Series Convergence Tests: How to Choose

To choose a convergence test for a series, look at the form of its terms. Start with the divergence test. Then check for a geometric, p-series or telescoping series. Use the ratio test for factorials, the root test for nth powers, a comparison test for fractions of powers of n, and the alternating series test when signs alternate.

Before this lesson: Divergence Test, p-Series, Limit Comparison Test, Alternating Series Test, Ratio Test and Root Test

Why the form of the terms picks the test

Each convergence test is built for one kind of term. The ratio test is easy when terms have factorials, because they cancel. The root test is easy when the whole term is an nth power. A comparison test is easy when the term is a fraction of powers of n, because it acts like a p-series.

So the skill is not doing a test. It is spotting the pattern in the term, then picking the test that fits it. The flowchart asks the questions in a good order.

A flowchart of eight yes-or-no questions, asked in order. If the limit of aₙ is not 0, the series diverges by the divergence test. If it is geometric or a p-series, use their rules. If the terms telescope, find the partial sums. Factorials or exponentials: ratio test. Whole term to the nth power: root test. Alternating signs: test the absolute values first, then the alternating series test. A fraction of powers of n: direct or limit comparison with a p-series. Positive, decreasing and easy to integrate: integral test. — Ask the questions from the top. Stop at the first yes and use that test.

Series convergence tests at a glance

Test	Use it when	What it tells you
Divergence test	Always, first	If $\lim a_n \ne 0$ , the series diverges. If the limit is 0, it tells you nothing.
Geometric series	$\sum ar^n$ , the variable is in the exponent	Converges when \|r\| < 1; see infinite geometric series.
p-series	$\sum \frac{1}{n^p}$ , the variable is in the base	Converges when p > 1.
Telescoping series	Terms split into a difference, like $\frac{1}{n} - \frac{1}{n + 1}$	Find the partial sums; see telescoping series.
Ratio test	Factorials or exponentials	L < 1: converges absolutely. L > 1: diverges. L = 1: no answer.
Root test	The whole term is raised to the nth power	Same as the ratio test.
Alternating series test	$\sum (-1)^n b_n$ with $b_n$ decreasing to 0	The series converges (maybe only conditionally).
Direct comparison test	Terms are clearly bigger or smaller than a known series	Smaller than convergent: converges. Bigger than divergent: diverges. See direct comparison test.
Limit comparison test	Fractions built from powers of n	If $\lim \frac{a_n}{b_n}$ is positive and finite, both series do the same thing.
Integral test	f(x) is positive, decreasing and easy to integrate	The series and the improper integral both converge or both diverge; see integral test.

A step-by-step strategy

Find $\lim a_n$ . If it is not 0, stop: the series diverges.
Check for a geometric series, a p-series or a telescoping series. These have direct answers.
If you see factorials or exponentials, use the ratio test. If the whole term is an nth power, use the root test.
If the signs alternate, first test $\sum |a_n|$ . If that diverges, use the alternating series test.
If the term is a fraction of powers of n, compare it with the p-series made from the highest powers.
If nothing else fits and f(x) is easy to integrate, use the integral test.

Choosing a convergence test: examples

Example 1: Try the divergence test first

Does $\sum_{n=1}^{\infty} \dfrac{n^2 + 1}{3n^2 - n}$ converge?

\lim_{n \to \infty} \dfrac{n^2 + 1}{3n^2 - n} = \lim_{n \to \infty} \dfrac{1 + \frac{1}{n^2}}{3 - \frac{1}{n}}

Check the limit of the terms first. Divide the top and bottom by n² (see limits at infinity).

\lim_{n \to \infty} \dfrac{1 + \frac{1}{n^2}}{3 - \frac{1}{n}} = \dfrac{1}{3}

1/n² and 1/n go to 0.

\dfrac{1}{3} \ne 0

The terms do not approach 0.

Answer: the series diverges by the divergence test. No other test is needed.

Example 2: A fraction of powers of n

Does $\sum_{n=1}^{\infty} \dfrac{n + 5}{n^3 + 2}$ converge?

b_n = \dfrac{n}{n^3} = \dfrac{1}{n^2}

The terms go to 0. Keep only the highest power on the top and on the bottom.

\dfrac{a_n}{b_n} = \dfrac{n + 5}{n^3 + 2} \cdot \dfrac{n^2}{1}

Divide aₙ by bₙ: multiply by the reciprocal of bₙ.

\dfrac{n + 5}{n^3 + 2} \cdot \dfrac{n^2}{1} = \dfrac{n^3 + 5n^2}{n^3 + 2}

Multiply n² into the top (see distributive property).

\lim_{n \to \infty} \dfrac{n^3 + 5n^2}{n^3 + 2} = \lim_{n \to \infty} \dfrac{1 + \frac{5}{n}}{1 + \frac{2}{n^3}} = 1

Divide the top and bottom by n³.

0 \lt 1 \lt \infty

The limit is positive and finite, so both series do the same thing.

\sum_{n=1}^{\infty} \dfrac{1}{n^2} \text{ converges}

It is a p-series with p = 2 > 1.

Answer: the series converges by the limit comparison test. The ratio test would give L = 1 here, which is no help.

Example 3: A factorial means the ratio test

Does $\sum_{n=1}^{\infty} \dfrac{(n + 1)2^n}{n!}$ converge?

\dfrac{a_{n+1}}{a_n} = \dfrac{(n + 2)2^{n+1}}{(n + 1)!} \cdot \dfrac{n!}{(n + 1)2^n}

The term has a factorial, so use the ratio test. Multiply by the reciprocal of aₙ.

\dfrac{(n + 2)2^{n+1}}{(n + 1)!} \cdot \dfrac{n!}{(n + 1)2^n} = 2 \cdot \dfrac{1}{n + 1} \cdot \dfrac{n + 2}{n + 1}

2ⁿ⁺¹ ÷ 2ⁿ = 2, and (n + 1)! = (n + 1) · n!.

2 \cdot \dfrac{1}{n + 1} \cdot \dfrac{n + 2}{n + 1} = \dfrac{2n + 4}{(n + 1)^2}

Multiply the fractions.

L = \lim_{n \to \infty} \dfrac{2n + 4}{n^2 + 2n + 1} = 0

Expand the bottom. It has the higher power of n.

0 \lt 1

Compare L with 1.

Answer: the series converges absolutely by the ratio test.

Example 4: Alternating signs

Does $\sum_{n=2}^{\infty} \dfrac{(-1)^n}{\ln n}$ converge? If so, how?

|a_n| = \dfrac{1}{\ln n}

The terms go to 0. Test for absolute convergence first.

\ln n \lt n \text{ for } n \ge 2

The natural log grows more slowly than n.

\dfrac{1}{\ln n} \gt \dfrac{1}{n}

Take reciprocals of positive numbers; the inequality flips.

\sum_{n=2}^{\infty} \dfrac{1}{n} \text{ diverges}

It is the harmonic series, so Σ|aₙ| diverges by direct comparison.

b_n = \dfrac{1}{\ln n} \text{ decreases to } 0

ln n increases without bound, so its reciprocal shrinks to 0.

Answer: by the alternating series test the series converges, but not absolutely. It converges conditionally.

Example 5: Energy lost by a fan (word problem)

After a fan is switched off, it loses $40ne^{-n^2}$ joules of energy in second n. Does the total energy lost over all seconds stay finite? The term is easy to integrate, so use the integral test.

f(x) = 40xe^{-x^2}

Match the term with a function of x. It is positive for x > 0.

f'(x) = 40e^{-x^2} + 40x(-2x)e^{-x^2}

Use the product rule and the chain rule.

40e^{-x^2} + 40x(-2x)e^{-x^2} = 40e^{-x^2}(1 - 2x^2)

Factor out 40e⁻ˣ².

40e^{-x^2}(1 - 2x^2) \lt 0 \text{ for } x \ge 1

1 − 2x² is negative when x ≥ 1, so f is decreasing.

\int_1^{\infty} 40xe^{-x^2}\,dx = \lim_{b \to \infty} \int_1^{b} 40xe^{-x^2}\,dx

Write the improper integral as a limit.

\int_1^{b} 40xe^{-x^2}\,dx = \left[-20e^{-x^2}\right]_1^{b}

Let u = −x², so du = −2x dx (see u-substitution).

\left[-20e^{-x^2}\right]_1^{b} = -20e^{-b^2} + 20e^{-1}

Evaluate at b and at 1.

\lim_{b \to \infty} \left(-20e^{-b^2} + 20e^{-1}\right) = \dfrac{20}{e} \approx 7.36

e⁻ᵇ² goes to 0.

Answer: the integral is finite, so the series converges by the integral test. The total energy lost is finite: adding the terms gives about 16.2 joules. (The ratio test also works here.)

Common mistakes

Mistake	Example of the error	Fix
Thinking terms going to 0 proves convergence	“1/n goes to 0, so $\sum \frac{1}{n}$ converges”	The divergence test can only show divergence. The harmonic series diverges.
Using the ratio test on powers of n	Ratio test on $\sum \frac{2n + 1}{n^3}$ gives L = 1	For fractions of powers of n, use limit comparison with a p-series.
Comparing in the wrong direction	“ $\frac{1}{n + 1} \lt \frac{1}{n}$ and $\sum \frac{1}{n}$ diverges, so $\sum \frac{1}{n + 1}$ diverges”	Smaller than a divergent series tells you nothing. Use limit comparison instead.
Stopping at the alternating series test	Saying $\sum \frac{(-1)^n}{n}$ converges absolutely	That test gives only convergence. Check $\sum \|a_n\|$ to decide absolute or conditional.
Mixing up geometric and p-series	Calling $\sum \frac{1}{2^n}$ a p-series with p = 2	Variable in the exponent: geometric. Variable in the base: p-series.
Ignoring the integral test conditions	Using it on $\sum \frac{\cos^2 n}{n^2}$	That f is not decreasing. Compare with $\frac{1}{n^2}$ instead.

Related methods

Which tests show absolute convergence?

The ratio and root tests look at |aₙ|, so when they show convergence, it is absolute. Any test you apply to $\sum |a_n|$ also shows absolute convergence. The alternating series test is the only test here that can show conditional convergence.

After you know a series converges

Many problems then ask for the sum or an estimate. Geometric and telescoping series have exact sums. For other series, you can estimate the sum with a partial sum and bound the error, which Unit 7 covers for alternating series and Taylor series.

Practice problems

For each series, choose a test, then decide whether it converges.

$\sum_{n=1}^{\infty} \dfrac{3^n}{4^n + 1}$
$\sum_{n=1}^{\infty} \dfrac{2n^2}{n^2 + 7}$
$\sum_{n=1}^{\infty} \dfrac{4}{n\sqrt{n}}$
$\sum_{n=1}^{\infty} \dfrac{n^4}{n!}$
$\sum_{n=1}^{\infty} \left(\dfrac{5n + 2}{2n + 7}\right)^n$
$\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} n}{n^2 + 4}$ (absolutely, conditionally or not at all?)
$\sum_{n=1}^{\infty} \dfrac{1}{n(n + 1)}$ (find the sum if it converges)
$\sum_{n=2}^{\infty} \dfrac{1}{n \ln n}$
$\sum_{n=2}^{\infty} \dfrac{\sqrt{n} + 3}{n^2 - 1}$
A model says a website gets $\dfrac{900n}{n^2 + 50}$ new sign-ups on day n. If the model held forever, would the total number of sign-ups stay finite?

Show answers

4^n + 1 \gt 4^n

A geometric-looking term: compare. A bigger bottom makes a smaller fraction.

\dfrac{3^n}{4^n + 1} \lt \dfrac{3^n}{4^n} = \left(\dfrac{3}{4}\right)^n

Compare term by term.

\sum_{n=1}^{\infty} \left(\dfrac{3}{4}\right)^n \text{ converges}

It is geometric with r = 3/4 < 1.

Answer: converges by direct comparison with a geometric series

\lim_{n \to \infty} \dfrac{2n^2}{n^2 + 7} = \lim_{n \to \infty} \dfrac{2}{1 + \frac{7}{n^2}}

Divergence test first. Divide the top and bottom by n².

\lim_{n \to \infty} \dfrac{2}{1 + \frac{7}{n^2}} = 2

7/n² goes to 0.

2 \ne 0

The terms do not approach 0.

Answer: diverges by the divergence test

\dfrac{4}{n\sqrt{n}} = \dfrac{4}{n^{3/2}}

n · n¹ᐟ² = n³ᐟ² (see exponent rules).

p = \dfrac{3}{2} \gt 1

It is 4 times a p-series.

Answer: converges (p-series with p = 3/2)

\dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)^4}{(n + 1)!} \cdot \dfrac{n!}{n^4}

Factorial: use the ratio test.

\dfrac{(n + 1)^4}{(n + 1)!} \cdot \dfrac{n!}{n^4} = \dfrac{(n + 1)^3}{n^4}

(n + 1)! = (n + 1) · n!, so cancel n! and one factor of n + 1.

L = \lim_{n \to \infty} \dfrac{(n + 1)^3}{n^4} = 0

The bottom has the higher power of n.

Answer: L = 0 < 1, so it converges absolutely by the ratio test

\sqrt[n]{\left(\dfrac{5n + 2}{2n + 7}\right)^n} = \dfrac{5n + 2}{2n + 7}

The whole term is an nth power: use the root test.

L = \lim_{n \to \infty} \dfrac{5 + \frac{2}{n}}{2 + \frac{7}{n}} = \dfrac{5}{2}

Divide the top and bottom by n.

Answer: L = 5/2 > 1, so it diverges by the root test

\lim_{n \to \infty} \dfrac{n}{n^2 + 4} \cdot \dfrac{n}{1} = \lim_{n \to \infty} \dfrac{n^2}{n^2 + 4} = 1

Test Σ|aₙ| first, with limit comparison against 1/n.

\sum_{n=1}^{\infty} \dfrac{1}{n} \text{ diverges}

So Σ|aₙ| diverges: no absolute convergence.

f'(x) = \dfrac{(x^2 + 4) - x \cdot 2x}{(x^2 + 4)^2}

Check that bₙ = n/(n² + 4) decreases. Use the quotient rule.

\dfrac{(x^2 + 4) - x \cdot 2x}{(x^2 + 4)^2} = \dfrac{4 - x^2}{(x^2 + 4)^2}

Combine like terms on top.

\dfrac{4 - x^2}{(x^2 + 4)^2} \lt 0 \text{ for } x \gt 2

So bₙ decreases from n = 2 on.

\lim_{n \to \infty} \dfrac{n}{n^2 + 4} = 0

The bottom has the higher power, so the alternating series test applies.

Answer: converges conditionally

\dfrac{1}{n(n + 1)} = \dfrac{1}{n} - \dfrac{1}{n + 1}

Split with partial fraction decomposition.

S_N = 1 - \dfrac{1}{N + 1}

The middle terms cancel: the series telescopes.

\lim_{N \to \infty} \left(1 - \dfrac{1}{N + 1}\right) = 1

1/(N + 1) goes to 0.

Answer: converges; its sum is 1

\int_2^{b} \dfrac{dx}{x \ln x} = \left[\ln(\ln x)\right]_2^{b}

f(x) = 1/(x ln x) is positive and decreasing for x ≥ 2. Let u = ln x, du = dx/x.

\left[\ln(\ln x)\right]_2^{b} = \ln(\ln b) - \ln(\ln 2)

Evaluate at b and at 2.

\lim_{b \to \infty} \left(\ln(\ln b) - \ln(\ln 2)\right) = \infty

ln(ln b) grows without bound.

Answer: diverges by the integral test

b_n = \dfrac{\sqrt{n}}{n^2} = \dfrac{1}{n^{3/2}}

Keep the highest powers on the top and bottom.

\dfrac{a_n}{b_n} = \dfrac{\sqrt{n} + 3}{n^2 - 1} \cdot n^{3/2} = \dfrac{n^2 + 3n^{3/2}}{n^2 - 1}

Multiply: √n · n³ᐟ² = n².

\lim_{n \to \infty} \dfrac{n^2 + 3n^{3/2}}{n^2 - 1} = \lim_{n \to \infty} \dfrac{1 + \frac{3}{\sqrt{n}}}{1 - \frac{1}{n^2}} = 1

Divide the top and bottom by n².

\sum_{n=2}^{\infty} \dfrac{1}{n^{3/2}} \text{ converges}

p = 3/2 > 1.

Answer: converges by limit comparison

10.

\dfrac{a_n}{b_n} = \dfrac{900n}{n^2 + 50} \cdot \dfrac{n}{1} = \dfrac{900n^2}{n^2 + 50}

A fraction of powers of n: compare with bₙ = n/n² = 1/n.

\lim_{n \to \infty} \dfrac{900n^2}{n^2 + 50} = \lim_{n \to \infty} \dfrac{900}{1 + \frac{50}{n^2}} = 900

Divide the top and bottom by n².

\sum_{n=1}^{\infty} \dfrac{1}{n} \text{ diverges}

900 is positive and finite, so both series diverge.

Answer: no; in the model the total grows without bound, although very slowly

Convergence tests FAQ

What is the best test for convergence of a series?

No single test works for every series. Start with the divergence test, then match the form of the terms: factorials to the ratio test, nth powers to the root test, fractions of powers of n to limit comparison, and alternating signs to the alternating series test.

What order should I try convergence tests in?

Divergence test; then geometric, p-series and telescoping; then the ratio or root test; then comparison tests; then the alternating series test. Save the integral test for terms you can integrate.

When should I use a comparison test instead of the ratio test?

Use a comparison test when the terms are fractions built from powers of n, like (n + 5)/(n³ + 2). The ratio test gives L = 1 for these, so it gives no answer.

Which convergence tests show absolute convergence?

The ratio and root tests do, when L < 1. So does any test applied to Σ|aₙ|. The alternating series test shows only convergence, so check Σ|aₙ| separately.

Summary

Always try the divergence test first. It can only prove divergence.
Geometric, p-series and telescoping series have direct answers.
Factorials or exponentials: ratio test. Whole term to the nth power: root test.
Fractions of powers of n: compare with the p-series made from the highest powers.
Alternating signs: test Σ|aₙ| first, then the alternating series test.
Use the integral test when f(x) is positive, decreasing and easy to integrate.

Previous lesson: Root Test

Next lesson: Taylor Polynomials

Back to the Calculus 2 Index