10.3 the nth term test for divergence: How to Use It

If the limit of the nth term of a series is not zero — or fails to exist — then the series diverges. That is the core statement of 10.3 the nth term test for divergence. Formally, for a series (\sum_{n=1}^{\infty} a_n), if (\lim_{n \to \infty} a_n \neq 0) (including the case where the limit does not exist), then the series diverges. The critical caveat is that the test only proves divergence; it can never prove convergence, because a limit of zero is inconclusive. The test is quick to apply and usually the first thing to check.

What the Test Says

The nth term test for divergence (also called the divergence test or the term test) examines the individual terms of a series rather than the partial sums.

Formal statement. Given an infinite series (\sum_{n=1}^{\infty} a_n):

• If (\lim_{n \to \infty} a_n \neq 0), then the series diverges.
• If the limit does not exist, the series diverges.
• If (\lim_{n \to \infty} a_n = 0), the test is inconclusive — you must use another test.

Why the Test Works

The intuition comes straight from the definition of a convergent series. If (\sum a_n) converges to a finite sum (S), then its partial sums (S_N = a_1 + a_2 + \cdots + a_N) approach (S). Since (S_N) and (S_{N-1}) both approach (S), their difference must approach zero:

[ a_N = S_N - S_{N-1} \to S - S = 0. ]

So in any convergent series, the individual terms must shrink toward zero. The OpenStax calculus sequence develops this idea as the “divergence test.” The contrapositive is what gives the test its power: if the terms fail to approach zero, convergence is impossible.

What the Test Cannot Do

The converse is false. A series whose terms approach zero can still diverge. The classic counterexample is the harmonic series (\sum_{n=1}^{\infty} \frac{1}{n}), where (\frac{1}{n} \to 0) but the series diverges. Because of cases like this, a limit of zero tells you nothing — you move on to the integral test, comparison tests, or the alternating series test. MIT OpenCourseWare notes this distinction in its single-variable calculus materials: the test is a one-way detector for divergence only.

How to Apply It

Applying the test is mechanical once you identify the general term.

1. Identify the general term (a_n). Strip away the summation symbol and write the expression in terms of (n).
2. Take the limit as (n \to \infty). Evaluate (\lim_{n \to \infty} a_n) using the standard limit laws: compare leading degrees, use L’Hôpital’s rule for indeterminate forms, or split rational expressions.
3. Compare the limit to zero. If the limit is any nonzero real number, or does not exist, stop — the series diverges.
4. If the limit equals zero, change tools. The test is inconclusive; apply another convergence test.

For rational terms (\frac{P(n)}{Q(n)}), divide numerator and denominator by the highest power of (n) in the denominator. For terms with exponentials such as (\frac{2^n}{n^2}), recall that exponentials with base greater than 1 grow faster than any polynomial. For terms with ((-1)^n), the limit cannot exist unless the magnitude vanishes.

Worked Example

Test (\sum_{n=1}^{\infty} \frac{3n}{n+5}).

[ \lim_{n \to \infty} \frac{3n}{n+5} = \lim_{n \to \infty} \frac{3}{1 + 5/n} = 3 \neq 0. ]

Since the terms approach 3 rather than 0, the series diverges. This makes sense: the terms stay near 3 forever, so the partial sums grow without bound.

Contrast this with the harmonic series (\sum_{n=1}^{\infty} \frac{1}{n}). Here (\frac{1}{n} \to 0), so the test is inconclusive — yet the harmonic series actually diverges. Proving that requires the integral test or a grouping argument, not the nth term test. Keep the outcomes straight:

Common Pitfalls

Treating it as a convergence test. The single most frequent error is concluding that a series converges because (\lim a_n = 0). It does not. A zero limit only means the test failed to detect divergence; the series may converge or diverge. Always reach for a second test in that case.

Misreading “not zero” as “positive.” The limit does not have to be positive to force divergence — it just has to be nonzero or nonexistent. A limit of (-2) diverges just as surely as a limit of (2).

Stopping too early on oscillating limits. For terms like (\sin n) or ((-1)^n), the limit simply may not exist, which itself is the divergence condition. State that clearly rather than forcing a numeric value.

Confusing the general term with the partial sum. The test looks at (a_n), the nth term of the sequence of terms, not (S_n), the nth partial sum. Mixing these up leads to incorrect limit evaluations.

Skipping the test when it is obviously fastest. On exams, a series with terms like (\frac{n^3+1}{2n^3-n}) should be flagged immediately: the limit is (\frac{1}{2} \neq 0), so it diverges in one line.

Key Takeaways

• The nth term test for divergence states that if (\lim_{n \to \infty} a_n \neq 0) or does not exist, then (\sum a_n) diverges.
• A limit of zero proves nothing — the test is inconclusive, and you must try another convergence test.
• To apply it: identify the general term, take its limit as (n \to \infty), and compare the result to zero.
• The harmonic series (\sum \frac{1}{n}) has a zero limit yet diverges; the test cannot detect this.
• Never interpret a zero limit as evidence of convergence; the test only ever concludes divergence.
• It is usually the fastest first check, so apply it before reaching for integral, comparison, or ratio tests.