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Limit Test for Divergence

Limit Test for Divergence

A convergence test that uses the fact that the terms of a convergent series must have a limit of zero.

 

If lim(n→∞) aₙ does not exist or lim(n→∞) aₙ ≠ 0, the series Σaₙ diverges. Otherwise the test is inconclusive.

 

 

See also

Divergent series

Key Formula

If limnan0, then n=1an diverges.\text{If } \lim_{n \to \infty} a_n \neq 0, \text{ then } \sum_{n=1}^{\infty} a_n \text{ diverges.}
Where:
  • ana_n = The nth term (general term) of the series
  • nn = The index of summation, running from 1 to infinity
  • n=1an\sum_{n=1}^{\infty} a_n = The infinite series whose convergence is being tested

Worked Example

Problem: Determine whether the series n=13n4n+5\sum_{n=1}^{\infty} \frac{3n}{4n+5} converges or diverges using the Limit Test for Divergence.
Step 1: Identify the general term of the series.
an=3n4n+5a_n = \frac{3n}{4n+5}
Step 2: Compute the limit of the general term as n approaches infinity. Divide numerator and denominator by n.
limn3n4n+5=limn34+5n\lim_{n \to \infty} \frac{3n}{4n+5} = \lim_{n \to \infty} \frac{3}{4 + \frac{5}{n}}
Step 3: As n approaches infinity, the fraction 5/n approaches 0, so the limit simplifies.
limn34+5n=34+0=34\lim_{n \to \infty} \frac{3}{4 + \frac{5}{n}} = \frac{3}{4+0} = \frac{3}{4}
Step 4: Since the limit is 3/4, which is not equal to 0, the Limit Test for Divergence tells us the series diverges.
340    n=13n4n+5 diverges\frac{3}{4} \neq 0 \implies \sum_{n=1}^{\infty} \frac{3n}{4n+5} \text{ diverges}
Answer: The series diverges by the Limit Test for Divergence because limnan=340\lim_{n \to \infty} a_n = \frac{3}{4} \neq 0.

Another Example

This example demonstrates the most common pitfall: when the limit equals 0, the test tells you nothing. It shows that the Divergence Test can only prove divergence, never convergence.

Problem: Apply the Limit Test for Divergence to the series n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2}.
Step 1: Identify the general term of the series.
an=1n2a_n = \frac{1}{n^2}
Step 2: Compute the limit of the general term as n approaches infinity.
limn1n2=0\lim_{n \to \infty} \frac{1}{n^2} = 0
Step 3: The limit equals 0. The Divergence Test is therefore inconclusive — it does NOT prove convergence. You would need a different test (such as the p-series test with p = 2 > 1) to confirm that this series converges.
limnan=0    test is inconclusive\lim_{n \to \infty} a_n = 0 \implies \text{test is inconclusive}
Answer: The Limit Test for Divergence is inconclusive here because the limit equals 0. A separate test must be used to determine convergence.

Frequently Asked Questions

Can the Divergence Test prove that a series converges?
No. The Divergence Test can only prove divergence. If the limit of the nth term equals zero, the test is inconclusive — the series might converge or might diverge. For example, 1/n\sum 1/n has terms approaching 0, yet it diverges (harmonic series), while 1/n2\sum 1/n^2 also has terms approaching 0 and converges.
What is the difference between the Divergence Test and the Limit Comparison Test?
The Divergence Test checks whether limnan0\lim_{n \to \infty} a_n \neq 0 to conclude divergence. The Limit Comparison Test compares the ratio limnan/bn\lim_{n \to \infty} a_n / b_n to a known benchmark series bn\sum b_n and can prove either convergence or divergence. The Divergence Test is a quick first check, while the Limit Comparison Test is a more powerful tool used after the Divergence Test is inconclusive.
When should you use the Divergence Test?
Use it as the very first step whenever you are asked to determine whether a series converges or diverges. It is fast and easy to apply. If the limit of the general term is clearly nonzero (or does not exist), you are done — the series diverges. If the limit is zero, move on to other tests like the Ratio Test, Integral Test, or Comparison Test.

Divergence Test vs. Integral Test

Divergence TestIntegral Test
What it testsWhether lim aₙ ≠ 0Whether ∫ f(x) dx converges, where f(n) = aₙ
Can prove convergence?No — only divergenceYes — can prove convergence or divergence
RequirementsOnly need to evaluate a limitf(x) must be positive, continuous, and decreasing
DifficultyUsually very quickRequires evaluating an improper integral
When to useAlways try first as a quick checkWhen terms resemble a function you can integrate

Why It Matters

The Divergence Test is typically the first test students learn and the first test they should apply in any series convergence problem, whether in AP Calculus BC or a college Calculus II course. It provides a fast way to eliminate series that obviously diverge before investing effort in more complex tests. Understanding this test also reinforces the crucial distinction between a necessary condition (terms going to zero) and a sufficient condition (the series actually converging).

Common Mistakes

Mistake: Concluding that a series converges because the limit of aₙ is 0.
Correction: A limit of 0 is necessary for convergence but not sufficient. The harmonic series 1/n\sum 1/n is the classic counterexample: its terms approach 0, yet it diverges. When the limit is 0, the Divergence Test is inconclusive, and you must use another test.
Mistake: Forgetting to check the Divergence Test before applying more complex tests.
Correction: Always apply the Divergence Test first. If the limit is nonzero or does not exist, you can immediately conclude divergence without needing the Ratio Test, Comparison Test, or any other method.

Related Terms

  • Convergence TestsFamily of tests that includes the Divergence Test
  • Convergent SeriesA series whose terms must satisfy lim aₙ = 0
  • Divergent SeriesWhat the test concludes when the limit is nonzero
  • LimitThe core operation evaluated in this test
  • TermThe individual aₙ whose limit is computed
  • ZeroThe critical threshold value in the test
  • Harmonic SeriesKey counterexample where lim aₙ = 0 but series diverges