Calculus II · Unit 4A · lesson

The Root Test

Concept

Learning objectives

use nth roots to detect terms raised to the nth power and compare the Root Test with the Ratio Test.

The Root Test

Explanation

nth powers reveal themselves under nth roots

When an entire expression is raised to the $n$ th power, taking an $n$ th root removes that outer exponent and exposes the effective geometric ratio. The Root Test is therefore natural for terms such as $((2n+1)/(3n+2))^n$ and less natural for plain rational functions.

The conclusion mirrors the Ratio Test. A limit below one gives absolute convergence; above one gives divergence; equal to one is inconclusive. In more advanced analysis, the test is often stated with a limit superior because the ordinary limit need not exist. Calculus examples are usually designed so the simpler limit exists.

Bridge

Extract the effective exponential base

The Root Test asks for the long-run base hidden inside $|a_n|$ by examining $\sqrt[n]{|a_n|}$ . It is especially effective when the entire term is raised to the $n$ th power or contains several exponential factors.

Its conclusions parallel the Ratio Test: a limit below one gives absolute convergence, a limit above one gives divergence, and a limit equal to one is inconclusive. Simplify the $n$ th root before taking the limit; that is where the method earns its keep.

Proof idea

The extracted base controls the whole term

If the root limit is below one, choose $r<1$ above the limit. Eventually $\sqrt[n]{|a_n|}<r$ , hence $|a_n|<r^n$ , and geometric comparison proves absolute convergence.

Concept

Root Test

Let

L=\lim_{n\to\infty}\sqrt[n]{|a_n|}.

If $L<1$ , the series converges absolutely. If $L>1$ , it diverges. If $L=1$ , the test is inconclusive.

Guided walkthrough

An nth-power series

For

a_n=\left(\frac{2n+1}{3n+2}\right)^n,

we have

\sqrt[n]{a_n}=\frac{2n+1}{3n+2}\to\frac23<1.

Therefore the series converges.

Worked example

An nth power reveals its own base

Test

\sum_{n=1}^{\infty}\left(\frac{2n+1}{3n+4}\right)^n.

Then

\sqrt[n]{|a_n|}=\frac{2n+1}{3n+4}\to\frac23.

Since $2/3<1$ , the series converges absolutely by the Root Test.

Common mistake

Take the root of the whole magnitude

The test uses $\sqrt[n]{|a_n|}$ . Applying the root to only one factor can destroy the expression’s effective base.

Interactive checku4a-root_test-01

Find the Root-Test limit for $a_n=((2n+1)/(3n+2))^n$ .

Your work stays on this device. No account or AI grader is used.

Show hint

The nth root cancels the outer power.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

Apply the Root Test to $\sum[(n+1)/(2n)]^n$ .

Exercise

Explain why it is inconclusive for $\sum1/n^3$ .

Exercise

Compare the algebra required by root and ratio tests for $(n/(n+1))^{n^2}$ .

Exercise

State the limit-superior version informally.

After the explanation

Use the section idea

Reading lens

Separate sign behavior from magnitude, then use ratios or roots when powers and factorials dominate.

Mental model

Absolute convergence controls magnitude strongly enough to imply convergence; conditional convergence relies on cancellation.

Decision

Test absolute values first when practical, use the alternating-series hypotheses explicitly, and reserve ratio or root tests for matching algebraic structure.

Common trap

Calling any alternating-looking series convergent or treating a ratio/root limit of one as a verdict.

Check yourself

Can you state whether convergence is absolute, conditional, divergent, or still undecided?

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.