Why doesn't the ratio test tell me what happens at the endpoints?

At an endpoint x=c± R the ratio-test limit equals exactly 1, and the ratio test is explicitly inconclusive when L=1. That is why you must substitute each endpoint back into the series and use a different tool — the alternating series test, p-series test, or a comparison — to decide convergence there.

How do I choose brackets versus parentheses in the interval of convergence?

Use a square bracket when that endpoint's series converges (so the point is included) and a parenthesis when it diverges (so the point is excluded). Test each endpoint separately; the left and right ends can require different brackets, giving intervals like [a,b) or (a,b].

What does a radius of convergence of 0 or infinity mean?

R=0 means the series converges only at its center x=c and nowhere else — the interval is just the single point \c\. R=∞ means the ratio-test limit is 0 for every x, so the series converges for all real numbers and the interval is (-∞,∞) with no endpoints to check.

Is convergence inside the interval always absolute?

Yes. The ratio test works with absolute values, so on the open interval (c-R,c+R) the series converges absolutely. Conditional convergence can only occur at an endpoint, such as when substituting a value produces the alternating harmonic series.

AP-CALCBC-10.13

U10.13 Power Series — Radius and Interval of Convergence

Master finding the radius R and interval of convergence for power series in AP Calc BC using the ratio test plus careful endpoint testing.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U10.13 Power Series — Radius and Interval of Convergence, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U10.13 Power Series — Radius and Interval of Convergence free

What this lesson covers

A power series looks like an infinite polynomial, and the first question we always ask is: for which values of

x

does it actually add up to a finite number? That set of

x

-values is the interval of convergence, and its half-width is the radius of convergence

R

. This lesson shows you the reliable machine for finding both: run the ratio test on the general term, solve the resulting inequality for a radius, then test each endpoint separately using the convergence tests you already know. Endpoint analysis is where most points are won or lost on the exam, so we drill it carefully.

What a Power Series Is and Why R Exists

A power series centered at

c

has the form

\sum_{n=0}^{\infty} a_n (x-c)^n

, where the

a_n

are fixed coefficients and

x

is a variable. When you plug in a specific number for

x

, the series becomes an ordinary infinite series of constants that either converges or diverges.

A fundamental theorem guarantees that exactly one of three things happens. The series converges only at

x=c

(then

R=0

); it converges for all real

x

(then

R=\infty

); or there is a positive number

R

such that the series converges for

|x-c|<R

and diverges for

|x-c|>R

. This is why we call

R

the radius of convergence: convergence is symmetric about the center

c

, spanning a distance

R

in each direction.

The interval of convergence (IOC) is the complete set of

x

where the series converges. Its interior is always

(c-R, c+R)

, but the two endpoints

x=c-R

and

x=c+R

must be checked one at a time, because the ratio test gives no information there. That is the single most important idea in this topic: the ratio test finds

R

, but it never decides the endpoints for you.

Using the Ratio Test to Find R

The ratio test is the standard tool. Form the limit of the absolute value of consecutive terms:

L = \lim_{n\to\infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n} \right| = |x-c|\cdot \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.

The series converges absolutely when

L<1

and diverges when

L>1

. So set the coefficient limit equal to some value

\ell

; then

L = \ell\,|x-c| < 1

gives

|x-c| < \frac{1}{\ell}

, which means

R = \frac{1}{\ell}

.

Three outcomes match the three cases from the theorem:

Coefficient limit $\ell$	Radius $R$	Meaning
$0$	$\infty$	converges for all $x$
finite $>0$	$1/\ell$	converges on an interval
$\infty$	$0$	converges only at $x=c$

Keep the factor

|x-c|

intact through the algebra; students often lose it or forget the absolute value. Factorials simplify by cancellation (

\frac{n!}{(n+1)!}=\frac{1}{n+1}

), and powers like

3^n

simplify because

\frac{3^n}{3^{n+1}}=\frac{1}{3}

. Once

L<1

is solved for

|x-c|<R

, you have found the radius and the open interval instantly.

Endpoint Analysis: The Part That Wins Points

After finding

R

, the open interval

(c-R, c+R)

is guaranteed. Now substitute each endpoint back into the original power series and analyze the resulting constant series with a test other than the ratio test (the ratio test always gives

L=1

at an endpoint, which is inconclusive).

Common endpoint outcomes: plugging in an endpoint often produces a

p

-series like

\sum \frac{1}{n^p}

, a harmonic series

\sum \frac{1}{n}

(diverges), or an alternating series

\sum \frac{(-1)^n}{n}

(converges by the alternating series test). Decide convergence at each endpoint independently, because the two ends can behave differently.

Then write the IOC using correct bracket notation: a square bracket for an endpoint that converges and is included, a parenthesis for one that diverges and is excluded. For example,

[c-R, c+R)

means the left endpoint converges and the right diverges.

Endpoint result	Include it?	Bracket
series converges	yes	$[$ or $]$
series diverges	no	$($ or $)$

A frequent exam error is stopping after finding

R

without testing endpoints, or using the wrong bracket. Always show the substituted series and name the test you used — free-response graders require that justification.

How the Exam Tests This and Common Traps

On multiple choice, you may be asked directly for

R

, for the IOC, or for whether a specific value of

x

lies inside the interval. On free response, this often appears within a Taylor/Maclaurin problem: you build a series, then must state its interval of convergence with full justification.

Watch these traps. First, when the center is

c\neq 0

, the interval is centered at

c

, not at

0

; do not assume symmetry about the origin. Second, if the series has a coefficient like

\frac{1}{n^2}

, both endpoints likely converge absolutely (giving a closed interval), whereas

\frac{1}{n}

typically gives one convergent and one divergent endpoint. Third, remember

R=\infty

means the IOC is

(-\infty,\infty)

with no endpoints to check — common for series like

e^x

.

Another subtlety: the ratio test measures absolute convergence, so inside the open interval convergence is always absolute. Conditional convergence, if it appears, happens only at an endpoint. Finally, a series that is missing terms (for example only even powers,

(x)^{2n}

) still works with the ratio test — just carry the exponent honestly, since

|x|^{2n}

leads to

|x|^2

and

R=\sqrt{1/\ell}

A Clean Step-by-Step Procedure

Use the same routine every time so you never skip endpoints.

Step	Action
1	Identify $a_n$ and the center $c$ from $\sum a_n(x-c)^n$ .
2	Compute $L=\lim_{n\to\infty}\left\|\frac{a_{n+1}}{a_n}\right\|\cdot\|x-c\|$ .
3	Set $L<1$ and solve for $\|x-c\|<R$ to read off $R$ .
4	Write the open interval $(c-R,\,c+R)$ .
5	Substitute $x=c-R$ ; test that constant series.
6	Substitute $x=c+R$ ; test that constant series.
7	Combine into the IOC with correct brackets.

This discipline matters because partial credit on the FRQ is tied to showing the ratio-test setup, the radius, and each endpoint test by name. Even if you handle the algebra quickly in your head, write out the substituted endpoint series and state "converges by the alternating series test" or "diverges by the p-series test with

p=1

." That single sentence per endpoint is what turns a correct answer into full credit.

Key terms

Power series.: An infinite series of the form $\sum a_n(x-c)^n$ whose partial sums behave like polynomials in $x$ ; convergence depends on the value of $x$ .
Center ( $c$ ).: The value about which the power series is built; the interval of convergence is always symmetric around $c$ before endpoints are checked.
Radius of convergence ( $R$ ).: The distance from the center within which the series converges: it converges for $|x-c|<R$ and diverges for $|x-c|>R$ .
Interval of convergence (IOC).: The full set of $x$ -values for which the series converges, equal to $(c-R,c+R)$ plus whichever endpoints pass their own convergence test.
Ratio test.: A test computing $L=\lim|a_{n+1}/a_n|$ (times $|x-c|$ here); convergence when $L<1$ , divergence when $L>1$ , inconclusive when $L=1$ .
Endpoint analysis.: Substituting $x=c-R$ and $x=c+R$ into the series and testing each resulting constant series separately, since the ratio test is inconclusive there.
Absolute convergence.: Convergence of $\sum|a_n(x-c)^n|$ ; guaranteed everywhere inside the open interval by the ratio test.

Worked example

Find the radius and interval of convergence of

\sum_{n=1}^{\infty} \frac{(x-2)^n}{n\,3^n}

Identify

a_n=\frac{1}{n\,3^n}

and center

c=2

.

Apply the ratio test to the full term

b_n=\frac{(x-2)^n}{n\,3^n}

L=\lim_{n\to\infty}\left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\,3^n}{(x-2)^n}\right| = |x-2|\lim_{n\to\infty}\frac{n}{(n+1)\cdot 3} = \frac{|x-2|}{3}.

Set

L<1

\frac{|x-2|}{3}<1

, so

|x-2|<3

. Thus

R=3

and the open interval is

(2-3,\,2+3)=(-1,5)

.

Now test the endpoints. At

x=-1

: substitute to get

\sum \frac{(-3)^n}{n\,3^n}=\sum \frac{(-1)^n}{n}

. This is the alternating harmonic series, which converges by the alternating series test, so

x=-1

is included.

At

x=5

: substitute to get

\sum \frac{3^n}{n\,3^n}=\sum \frac{1}{n}

. This is the harmonic series (

p=1

), which diverges, so

x=5

is excluded.

Combine: the interval of convergence is

[-1,5)

with radius

R=3

Practice questions

What is the radius of convergence of

\sum_{n=0}^{\infty} \frac{(x+4)^n}{n!}

$R=0$
$R=1$
$R=4$
$R=\infty$

Answer: $R=\infty$

Apply the ratio test:

\lim_{n\to\infty}\left|\frac{(x+4)^{n+1}}{(n+1)!}\cdot\frac{n!}{(x+4)^n}\right| = |x+4|\lim_{n\to\infty}\frac{1}{n+1}=0

for every

x

. Since

L=0<1

regardless of

x

, the series converges for all real numbers, so

R=\infty

and the IOC is

(-\infty,\infty)

. Factorials in the denominator overwhelm any power, which is why series like this converge everywhere.

Determine the radius and interval of convergence of

\sum_{n=1}^{\infty} \frac{(x-1)^n}{n^2}

, justifying each endpoint.

Answer: $R=1$ ; interval of convergence is $[0,2]$ .

Ratio test:

\lim_{n\to\infty}\left|\frac{(x-1)^{n+1}}{(n+1)^2}\cdot\frac{n^2}{(x-1)^n}\right|=|x-1|\lim\frac{n^2}{(n+1)^2}=|x-1|

. Setting

|x-1|<1

gives

R=1

and open interval

(0,2)

. At

x=0

\sum\frac{(-1)^n}{n^2}

converges absolutely (comparison to the

p

-series

p=2

). At

x=2

\sum\frac{1}{n^2}

converges as a

p

-series with

p=2>1

. Both endpoints converge, so the IOC is the closed interval

[0,2]

A power series centered at

c=3

has radius of convergence

R=2

. Its series diverges at

x=1

and converges at

x=5

. Which interval could be its interval of convergence?

$(1,5)$
$(1,5]$
$[1,5]$
$[1,5)$

Answer: $(1,5]$

With center

3

and

R=2

, the endpoints are

x=3-2=1

and

x=3+2=5

. The series diverges at

x=1

, so that endpoint is excluded (parenthesis), and it converges at

x=5

, so that endpoint is included (square bracket). The correct notation is therefore

(1,5]

. This illustrates that the two endpoints are tested independently and can behave differently.

FAQ

Why doesn't the ratio test tell me what happens at the endpoints?: At an endpoint $x=c\pm R$ the ratio-test limit equals exactly $1$ , and the ratio test is explicitly inconclusive when $L=1$ . That is why you must substitute each endpoint back into the series and use a different tool — the alternating series test, $p$ -series test, or a comparison — to decide convergence there.
How do I choose brackets versus parentheses in the interval of convergence?: Use a square bracket when that endpoint's series converges (so the point is included) and a parenthesis when it diverges (so the point is excluded). Test each endpoint separately; the left and right ends can require different brackets, giving intervals like $[a,b)$ or $(a,b]$ .
What does a radius of convergence of 0 or infinity mean?: $R=0$ means the series converges only at its center $x=c$ and nowhere else — the interval is just the single point $\{c\}$ . $R=\infty$ means the ratio-test limit is $0$ for every $x$ , so the series converges for all real numbers and the interval is $(-\infty,\infty)$ with no endpoints to check.
Is convergence inside the interval always absolute?: Yes. The ratio test works with absolute values, so on the open interval $(c-R,c+R)$ the series converges absolutely. Conditional convergence can only occur at an endpoint, such as when substituting a value produces the alternating harmonic series.

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