AP Calculus BC Quiz: Power Series Radius Interval Of Convergence

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Power Series Radius Interval Of ConvergenceQuestion 1 of 20

What is the radius of convergence of $\sum_{n=0}^{\infty}\dfrac{(x-4)^{2n}}{5^n}$ ?

R=\sqrt{5}

R=5

R=\tfrac{1}{\sqrt{5}}

R=\tfrac{1}{5}

R=\infty

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AP Calculus BC Quiz

AP Calculus BC Quiz: Power Series Radius Interval Of Convergence

Practice Power Series Radius Interval Of Convergence in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Power Series Radius Interval Of Convergence, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the radius of convergence of $\sum_{n=0}^{\infty}\dfrac{(x-4)^{2n}}{5^n}$ ?

$R=\sqrt{5}$ (correct answer)
$R=5$
$R=\tfrac{1}{\sqrt{5}}$
$R=\tfrac{1}{5}$
$R=\infty$

Explanation: The skill being tested here is finding the radius of convergence for a power series. Apply the ratio test to (x-4)^{2n}/5^n, giving a limit of (x-4)^2/5, which is less than 1 when |x-4| < √5. This indicates the radius is √5. The even powers effectively make it a geometric series in terms of (x-4)^2/5. A tempting distractor like R=1/√5 fails by incorrectly applying the root test without accounting for the exponent. Always remember to use the ratio or root test for the radius and separately test endpoints for conditional convergence in power series problems.

Question 2

Find the radius of convergence of $\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^2}\,x^n$ .

$R=0$
$R=\tfrac{1}{4}$ (correct answer)
$R=\tfrac{1}{2}$
$R=1$
$R=4$

Explanation: The skill being tested here is finding the radius of convergence for a power series. Apply the ratio test to the coefficients (2n)!/(n!)^2 x^n, resulting in a limit of 4|x| as n approaches infinity. The series converges when 4|x| < 1, so |x| < 1/4, giving a radius of 1/4. This central binomial coefficient series is known to have this radius, confirming the calculation. A tempting distractor like R=1/2 fails because it miscalculates the limit by overlooking the factorial growth rates properly. Always remember to use the ratio or root test for the radius and separately test endpoints for conditional convergence in power series problems.

Question 3

Find the radius of convergence of $\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\,n^2}$ in a pure convergence example.

$R=0$
$R=1$ (correct answer)
$R=2$
$R=\frac12$
$R=\infty$

Explanation: The skill here is finding the radius and interval of convergence for a power series. To find the radius, apply the ratio test:

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

, so

R=1

. Since the question asks only for the radius, we stop here. The n^3 in the denominator gives

R=1

. A tempting distractor is

R=\infty

, but this fails because the polynomial growth does not outpace the exponential. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.

Question 4

What is the radius of convergence of $\sum_{n=0}^{\infty}\frac{(x+1)^{n}}{7^{n}}$ for a geometric-series approximation?

$R=7$ (correct answer)
$R=1$
$R=\frac17$
$R=8$
$R=\infty$

Explanation: The skill being tested is finding the radius of convergence for a power series.

\sum (x+1)^n / 7^n

, geometric,

|x+1|/7 < 1

R=7

. Simple ratio. No endpoints. Tempting

R=1/7

inverts. Apply geometric series formula directly.

Question 5

Find the interval of convergence for $\sum_{n=0}^{\infty}\frac{(x-2)^n}{3^n}$ used to approximate a response curve.

$(-1,5)$ (correct answer)
$[-1,5]$
$(-1,5]$
$[-1,5)$
$(-\infty,\infty)$

Explanation: The skill being tested is finding the interval of convergence for a power series.

\sum \frac{(x-2)^n}{3^n}

, geometric with

r=\frac{x-2}{3}

|r|<1

|x-2|<3

(-1,5)

. Endpoints: at -1,

\sum \left[ \frac{-3}{3} \right]^n = \sum (-1)^n

, alternates but

|

term

|=1

not to 0, diverges. At 5,

\sum \left( \frac{3}{3} \right)^n=\sum 1

, diverges. So open

(-1,5)

. Yes A. Tempting

[-1,5]

fails as terms don't go to 0. Verify term limit to 0 at endpoints.

Question 6

Determine the radius of convergence of $\sum_{n=1}^{\infty}\frac{3^n(x-1)^n}{n\,4^n}$ for a weighted series model.

$R=\frac{3}{4}$
$R=\frac{4}{3}$ (correct answer)
$R=4$
$R=3$
$R=\infty$

Explanation: The skill here is finding the radius and interval of convergence for a power series. To find the radius, apply the ratio test: the limit as n approaches infinity of

\left| \frac{3^{n+1} (x-1)^{n+1}}{(n+1) 4^{n+1}} \right| / \left| \frac{3^n (x-1)^n}{n 4^n} \right| = |x-1| \cdot \frac{3}{4} \cdot \frac{n}{n+1} \to |x-1| \cdot \frac{3}{4} < 1

, so

|x-1| < \frac{4}{3}

, giving

R = \frac{4}{3}

. Since the question asks only for the radius, we stop here. The weighted ratio

\frac{3}{4}

inverts to

R = \frac{4}{3}

. A tempting distractor is

R = \frac{3}{4}

, but this fails because R is the reciprocal of the limit coefficient. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.

Question 7

Determine the interval of convergence for $\sum_{n=1}^{\infty}\frac{(x+1)^n}{n^2}$ in a convergence test demo.

$(-2,0)$
$[-2,0]$ (correct answer)
$(-2,0]$
$[-2,0)$
$(-\infty,\infty)$

Explanation: The skill being tested is finding the interval of convergence for a power series. Center at -1, ratio

|x+1|

lim

\frac{1}{n+1}

wait no, it's

\sum \frac{(x+1)^n}{n^2}

, ratio

| \frac{a_{n+1}}{a_n} | = |x+1| \cdot \frac{n^2}{(n+1)^2} \to |x+1| < 1

? Wait, lim

= |x+1|

, so converges when

|x+1| < 1

, R=1, open

(-2,0)

. But check endpoints: at x=-2,

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

, alternates, converges absolutely actually since

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

converges. At x=0,

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

converges. So

[-2,0]

. Yes. Tempting

(-2,0)

fails to recognize p-series convergence at endpoints. Use absolute convergence tests at boundaries for inclusion.

Question 8

What is the radius of convergence of $\sum_{n=1}^{\infty}\frac{n^2(x-1)^n}{5^n}$ for a local series model?

$R=5$ (correct answer)
$R=\frac15$
$R=1$
$R=0$
$R=\infty$

Explanation: The skill being tested is finding the radius of convergence for a power series. Ratio test: lim |n^2 (x-1)^n / 5^n * 5^{n+1} / ((n+1)^2 (x-1)^{n+1}) | wait, actually lim |(

a_{n+1}

/a_n)| = |x-1|/5 * lim (n+1)^2 / n^2 = |x-1|/5 <1, so |x-1|<5, R=5. The n^2 grows polynomially, overpowered by exponential 5^n. No endpoints here as question is radius only. Tempting R=1/5 might come from inverting the limit incorrectly. Remember to simplify the limit properly in ratio test for accurate radius in various series.

Question 9

Determine the radius of convergence of $\sum_{n=2}^{\infty}\frac{(x-3)^n}{(n-1)2^n}$ for an iterative method.

$R=1$
$R=2$ (correct answer)
$R=3$
$R=\frac12$
$R=\infty$

Explanation: The skill here is finding the radius and interval of convergence for a power series. To find the radius, apply the ratio test: the limit as n approaches infinity of

|(x-3)^{n+1}/(n 2^{n+1})| / |(x-3)^n/((n-1) 2^n)| =

|x-3|/2 \cdot ((n-1)/n) \rightarrow |x-3|/2 < 1

, so

|x-3| < 2

, giving

R=2

. Since the question asks only for the radius, we stop here, but endpoints would need checking for the full interval. The shift in indexing from n=2 does not affect the limit. A tempting distractor is

R=1/2

, but this fails because the 2^n in the denominator gives

R=2$. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.

Question 10

Determine the radius of convergence of $\sum_{n=1}^{\infty} \dfrac{(x-5)^n}{n\,4^n}$ .

$R=4$ (correct answer)
$R=\tfrac{1}{4}$
$R=1$
$R=0$
$R=\infty$

Explanation: Finding the radius of convergence requires applying the ratio test to this power series. We compute

\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty}|\frac{(x-5)n}{4(n+1)}| = \frac{|x-5|}{4}\lim_{n\to\infty}\frac{n}{n+1} = \frac{|x-5|}{4}

. The series converges when

\frac{|x-5|}{4} < 1

, which means

|x-5| < 4

, giving radius

R = 4

. The interval before checking endpoints would be

(1, 9)

. A common mistake is to think the denominator

4^n

means

R = \frac{1}{4}

, but the ratio test shows the radius equals the reciprocal of the coefficient in the limit. Remember that for series of the form

\sum \frac{(x-c)^n}{a^n}

, the radius of convergence is

R = a

Question 11

What is the radius of convergence of $\sum_{n=1}^{\infty} \dfrac{n^2(x-1)^n}{3^n}$ ?

$R=\tfrac{1}{3}$
$R=3$ (correct answer)
$R=1$
$R=0$
$R=\infty$

Explanation: Finding the radius of convergence requires applying the ratio test to this power series. We compute

\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty}|\frac{(n+1)^2(x-1)}{3n^2}| = \frac{|x-1|}{3}\lim_{n\to\infty}\frac{(n+1)^2}{n^2} = \frac{|x-1|}{3}

. The series converges when

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

, which means

|x-1| < 3

, so the radius is

R = 3

. The presence of

n^2

in the numerator doesn't change the radius since

\lim_{n\to\infty}\frac{(n+1)^2}{n^2} = 1

. A common error is thinking the

n^2

factor affects the radius, but polynomial factors in

n

don't change the ratio test limit. Remember that the radius depends on the exponential growth rate, not polynomial factors in the coefficients.

Question 12

Determine the radius of convergence of $\sum_{n=0}^{\infty}\dfrac{(x-2)^n}{n!}$ .

$R=0$
$R=1$
$R=2$
$R=\infty$ (correct answer)
$R=\dfrac{1}{2}$

Explanation: This problem involves finding the radius of convergence for the exponential series. The series

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

is the Taylor series for

e^{x-2}

. Using the ratio test,

\lim_{n\to\infty}\left|\frac{(x-2)\cdot n!}{(n+1)!}\right| = \lim_{n\to\infty}\frac{|x-2|}{n+1} = 0

for any finite

x

. Since this limit is always 0 < 1, the series converges for all real

x

, giving radius

R=\infty

. Students might think

R=1

by analogy with geometric series, but factorial growth in the denominator overwhelms any polynomial growth in the numerator. The exponential function's Taylor series converges everywhere, illustrating that some power series have infinite radius of convergence.

Question 13

Determine the radius of convergence of $\sum_{n=0}^{\infty}\frac{(2x+1)^n}{n!}$ .

$R=\tfrac{1}{2}$
$R=1$
$R=2$
$R=0$
$R=\infty$ (correct answer)

Explanation: Recognizing special power series patterns is crucial for finding convergence properties efficiently. The series

\sum_{n=0}^{\infty}\frac{(2x+1)^n}{n!}

has the factorial

n!

in the denominator, which is characteristic of the exponential function series. Applying the ratio test:

\lim_{n\to\infty}\left|\frac{(2x+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{(2x+1)^n}\right| = \lim_{n\to\infty}\frac{|2x+1|}{n+1} = 0

for all

x

. Since this limit is always less than 1, the series converges for all real

x

, giving

R=\infty

. Students might be tempted to think the coefficient 2 affects the radius, but factorial growth dominates any polynomial or exponential factors. Remember that series with

n!

in the denominator always converge everywhere.

Question 14

For $P(x)=\sum_{n=1}^{\infty}\frac{(x-4)^n}{n^2}$ , what is the interval of convergence?

$(-\infty,\infty)$
$(-1,9)$
$[3,5]$ (correct answer)
$[3,5)$
$(3,5)$

Explanation: The skill here is finding the interval of convergence for a power series. To determine the radius, the ratio test gives lim |x-4| n^2 / (n+1)^2 = |x-4| < 1, so radius R=1 and open interval 3 < x < 5. Check endpoint x=3: sum (-1)^n / n^2 converges absolutely by p-series with p=2 >1. At x=5: sum 1/n^2 also converges by p-series. A tempting distractor is (3,5), failing because it excludes the endpoints where the series actually converges. A transferable convergence-interval strategy is to first compute the radius using the ratio or root test, then evaluate convergence at each endpoint using tests like p-series, alternating series, or comparison.

Question 15

A concentration is approximated by $C(x)=\sum_{n=0}^{\infty}\frac{(2x-1)^n}{7^n}$ ; what is the interval of convergence?

$\left(-\infty,\infty\right)$
$\left(-3,4\right)$ (correct answer)
$\left[-3,4\right]$
$\left(-3,4\right]$
$\left[-3,4\right)$

Explanation: The skill here is finding the interval of convergence for a power series. This is a geometric series with ratio

r = \frac{2x-1}{7}

, converging when

|r| < 1

, so

|2x-1| < 7

, yielding interval

-3 < x < 4

and radius

\frac{7}{2}

centered at

x=\frac{1}{2}

. At endpoints

x=-3

and

x=4

|r|=1

, so the geometric series diverges. No convergence at endpoints. A tempting distractor is

[-3,4]

, failing because geometric series diverge at

|r|=1

. A transferable convergence-interval strategy is to first compute the radius using the ratio or root test, then evaluate convergence at each endpoint using tests like p-series, alternating series, or comparison.

Question 16

For $h(x)=\sum_{n=1}^{\infty}\frac{(x-3)^n}{n}$ , what is the interval of convergence?

$(-\infty,\infty)$
$[2,4]$
$(2,4)$
$[2,4)$ (correct answer)
$(2,4]$

Explanation: The skill here is finding the interval of convergence for a power series. To determine the radius, the ratio test yields

\lim |x-3| \frac{n}{n+1} = |x-3| < 1

, so R=1 and open interval

2 < x < 4

. Check endpoint x=2:

\sum (-1)^n / n

converges by alternating series test. At x=4:

\sum 1/n

diverges as harmonic series. A tempting distractor is

(2,4)

, failing because it excludes x=2 where the series converges. A transferable convergence-interval strategy is to first compute the radius using the ratio or root test, then evaluate convergence at each endpoint using tests like p-series, alternating series, or comparison.

Question 17

Determine the radius of convergence of $\sum_{n=1}^{\infty}\frac{(x-3)^n}{n\,(-2)^n}$ for a series-based filter.

$R=2$ (correct answer)
$R=-2$
$R=\frac12$
$R=3$
$R=\infty$

Explanation: The skill being tested is finding the radius of convergence for a power series. Term

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

, ratio

\frac{|x-3|}{|-2|} \cdot \lim \frac{n}{n+1} = \frac{|x-3|}{2} < 1

, R=2. The negative in denominator doesn't affect radius. No endpoints. Tempting R=-2 invalid as radius positive. Include absolute value in ratio for correct R.

Question 18

What is the interval of convergence for $\sum_{n=0}^{\infty}\frac{(x-4)^n}{2^{2n}}$ in a geometric-series calibration?

$(0,8)$ (correct answer)
[0,8]
[0,8)
(0,8]
$(-\infty,\infty)$

Explanation: The skill here is finding the interval of convergence for a power series. Recognize this as a geometric series

\sum [(x-4)/4]^n

, which converges when

|(x-4)/4| < 1

, so

|x-4| < 4

, giving open interval

(0, 8)

. Then, check the endpoints: at x=0, r=-1,

\sum (-1)^n

diverges; at x=8, r=1,

\sum 1^n

diverges. Neither endpoint converges. A tempting distractor is [0,8], but this fails because geometric series diverge at

|r|=1

. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.

Question 19

What is the radius of convergence of $\sum_{n=1}^{\infty}\frac{(x+3)^n}{n\,12^n}$ for a log-series computation?

$R=12$ (correct answer)
$R=\frac{1}{12}$
$R=3$
$R=9$
$R=\infty$

Explanation: The skill here is finding the radius and interval of convergence for a power series. To find the radius, apply the ratio test: the limit as n approaches infinity of

\left| \frac{(x+3)^{n+1}}{(n+1) 12^{n+1}} \right| / \left| \frac{(x+3)^n}{n 12^n} \right| = \frac{|x+3|}{12} < 1

, so

|x+3| < 12

, giving

R=12

. Since the question asks only for the radius, we stop here. The large 12^n gives a large radius. A tempting distractor is

R=\frac{1}{12}

, but this fails because R is the reciprocal of the coefficient in the limit. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.

Question 20

Determine the interval of convergence for $\sum_{n=0}^{\infty}(-1)^n\dfrac{(x-2)^n}{4^n}$ .

$(-2,6)$ (correct answer)
$[-2,6]$
$(-2,6]$
$[-2,6)$
$(-\infty,\infty)$

Explanation: The skill being tested here is finding the interval of convergence for a power series. This is a geometric series with ratio

r = -(x-2)/4

, converging when

|r| < 1

, or

|x-2| < 4

, giving

(-2,6)

. At the endpoints

x=-2

and

x=6

, the series oscillates and diverges as

|r|=1

. Geometric series only converge inside the open interval. A tempting distractor like

[-2,6]

fails because it includes endpoints where the terms don't approach zero. Always remember to use the ratio or root test for the radius and separately test endpoints for conditional convergence in power series problems.