AP Calculus BC Quiz: Ratio Test For Convergence
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Ratio Test For ConvergenceQuestion 1 of 20

Use the ratio test on ∑n=1∞4n(2n)!\sum_{n=1}^{\infty} \frac{4^n}{(2n)!} to determine convergence behavior.

Diverges because the ratio-test limit equals 44
Inconclusive because the ratio-test limit equals 11
Converges absolutely because the ratio-test limit equals 00
Diverges because the ratio-test limit equals ∞\infty
Converges absolutely because the ratio-test limit equals 14\tfrac{1}{4}
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AP Calculus BC Quiz

AP Calculus BC Quiz: Ratio Test For Convergence

Practice Ratio Test For 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 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.

All questions

Question 1

Use the ratio test on ∑n=1∞4n(2n)!\sum_{n=1}^{\infty} \frac{4^n}{(2n)!} to determine convergence behavior.

  1. Diverges because the ratio-test limit equals 44
  2. Inconclusive because the ratio-test limit equals 11
  3. Converges absolutely because the ratio-test limit equals 00 (correct answer)
  4. Diverges because the ratio-test limit equals ∞\infty
  5. Converges absolutely because the ratio-test limit equals 14\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 2

For ∑n=1∞3nn!\sum_{n=1}^{\infty} \frac{3^n}{n!}, what does the ratio test conclude about convergence?

  1. Diverges because the ratio-test limit equals 33
  2. Converges absolutely because the ratio-test limit equals 00 (correct answer)
  3. Inconclusive because the ratio-test limit equals 11
  4. Diverges because the ratio-test limit equals ∞\infty
  5. Converges absolutely because the ratio-test limit equals 13\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 3

A series for an algorithm's error is ∑n=1∞n2n\sum_{n=1}^{\infty} \frac{n}{2^n}; decide convergence using the ratio test.

  1. Diverges because lim⁡n→∞∣an+1an∣=2\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=2.
  2. Converges because lim⁡n→∞∣an+1an∣=12<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1. (correct answer)
  3. Inconclusive because lim⁡n→∞∣an+1an∣=1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Diverges because lim⁡n→∞∣an+1an∣=12<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1.
  5. Converges because lim⁡n→∞∣an+1an∣=0\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 4

Apply the ratio test to decide the behavior of ∑n=1∞n!5n\sum_{n=1}^{\infty} \frac{n!}{5^n}.

  1. Converges because lim⁡∣an+1an∣=15<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1.
  2. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  3. Diverges because lim⁡∣an+1an∣=∞>1\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1. (correct answer)
  4. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1.
  5. Diverges because lim⁡∣an+1an∣=15<1\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 ∑n=1∞n!5n\sum_{n=1}^{\infty} \frac{n!}{5^n}, we calculate ∣an+1an∣=(n+1)!/5n+1n!/5n=(n+1)!cdot5nn!cdot5n+1=n+15\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→∞n \to \infty, this ratio approaches ∞\infty, which is greater than 1, so the series diverges. Choice A incorrectly suggests the limit is 15\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 5

Use the ratio test to determine whether ∑n=1∞n2n\sum_{n=1}^{\infty} \frac{n}{2^n} converges or diverges.​

  1. Diverges because lim⁡∣an+1an∣=2>1\lim \left|\frac{a_{n+1}}{a_n}\right|=2>1.
  2. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  3. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1.
  4. Converges because lim⁡∣an+1an∣=12<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1. (correct answer)
  5. Diverges because lim⁡∣an+1an∣=12<1\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 ∑n=1∞n2n\sum_{n=1}^{\infty} \frac{n}{2^n}, we compute ∣an+1an∣=(n+1)/2n+1n/2n=n+12n=12⋅n+1n\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→∞n \to \infty, this approaches 12⋅1=12<1\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 6

Use the ratio test to determine convergence of ∑n=1∞n4n!\sum_{n=1}^{\infty} \frac{n^4}{n!}.​

  1. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  2. Diverges because lim⁡∣an+1an∣=∞>1\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1.
  3. Converges because lim⁡∣an+1an∣=∞>1\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1.
  4. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1. (correct answer)
  5. Diverges because lim⁡∣an+1an∣=0<1\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 ∑n=1∞n4n!\sum_{n=1}^{\infty} \frac{n^4}{n!}, we calculate ∣an+1an∣=(n+1)4/(n+1)!n4/n!=(n+1)4⋅n!n4⋅(n+1)!=(n+1)4n4(n+1)=(n+1)3n4=1n⋅(n+1n)3\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→∞n \to \infty, this approaches 0<10 < 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 7

Apply the ratio test to ∑n=1∞5nn5\sum_{n=1}^{\infty} \frac{5^n}{n^5} and select the correct conclusion.

  1. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1.
  2. Diverges because lim⁡∣an+1an∣=5>1\lim \left|\frac{a_{n+1}}{a_n}\right|=5>1. (correct answer)
  3. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Converges because lim⁡∣an+1an∣=15<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1.
  5. Diverges because lim⁡∣an+1an∣=15<1\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 ∑n=1∞5nn5\sum_{n=1}^{\infty} \frac{5^n}{n^5}, we calculate ∣an+1an∣=5n+1/(n+1)55n/n5=5⋅(nn+1)5\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→∞n \to \infty, this becomes 5⋅15=5>15 \cdot 1^5 = 5 > 1, so the series diverges. Choice D incorrectly suggests the limit is 15\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 8

Use the ratio test to analyze ∑n=1∞7n(2n)!\sum_{n=1}^{\infty} \frac{7^n}{(2n)!} for convergence.

  1. Diverges because the ratio-test limit is 7>17>1
  2. Inconclusive because the ratio-test limit is 11
  3. Converges absolutely because the ratio-test limit is 0<10<1 (correct answer)
  4. Diverges because the ratio-test limit is 17>1\tfrac{1}{7}>1
  5. Converges absolutely because the ratio-test limit is 17<1\tfrac{1}{7}<1
Explanation: This problem involves ratio test with factorial denominators. For ∑n=1∞7n(2n)!\sum_{n=1}^{\infty} \frac{7^n}{(2n)!}, we find lim⁡n→∞∣an+1an∣=lim⁡n→∞7n+1/(2n+2)!7n/(2n)!=lim⁡n→∞7(2n+2)(2n+1)=0\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)!(2n+2)! = (2n+2)(2n+1)(2n)! creates quadratic growth in the denominator. Double factorial (2n)!(2n)! grows much faster than single factorial n!n!, dominating any exponential numerator.

Question 9

Using the ratio test, determine convergence of ∑n=1∞(−1)nn26n\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{6^n}.

  1. Converges because lim⁡∣an+1an∣=16<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{6}<1. (correct answer)
  2. Diverges because lim⁡∣an+1an∣=6>1\lim \left|\frac{a_{n+1}}{a_n}\right|=6>1.
  3. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Converges because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  5. Diverges because lim⁡∣an+1an∣=16<1\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 ∑n=1∞(−1)nn26n\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{6^n}, we apply the ratio test to the absolute values: ∣an+1an∣=(n+1)2/6n+1n2/6n=(n+1)26n2=16⋅(n+1n)2\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→∞16⋅12=16<1\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 10

For ∑n=1∞n!nn\sum_{n=1}^{\infty} \frac{n!}{n^n}, use the ratio test to determine whether it converges.

  1. Diverges because lim⁡n→∞∣an+1an∣=1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1.
  2. Converges because lim⁡n→∞∣an+1an∣=1e<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{e}<1. (correct answer)
  3. Converges because lim⁡n→∞∣an+1an∣=1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Diverges because lim⁡n→∞∣an+1an∣=1e<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{e}<1.
  5. Inconclusive because lim⁡n→∞∣an+1an∣=e\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 11

A damping series is ∑n=1∞(−2)nn!\sum_{n=1}^{\infty} \frac{(-2)^n}{n!}; apply the ratio test to decide convergence.

  1. Diverges because lim⁡n→∞∣an+1an∣=2>1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=2>1.
  2. Inconclusive because lim⁡n→∞∣an+1an∣=1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1.
  3. Converges because lim⁡n→∞∣an+1an∣=0<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1. (correct answer)
  4. Diverges because lim⁡n→∞∣an+1an∣=0<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1.
  5. Converges because lim⁡n→∞∣an+1an∣=∞\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 12

In a finance model, ∑n=1∞(n+2)2nn!\sum_{n=1}^{\infty} \frac{(n+2)2^n}{n!} appears; use the ratio test for convergence.

  1. Converges because lim⁡n→∞∣an+1an∣=0<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1. (correct answer)
  2. Diverges because lim⁡n→∞∣an+1an∣=2>1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=2>1.
  3. Inconclusive because lim⁡n→∞∣an+1an∣=1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Diverges because lim⁡n→∞∣an+1an∣=0<1\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1.
  5. Converges because lim⁡n→∞∣an+1an∣=∞\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 13

Apply the ratio test to ∑n=1∞5nn5\sum_{n=1}^{\infty} \frac{5^n}{n^5} and select the correct conclusion.​

  1. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1.
  2. Diverges because lim⁡∣an+1an∣=5>1\lim \left|\frac{a_{n+1}}{a_n}\right|=5>1. (correct answer)
  3. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Converges because lim⁡∣an+1an∣=15<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{5}<1.
  5. Diverges because lim⁡∣an+1an∣=15<1\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 ∑n=1∞5nn5\sum_{n=1}^{\infty} \frac{5^n}{n^5}, we calculate ∣an+1an∣=5n+1/(n+1)55n/n5=5⋅(nn+1)5\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→∞n \to \infty, this becomes 5⋅15=5>15 \cdot 1^5 = 5 > 1, so the series diverges. Choice D incorrectly suggests the limit is 15\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

Use the ratio test to determine convergence of ∑n=1∞2nn3\sum_{n=1}^{\infty} \frac{2^n}{n^3}.

  1. Diverges because lim⁡∣an+1an∣=2>1\lim \left|\frac{a_{n+1}}{a_n}\right|=2>1. (correct answer)
  2. Converges because lim⁡∣an+1an∣=12<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1.
  3. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1.
  4. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  5. Diverges because lim⁡∣an+1an∣=0<1\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 ∑n=1∞2nn3\sum_{n=1}^{\infty} \frac{2^n}{n^3}, we compute ∣an+1an∣=2n+1/(n+1)32n/n3=2n3(n+1)3=2cdot(nn+1)3\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→∞2cdot(nn+1)3=2cdot13=2>1\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 12\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

Decide convergence of ∑n=1∞(n+1)!3n\sum_{n=1}^{\infty} \frac{(n+1)!}{3^n} using the ratio test.​

  1. Converges because lim⁡∣an+1an∣=13<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3}<1.
  2. Diverges because lim⁡∣an+1an∣=∞>1\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1. (correct answer)
  3. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1.
  5. Diverges because lim⁡∣an+1an∣=13<1\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 ∑n=1∞(n+1)!3n\sum_{n=1}^{\infty} \frac{(n+1)!}{3^n}, we calculate ∣an+1an∣=(n+2)!/3n+1(n+1)!/3n=(n+2)!(n+1)!⋅3n3n+1=n+23\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→∞n+23=∞>1\lim_{n \to \infty} \frac{n+2}{3} = \infty > 1, so the series diverges. Choice A incorrectly suggests convergence with limit 13\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

Decide convergence of ∑n=1∞(n+1)!3n\sum_{n=1}^{\infty} \frac{(n+1)!}{3^n} using the ratio test.

  1. Converges because lim⁡∣an+1an∣=13<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3}<1.
  2. Diverges because lim⁡∣an+1an∣=∞>1\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1. (correct answer)
  3. Inconclusive because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1.
  5. Diverges because lim⁡∣an+1an∣=13<1\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 ∑n=1∞(n+1)!3n\sum_{n=1}^{\infty} \frac{(n+1)!}{3^n}, we calculate ∣an+1an∣=(n+2)!/3n+1(n+1)!/3n=(n+2)!(n+1)!⋅3n3n+1=n+23\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→∞n+23=∞>1\lim_{n \to \infty} \frac{n+2}{3} = \infty > 1, so the series diverges. Choice A incorrectly suggests convergence with limit 13\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

A power series is ∑n=1∞nn+1\sum_{n=1}^{\infty} \frac{n}{n+1}; what does the ratio test conclude?

  1. Converges absolutely because the ratio-test limit is 0<10<1
  2. Diverges because the ratio-test limit is ∞\infty
  3. Inconclusive because the ratio-test limit is 11 (correct answer)
  4. Diverges because the ratio-test limit is 12>1\tfrac{1}{2}>1
  5. Converges absolutely because the ratio-test limit is 12<1\tfrac{1}{2}<1
Explanation: This question examines ratio test application to a simple rational series. For ∑n=1∞nn+1\sum_{n=1}^{\infty} \frac{n}{n+1}, we calculate lim⁡n→∞∣an+1an∣=lim⁡n→∞(n+1)/(n+2)n/(n+1)=lim⁡n→∞(n+1)2n(n+2)=1\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 12\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→∞nn+1=1≠0\lim_{n \to \infty} \frac{n}{n+1} = 1 \neq 0.

Question 18

Use the ratio test on ∑n=1∞(−1)nn33n\sum_{n=1}^{\infty} \frac{(-1)^n n^3}{3^n} to decide convergence.

  1. Diverges because the ratio-test limit is 3>13>1
  2. Converges absolutely because the ratio-test limit is 13<1\tfrac{1}{3}<1 (correct answer)
  3. Inconclusive because the ratio-test limit is 11
  4. Diverges because the ratio-test limit is 13>1\tfrac{1}{3}>1
  5. Converges absolutely because the ratio-test limit is 0<10<1
Explanation: This question involves applying the ratio test to an alternating series. For ∑n=1∞(−1)nn33n\sum_{n=1}^{\infty} \frac{(-1)^n n^3}{3^n}, we use absolute values: lim⁡n→∞∣an+1an∣=lim⁡n→∞(n+1)3/3n+1n3/3n=lim⁡n→∞(n+1)33n3=13\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 13<1\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

Using the ratio test, determine whether ∑n=1∞3nn!\sum_{n=1}^{\infty} \frac{3^n}{n!} converges or diverges.​

  1. Diverges because lim⁡∣an+1an∣=3>1\lim \left|\frac{a_{n+1}}{a_n}\right|=3>1.
  2. Converges because lim⁡∣an+1an∣=0<1\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1. (correct answer)
  3. Diverges because lim⁡∣an+1an∣=1\lim \left|\frac{a_{n+1}}{a_n}\right|=1.
  4. Converges because lim⁡∣an+1an∣=13<1\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{3}<1.
  5. Inconclusive because lim⁡∣an+1an∣=1\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 ∑n=1∞3nn!\sum_{n=1}^{\infty} \frac{3^n}{n!}, we compute ∣an+1an∣=3n+1/(n+1)!3n/n!=3n+1cdotn!3ncdot(n+1)!=3n+1\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→∞n \to \infty, we get lim⁡n→∞3n+1=0<1\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

Consider ∑n=1∞(2n)!(n!)25n\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 5^n}; what does the ratio test imply?

  1. Converges absolutely because the ratio-test limit is 45<1\tfrac{4}{5}<1 (correct answer)
  2. Diverges because the ratio-test limit is 45>1\tfrac{4}{5}>1
  3. Inconclusive because the ratio-test limit is 11
  4. Diverges because the ratio-test limit is ∞\infty
  5. Converges absolutely because the ratio-test limit is 54<1\tfrac{5}{4}<1
Explanation: This problem requires careful ratio test computation with complex factorial expressions. For ∑n=1∞(2n)!(n!)25n\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 5^n}, we calculate lim⁡n→∞∣an+1an∣=lim⁡n→∞(2n+2)!/(n+1)!25n+1(2n)!/(n!)25n\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)!(2n+2)! = (2n+2)(2n+1)(2n)! and (n+1)!=(n+1)n!(n+1)! = (n+1)n!, this becomes lim⁡n→∞(2n+2)(2n+1)5(n+1)2=45\lim_{n \to \infty} \frac{(2n+2)(2n+1)}{5(n+1)^2} = \frac{4}{5}. Since 45<1\frac{4}{5} < 1, the series converges absolutely. Choice E incorrectly inverts to get 54\frac{5}{4}. When handling double factorials or squared factorials, track each term's growth rate carefully.