first?
Yes. It is the cheapest test, one limit, and when it fires it settles divergence instantly with no further work. Only when it returns inconclusive (terms going to zero) do you need the more elaborate integral, ratio, root, or comparison tests.
What tests should I use when this one is inconclusive?
For positive terms, try the integral test, comparison tests, or the ratio and root tests depending on the term's form. Factorials and exponentials suit the ratio test; polynomial decay suits comparison with a p-series; and the integral test handles terms resembling integrable functions.
Does the divergence test apply to alternating series?
Yes, as a first check: if the terms of an alternating series do not go to zero in absolute value, it diverges. If they do go to zero, the divergence test is inconclusive and the alternating series test takes over to decide convergence.
How large should n be to trust the term limit?
Large enough that the value has stabilized. Terms like n/(2n+1) approach their limit slowly, still 0.4995 at n = 1000 versus the true 0.5. The tool evaluates out to 10 million to reveal the genuine limiting behavior rather than a misleading small-n value.
Can a series with terms going to zero still be summed exactly?
Sometimes, if it converges. Σ 1/2ⁿ sums to 1 and Σ 1/n² sums to π²/6, both with vanishing terms. But vanishing terms alone never guarantee this; convergence must be established by another test before any exact sum is meaningful.