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Convergence Tests

Convergence Tests

Limit test for divergence
Integral test
Comparison test
Limit comparison test
Alternating series test
Ratio test
Root test

 

 

See also

Convergent series, divergent series, power series, power series convergence, nth partial sum, remainder of a series, series rules, series

Worked Example

Problem: Determine whether the series ∑n=1∞n3n\displaystyle\sum_{n=1}^{\infty} \frac{n}{3^n} converges or diverges by selecting and applying an appropriate convergence test.
Step 1: Choose a test: The general term involves nn in the numerator and an exponential 3n3^n in the denominator. When you see a ratio of successive terms that simplifies nicely—especially with exponentials or factorials—the Ratio Test is a strong choice.
Step 2: Set up the Ratio Test: The Ratio Test examines the limit of the absolute value of the ratio of consecutive terms:
L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|
Step 3: Compute the ratio: With an=n3na_n = \frac{n}{3^n}, substitute an+1=n+13n+1a_{n+1} = \frac{n+1}{3^{n+1}}:
an+1an=n+13n+1⋅3nn=n+13n\frac{a_{n+1}}{a_n} = \frac{n+1}{3^{n+1}} \cdot \frac{3^n}{n} = \frac{n+1}{3n}
Step 4: Evaluate the limit: Take the limit as n→∞n \to \infty:
L=lim⁡n→∞n+13n=lim⁡n→∞1+1n3=13L = \lim_{n \to \infty} \frac{n+1}{3n} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{3} = \frac{1}{3}
Step 5: Apply the conclusion: Since L=13<1L = \frac{1}{3} < 1, the Ratio Test guarantees that the series converges.
Answer: By the Ratio Test, ∑n=1∞n3n\displaystyle\sum_{n=1}^{\infty} \frac{n}{3^n} converges because L=13<1L = \frac{1}{3} < 1.

Another Example

Problem: Determine whether the series ∑n=1∞1n2+n\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + n} converges or diverges using the Comparison Test.
Step 1: Choose a test and a comparison series: The terms 1n2+n\frac{1}{n^2 + n} resemble 1n2\frac{1}{n^2} for large nn. Since ∑1n2\sum \frac{1}{n^2} is a known convergent pp-series (p=2>1p = 2 > 1), try the Comparison Test.
Step 2: Establish the inequality: For all n≥1n \geq 1, n2+n>n2n^2 + n > n^2, so:
0<1n2+n<1n20 < \frac{1}{n^2 + n} < \frac{1}{n^2}
Step 3: Conclude: Each term of our series is smaller than the corresponding term of a convergent series. By the Comparison Test, ∑1n2+n\sum \frac{1}{n^2 + n} also converges.
Answer: By the Comparison Test with ∑1n2\sum \frac{1}{n^2}, the series ∑n=1∞1n2+n\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + n} converges.

Frequently Asked Questions

How do I know which convergence test to use?
Start with the Limit Test for Divergence—it's quick and rules out series whose terms don't approach zero. If terms involve factorials or exponentials, try the Ratio Test. If the general term looks like a known series (geometric or pp-series), use the Comparison or Limit Comparison Test. For alternating signs, use the Alternating Series Test. If the nn-th root simplifies nicely, try the Root Test. Practice builds intuition for matching series structure to the right test.
Can a convergence test be inconclusive?
Yes. For example, the Ratio Test is inconclusive when L=1L = 1, meaning it cannot determine convergence or divergence. The Limit Test for Divergence can only prove divergence; if the limit of terms equals zero, the test tells you nothing. When one test is inconclusive, you must try a different test.

Comparison Test vs. Limit Comparison Test

The Comparison Test requires you to establish a direct inequality between your series and a known series term by term. The Limit Comparison Test instead computes the limit of the ratio of the two series' terms. If this limit is a finite positive number, both series share the same convergence behavior. The Limit Comparison Test is often easier when finding a clean inequality is difficult.

Why It Matters

Many problems in calculus, physics, and engineering require you to add infinitely many terms—for instance, representing functions as power series or Fourier series. Convergence tests tell you whether those infinite sums actually produce a meaningful finite value. Without them, you have no rigorous way to know if an infinite series can be safely used in a calculation.

Common Mistakes

Mistake: Assuming that if lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the series must converge.
Correction: Terms approaching zero is necessary but not sufficient for convergence. The harmonic series ∑1n\sum \frac{1}{n} has terms going to zero yet diverges. You must apply an additional test beyond the Limit Test for Divergence.
Mistake: Using the Ratio Test or Root Test on series where the limit equals 1 and concluding convergence or divergence.
Correction: When L=1L = 1, both the Ratio Test and Root Test are inconclusive. You need to switch to a different test, such as the Comparison Test or Integral Test, to reach a conclusion.

Related Terms