Calculus II · Unit 4B · lesson

Using the Ratio Test on Power Series

Concept

Learning objectives

derive an inequality in $x$ from the Ratio Test and identify the radius.

Using the Ratio Test on Power Series

Explanation

The Ratio Test turns coefficient growth into a distance condition

For many power series, the ratio of consecutive terms simplifies to a constant multiple of $|x-a|$ . The inequality $L<1$ then becomes the interior convergence condition. This is one of the most reusable calculations in Calculus II.

The ratio test is applied to the whole term, not only to coefficients. After solving the inequality, record the radius and open interval before touching endpoints. Keeping those stages separate prevents endpoint conclusions from being smuggled out of a test that is explicitly inconclusive there.

Bridge

The Ratio Test turns x into a distance inequality

When a power series contains factorials or complicated coefficients, the Ratio Test usually isolates a factor of $|x-a|$ . The resulting inequality $L(x)<1$ describes the open interval of absolute convergence.

The calculation finds the radius, not the complete interval. After solving the inequality, return to the original series at both endpoints. The ratio test often becomes inconclusive there precisely because its limiting ratio equals one.

Proof idea

Coefficient growth and distance multiply

The neighboring-term ratio often separates as

\left|\frac{c_{n+1}}{c_n}\right||x-a|.

The coefficient limit sets the allowable size of $|x-a|$ . This is the series analogue of balancing coefficient growth against powers of distance.

Concept

Generic ratio pattern

For $a_n=c_n(x-a)^n$ , compute

\left|\frac{a_{n+1}}{a_n}\right| =\left|\frac{c_{n+1}}{c_n}\right||x-a|.

Guided walkthrough

A factorial coefficient

For

\sum_{n=0}^{\infty}\frac{(x-1)^n}{n!},

the ratio is $|x-1|/(n+1)\to0$ for every real $x$ . Therefore $R=\infty$ .

Worked example

A factorial produces infinite radius

Find the radius of convergence of

\sum_{n=0}^{\infty}\frac{(x-2)^n}{n!}.

For $a_n=(x-2)^n/n!$ ,

\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x-2|}{n+1}\to0

for every real $x$ . Since the ratio limit is always below one, the series converges absolutely for all $x$ , so $R=\infty$ .

Common mistake

Do not replace endpoint tests with the ratio limit

If solving $L(x)<1$ gives $|x-a|<R$ , the points where $|x-a|=R$ still require direct substitution into the original series.

Interactive checku4b-ratio_test_for_power_series-01

Find the radius of $\sum (x-1)^n/n!$ .

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Show hint

The ratio includes a factor $1/(n+1)$ .

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Exercise

Find the radius of $\sum n!(x+2)^n$ .

Exercise

Find the radius of $\sum (x-3)^n/[n2^n]$ .

Exercise

Explain why the test is inconclusive at $|x-a|=R$ .

Exercise

Derive a radius formula when $|c_{n+1}/c_n|\to L$ .

After the explanation

Use the section idea

Reading lens

Find the radius from interior behavior, then test each boundary point as a separate numerical series.

Mental model

Distance from the center organizes the automatic interior and exterior behavior; endpoints remain independent decisions.

Decision

Use a ratio or root argument for the radius, convert it to an interval, and test both endpoints explicitly.

Common trap

Including or excluding both endpoints from the radius calculation alone.

Check yourself

Can you justify the radius and each endpoint with separate evidence?

Source & rights

Original instruction with traceable references.

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