AP Calculus BC Quiz: Ratio Test For Convergence

What this quiz covers

This quiz focuses on Ratio Test For 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.

Question 1

1. Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=2$.
2. Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1$. (correct answer)
3. Inconclusive because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1$.
5. Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0$.

Explanation: The ratio test is a key method for determining the convergence of infinite series, particularly those involving linear terms over exponentials. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for a_n = n / 2^n, the ratio |a_{n+1}/a_n| = ((n+1)/n) / 2. As n approaches infinity, this limit is 1/2, which is less than 1. Therefore, the ratio test indicates that the series converges. A tempting distractor might be choice A, which claims divergence with limit 2, but this misapplies the ratio by inverting it. Always remember that in the ratio test, for series like geometric with polynomial adjustments, a limit less than 1 ensures convergence.

Question 2

1. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1$.
2. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
3. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1$. (correct answer)
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1$.

Explanation: This problem requires applying the ratio test to determine series convergence. For $\sum_{n=1}^{\infty} \frac{n!}{5^n}$, we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!/5^{n+1}}{n!/5^n} = \frac{(n+1)! cdot 5^n}{n! cdot 5^{n+1}} = \frac{n+1}{5}$. As $n \to \infty$, this ratio approaches $\infty$, which is greater than 1, so the series diverges. Choice A incorrectly suggests the limit is $\frac{1}{5}$, likely by mishandling the factorial terms. The key insight is that factorial growth in the numerator eventually overwhelms any exponential base in the denominator.

Question 3

1. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=2>1$.
2. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
3. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1$. (correct answer)
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1$.

Explanation: This problem requires applying the ratio test to a series with linear and exponential terms. For $\sum_{n=1}^{\infty} \frac{n}{2^n}$, we compute $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{2n} = \frac{1}{2} \cdot \frac{n+1}{n}$. As $n \to \infty$, this approaches $\frac{1}{2} \cdot 1 = \frac{1}{2} < 1$, so the series converges. Choice A incorrectly claims the limit is 2, likely by mishandling the exponential terms. The exponential denominator with base 2 grows fast enough to overcome linear growth in the numerator.

Question 4

1. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
2. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1$.
3. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1$.
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$. (correct answer)
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.

Explanation: This problem requires applying the ratio test to a series with polynomial numerator and factorial denominator. For $\sum_{n=1}^{\infty} \frac{n^4}{n!}$, we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^4/(n+1)!}{n^4/n!} = \frac{(n+1)^4 \cdot n!}{n^4 \cdot (n+1)!} = \frac{(n+1)^4}{n^4(n+1)} = \frac{(n+1)^3}{n^4} = \frac{1}{n} \cdot \left(\frac{n+1}{n}\right)^3$. As $n \to \infty$, this approaches $0 < 1$, so the series converges. Choice B incorrectly suggests the limit is infinity and divergence, missing that factorial growth dominates polynomial growth. Factorials in denominators almost always lead to convergence.

Question 5

1. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.
2. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=5>1$. (correct answer)
3. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1$.
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1$.

Explanation: This problem involves applying the ratio test to a series with exponential numerator and polynomial denominator. For $\sum_{n=1}^{\infty} \frac{5^n}{n^5}$, we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{5^{n+1}/(n+1)^5}{5^n/n^5} = 5 \cdot \left(\frac{n}{n+1}\right)^5$. As $n \to \infty$, this becomes $5 \cdot 1^5 = 5 > 1$, so the series diverges. Choice D incorrectly suggests the limit is $\frac{1}{5}$ and convergence, missing that exponential growth with base 5 dominates fifth-degree polynomial growth. When the base of an exponential exceeds 1, it will eventually overpower any polynomial denominator.

Question 6

1. Diverges because the ratio-test limit is $7>1$
2. Inconclusive because the ratio-test limit is $1$
3. Converges absolutely because the ratio-test limit is $0<1$ (correct answer)
4. Diverges because the ratio-test limit is $\tfrac{1}{7}>1$
5. Converges absolutely because the ratio-test limit is $\tfrac{1}{7}<1$

Explanation: This problem involves ratio test with factorial denominators. For $\sum_{n=1}^{\infty} \frac{7^n}{(2n)!}$, we find $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{7^{n+1}/(2n+2)!}{7^n/(2n)!} = \lim_{n \to \infty} \frac{7}{(2n+2)(2n+1)} = 0$. Since the limit is 0 (less than 1), the series converges absolutely by the ratio test. Choice A incorrectly claims the limit is 7, missing that $(2n+2)! = (2n+2)(2n+1)(2n)!$ creates quadratic growth in the denominator. Double factorial $(2n)!$ grows much faster than single factorial $n!$, dominating any exponential numerator.

Question 7

1. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{6}<1$. (correct answer)
2. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=6>1$.
3. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{6}<1$.

Explanation: This question tests the ratio test on an alternating series with polynomial and exponential terms. For $\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{6^n}$, we apply the ratio test to the absolute values: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^2/6^{n+1}}{n^2/6^n} = \frac{(n+1)^2}{6n^2} = \frac{1}{6} \cdot \left(\frac{n+1}{n}\right)^2$. Taking the limit: $\lim_{n \to \infty} \frac{1}{6} \cdot 1^2 = \frac{1}{6} < 1$, so the series converges absolutely. Choice B incorrectly states the limit is 6, inverting the result. The alternating sign doesn't affect the ratio test calculation when we use absolute values.

Question 8

1. Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$.
2. Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{e}<1$. (correct answer)
3. Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{e}<1$.
5. Inconclusive because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=e$.

Explanation: The ratio test is a key method for determining the convergence of infinite series, particularly those involving factorials and exponential-like terms. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for a_n = n! / n^n, the ratio |a_{n+1}/a_n| = (n/(n+1))^n. As n approaches infinity, this limit is 1/e, which is less than 1. Therefore, the ratio test indicates that the series converges. A tempting distractor might be choice A, which claims divergence with limit 1, but this overlooks the precise limit involving e. Always remember that in the ratio test, recognizing limits like 1/e can confirm convergence for series balancing factorials and powers.

Question 9

1. Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=2>1$.
2. Inconclusive because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$.
3. Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1$. (correct answer)
4. Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1$.
5. Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty$.

Explanation: The ratio test is a key method for determining the convergence of infinite series, particularly those involving factorials or exponentials, even with alternating signs since we use absolute values. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for a_n = (-2)^n / n!, the ratio |a_{n+1}/a_n| = 2/(n+1). As n approaches infinity, this limit is 0, which is less than 1. Therefore, the ratio test indicates that the series converges absolutely. A tempting distractor might be choice A, which claims divergence with limit 2 > 1, but this neglects the factorial's growth. Always remember that in the ratio test, a limit of 0 strongly suggests convergence, especially for exponential series with factorial denominators.

Question 10

1. Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1$. (correct answer)
2. Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=2>1$.
3. Inconclusive because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1$.
5. Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty$.

Explanation: The ratio test is a key method for determining the convergence of infinite series, particularly those involving factorials and polynomials. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for a_n = (n+2) 2^n / n!, the ratio |a_{n+1}/a_n| = 2 (n+3) / ((n+1)(n+2)). As n approaches infinity, this limit is 0, which is less than 1. Therefore, the ratio test indicates that the series converges. A tempting distractor might be choice B, which claims divergence with limit 2 > 1, but this ignores the factorial's overpowering effect. Always remember that in the ratio test, when polynomials are multiplied by exponentials over factorials, the limit often approaches 0, confirming convergence.

Question 11

1. Diverges because the ratio-test limit equals $4$
2. Inconclusive because the ratio-test limit equals $1$
3. Converges absolutely because the ratio-test limit equals $0$ (correct answer)
4. Diverges because the ratio-test limit equals $\infty$
5. Converges absolutely because the ratio-test limit equals $\tfrac{1}{4}$

Explanation: The skill here is using the ratio test to determine the convergence of infinite series. To apply the ratio test, compute the limit L = lim_{n→∞} |a_{n+1}/a_n| where a_n = 4^n / (2n)!, so |a_{n+1}/a_n| = 4 / ((2n+1)(2n+2)). The limit is 0 as n approaches infinity. Since L = 0 < 1, the series converges absolutely. A tempting distractor is choice E, which claims convergence with limit 1/4, but this fails because the double factorial growth in the denominator produces 0, not 1/4. A transferable ratio-test strategy is to simplify the expression for |a_{n+1}/a_n| carefully, evaluate the limit, and conclude absolute convergence if less than 1, divergence if greater than 1, or inconclusive if equal to 1.

Question 12

1. Diverges because the ratio-test limit equals $3$
2. Converges absolutely because the ratio-test limit equals $0$ (correct answer)
3. Inconclusive because the ratio-test limit equals $1$
4. Diverges because the ratio-test limit equals $\infty$
5. Converges absolutely because the ratio-test limit equals $\tfrac{1}{3}$

Explanation: The skill here is using the ratio test to determine the convergence of infinite series. To apply the ratio test, compute the limit L = lim_{n→∞} |a_{n+1}/a_n| where a_n = 3^n / n!, so |a_{n+1}/a_n| = 3/(n+1). The limit is 0 as n approaches infinity. Since L = 0 < 1, the series converges absolutely. A tempting distractor is choice E, which suggests convergence with limit 1/3, but this fails because it incorrectly places the 3 in the denominator instead of the numerator in the ratio. A transferable ratio-test strategy is to simplify the expression for |a_{n+1}/a_n| carefully, evaluate the limit, and conclude absolute convergence if less than 1, divergence if greater than 1, or inconclusive if equal to 1.

Question 13

1. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.
2. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=5>1$. (correct answer)
3. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1$.
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1$.

Explanation: This problem involves applying the ratio test to a series with exponential numerator and polynomial denominator. For $\sum_{n=1}^{\infty} \frac{5^n}{n^5}$, we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{5^{n+1}/(n+1)^5}{5^n/n^5} = 5 \cdot \left(\frac{n}{n+1}\right)^5$. As $n \to \infty$, this becomes $5 \cdot 1^5 = 5 > 1$, so the series diverges. Choice D incorrectly suggests the limit is $\frac{1}{5}$ and convergence, missing that exponential growth with base 5 dominates fifth-degree polynomial growth. When the base of an exponential exceeds 1, it will eventually overpower any polynomial denominator.

Question 14

1. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=2>1$. (correct answer)
2. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1$.
3. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.
4. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.

Explanation: This question tests using the ratio test on a series with exponential and polynomial terms. For $\sum_{n=1}^{\infty} \frac{2^n}{n^3}$, we compute $\left|\frac{a_{n+1}}{a_n}\right| = \frac{2^{n+1}/(n+1)^3}{2^n/n^3} = \frac{2n^3}{(n+1)^3} = 2 cdot \left(\frac{n}{n+1}\right)^3$. Taking the limit: $\lim_{n \to \infty} 2 cdot \left(\frac{n}{n+1}\right)^3 = 2 cdot 1^3 = 2 > 1$, so the series diverges. Choice B incorrectly claims the limit is $\frac{1}{2}$, perhaps by inverting the ratio. Remember that exponential growth with base greater than 1 will eventually dominate any polynomial growth.

Question 15

1. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3}<1$.
2. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1$. (correct answer)
3. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3}<1$.

Explanation: This question tests the ratio test on a series with factorial and exponential terms. For $\sum_{n=1}^{\infty} \frac{(n+1)!}{3^n}$, we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+2)!/3^{n+1}}{(n+1)!/3^n} = \frac{(n+2)!}{(n+1)!} \cdot \frac{3^n}{3^{n+1}} = \frac{n+2}{3}$. Taking the limit: $\lim_{n \to \infty} \frac{n+2}{3} = \infty > 1$, so the series diverges. Choice A incorrectly suggests convergence with limit $\frac{1}{3}$, missing that the factorial growth dominates. Remember that factorial terms in numerators typically cause divergence unless balanced by stronger denominators.

Question 16

1. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3}<1$.
2. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1$. (correct answer)
3. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$.
5. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3}<1$.

Explanation: This question tests the ratio test on a series with factorial and exponential terms. For $\sum_{n=1}^{\infty} \frac{(n+1)!}{3^n}$, we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+2)!/3^{n+1}}{(n+1)!/3^n} = \frac{(n+2)!}{(n+1)!} \cdot \frac{3^n}{3^{n+1}} = \frac{n+2}{3}$. Taking the limit: $\lim_{n \to \infty} \frac{n+2}{3} = \infty > 1$, so the series diverges. Choice A incorrectly suggests convergence with limit $\frac{1}{3}$, missing that the factorial growth dominates. Remember that factorial terms in numerators typically cause divergence unless balanced by stronger denominators.

Question 17

1. Converges absolutely because the ratio-test limit is $0<1$
2. Diverges because the ratio-test limit is $\infty$
3. Inconclusive because the ratio-test limit is $1$ (correct answer)
4. Diverges because the ratio-test limit is $\tfrac{1}{2}>1$
5. Converges absolutely because the ratio-test limit is $\tfrac{1}{2}<1$

Explanation: This question examines ratio test application to a simple rational series. For $\sum_{n=1}^{\infty} \frac{n}{n+1}$, we calculate $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(n+1)/(n+2)}{n/(n+1)} = \lim_{n \to \infty} \frac{(n+1)^2}{n(n+2)} = 1$. Since the limit equals 1, the ratio test is inconclusive—it cannot determine convergence or divergence. Choice E incorrectly suggests the limit is $\frac{1}{2}$, perhaps from algebraic error. When the ratio test yields 1, you must use other tests; note that this series diverges since $\lim_{n \to \infty} \frac{n}{n+1} = 1 \neq 0$.

Question 18

1. Diverges because the ratio-test limit is $3>1$
2. Converges absolutely because the ratio-test limit is $\tfrac{1}{3}<1$ (correct answer)
3. Inconclusive because the ratio-test limit is $1$
4. Diverges because the ratio-test limit is $\tfrac{1}{3}>1$
5. Converges absolutely because the ratio-test limit is $0<1$

Explanation: This question involves applying the ratio test to an alternating series. For $\sum_{n=1}^{\infty} \frac{(-1)^n n^3}{3^n}$, we use absolute values: $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(n+1)^3/3^{n+1}}{n^3/3^n} = \lim_{n \to \infty} \frac{(n+1)^3}{3n^3} = \frac{1}{3}$. Since $\frac{1}{3} < 1$, the series converges absolutely by the ratio test. Choice A incorrectly claims the limit is 3, possibly from inverting the fraction. The ratio test examines absolute convergence, so the alternating sign doesn't affect the limit calculation.

Question 19

1. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=3>1$.
2. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$. (correct answer)
3. Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.
4. Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3}<1$.
5. Inconclusive because $\lim \left|\frac{a_{n+1}}{a_n}\right|=1$.

Explanation: This question tests the ratio test for convergence of infinite series. For the series $\sum_{n=1}^{\infty} \frac{3^n}{n!}$, we compute $\left|\frac{a_{n+1}}{a_n}\right| = \frac{3^{n+1}/(n+1)!}{3^n/n!} = \frac{3^{n+1} cdot n!}{3^n cdot (n+1)!} = \frac{3}{n+1}$. Taking the limit as $n \to \infty$, we get $\lim_{n \to \infty} \frac{3}{n+1} = 0 < 1$, so the series converges by the ratio test. Choice A incorrectly claims the limit is 3, likely by forgetting the factorial growth in the denominator. When applying the ratio test, always simplify the ratio completely before taking the limit, and remember that factorial growth dominates exponential growth.

Question 20

1. Converges absolutely because the ratio-test limit is $\tfrac{4}{5}<1$ (correct answer)
2. Diverges because the ratio-test limit is $\tfrac{4}{5}>1$
3. Inconclusive because the ratio-test limit is $1$
4. Diverges because the ratio-test limit is $\infty$
5. Converges absolutely because the ratio-test limit is $\tfrac{5}{4}<1$

Explanation: This problem requires careful ratio test computation with complex factorial expressions. For $\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 5^n}$, we calculate $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(2n+2)!/(n+1)!^2 5^{n+1}}{(2n)!/(n!)^2 5^n}$. After simplification using $(2n+2)! = (2n+2)(2n+1)(2n)!$ and $(n+1)! = (n+1)n!$, this becomes $\lim_{n \to \infty} \frac{(2n+2)(2n+1)}{5(n+1)^2} = \frac{4}{5}$. Since $\frac{4}{5} < 1$, the series converges absolutely. Choice E incorrectly inverts to get $\frac{5}{4}$. When handling double factorials or squared factorials, track each term's growth rate carefully.