Ratio Test for Convergence - AP Calculus BC

For the series $\sum_{n=1}^{\infty} \frac{3^n}{n!}$ , what does the ratio test conclude about convergence?
Answer: Converges absolutely because the ratio-test limit is $0<1$ This problem tests your ability to apply the ratio test for convergence. For the series $\sum_{n=1}^{\infty} \frac{3^n}{n!}$ , we compute $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{3^{n+1}/(n+1)!}{3^n/n!} = \lim_{n \to \infty} \frac{3}{n+1} = 0$ . Since this limit equals 0, which is less than 1, the ratio test tells us the series converges absolutely. Choice A incorrectly claims the limit is 3, likely from forgetting that $(n+1)! = (n+1) \cdot n!$ in the denominator. When applying the ratio test, always simplify the ratio $\frac{a_{n+1}}{a_n}$ completely before taking the limit.
Using the ratio test, determine whether $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$ converges or diverges.
Answer: Converges absolutely because the ratio-test limit is $\tfrac{1}{2}<1$ This question requires using the ratio test to analyze convergence. For $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$ , we calculate $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(n+1)^2/2^{n+1}}{n^2/2^n} = \lim_{n \to \infty} \frac{(n+1)^2}{2n^2}$ . Expanding and simplifying gives $\lim_{n \to \infty} \frac{n^2 + 2n + 1}{2n^2} = \frac{1}{2}$ . Since $\frac{1}{2} < 1$ , the series converges absolutely by the ratio test. Choice B incorrectly states the limit is 2, possibly from inverting the fraction. Remember that exponential growth in the denominator typically dominates polynomial growth in the numerator.
A sensor model uses $\sum_{n=1}^{\infty} \frac{n!}{2^n}$ . What does the ratio test say about convergence?
Answer: Diverges because the ratio-test limit equals $\infty$ 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 $a_n$ = n! / $2^n$ , so | $a_{n+1}$ / $a_n$ | = (n+1)/2. The limit is ∞ as n approaches infinity. Since L = ∞ > 1, the series diverges. A tempting distractor is choice E, which claims divergence with limit 0, but this fails because the factorial in the numerator grows faster than the exponential in the denominator, leading to ∞, not 0. 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.
A damping series is $\sum_{n=1}^{\infty} \frac{(-2)^n}{n!}$ ; apply the ratio test to decide convergence.
Answer: Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1$ . 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.
Use the ratio test on $\sum_{n=1}^{\infty} \frac{(-1)^n n^3}{3^n}$ to decide convergence.
Answer: Converges absolutely because the ratio-test limit is $\tfrac{1}{3}<1$ 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.
Apply the ratio test to $\sum_{n=1}^{\infty} \frac{5^n}{n^5}$ and select the correct conclusion.
Answer: Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=5>1$ . 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.
A power series is $\sum_{n=1}^{\infty} \frac{n}{n+1}$ ; what does the ratio test conclude?
Answer: Inconclusive because the ratio-test limit is $1$ 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 $\lim_{n \to \infty} \frac{n}{n+1} = 1 \neq 0$ .
For $\sum_{n=1}^{\infty} \frac{3^n}{n!}$ , what does the ratio test conclude about convergence?
Answer: Converges absolutely because the ratio-test limit equals $0$ 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.
Consider $\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 5^n}$ ; what does the ratio test imply?
Answer: Converges absolutely because the ratio-test limit is $\tfrac{4}{5}<1$ 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.
A power-series coefficient sum is $\sum_{n=1}^{\infty} \frac{3^n}{n^3}$ ; determine convergence using the ratio test.
Answer: Diverges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=3>1$ . The ratio test is a key method for determining the convergence of infinite series, particularly those involving exponentials over polynomials. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for $a_n$ = $3^n$ / $n^3$ , the ratio | $a_{n+1}$ / $a_n$ | = 3 * $(n/(n+1))^3$ . As n approaches infinity, this limit is 3, which is greater than 1. Therefore, the ratio test indicates that the series diverges. A tempting distractor might be choice E, which claims convergence with limit 1/3 < 1, but this inverts the exponential and polynomial roles. Always remember that in the ratio test, if the exponential base is greater than 1 and dominates the polynomial, the series diverges.
Apply the ratio test to $\sum_{n=1}^{\infty} \frac{n!}{4^n}$ to determine convergence or divergence.
Answer: Diverges because the ratio-test limit is $\infty>1$ This question tests ratio test application with factorial terms. For $\sum_{n=1}^{\infty} \frac{n!}{4^n}$ , we find $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(n+1)!/4^{n+1}}{n!/4^n} = \lim_{n \to \infty} \frac{n+1}{4} = \infty$ . Since this limit is infinity (greater than 1), the series diverges by the ratio test. Choice A incorrectly suggests the limit is $\frac{1}{4}$ , missing that factorial growth dominates exponential growth. Remember that $n!$ grows faster than any exponential function $a^n$ for fixed $a$ .
Apply the ratio test to $\sum_{n=1}^{\infty} \frac{n!}{n^n}$ to determine convergence.
Answer: Converges absolutely because the ratio-test limit is $\tfrac{1}{e}<1$ This question tests ratio test application to Stirling-type series. For $\sum_{n=1}^{\infty} \frac{n!}{n^n}$ , we compute $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} = \lim_{n \to \infty} \frac{(n+1) \cdot n^n}{(n+1)^{n+1}} = \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n$ . This limit equals $\lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^n = \frac{1}{e}$ . Since $\frac{1}{e} < 1$ , the series converges absolutely. Choice C incorrectly claims the limit is $e$ , possibly from mishandling the exponent. The key insight is recognizing $\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$ implies $\lim_{n \to \infty} (1 - \frac{1}{n})^n = \frac{1}{e}$ .
For $\sum_{n=1}^{\infty} \frac{n!}{n^n}$ , use the ratio test to determine whether it converges.
Answer: Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{e}<1$ . 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.
Use the ratio test on $\sum_{n=1}^{\infty} \frac{n^2}{5^n}$ to determine whether the series converges.
Answer: Converges absolutely because the ratio-test limit equals $\tfrac{1}{5}$ 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 $a_n$ = $n^2$ / $5^n$ , so | $a_{n+1}$ / $a_n$ | = $((n+1)/n)^2$ / 5. The limit is 1/5 as n approaches infinity. Since L = 1/5 < 1, the series converges absolutely. A tempting distractor is choice E, which claims divergence with limit 1/5, but this fails because when L < 1, the test indicates convergence, not divergence. 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.
Decide convergence of $\sum_{n=1}^{\infty} \frac{(n+1)!}{3^n}$ using the ratio test.
Answer: Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\infty>1$ . 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.
Use the ratio test to determine convergence of $\sum_{n=1}^{\infty} \frac{2^n}{n^3}$ .
Answer: Diverges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=2>1$ . 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.
Use the ratio test to determine whether $\sum_{n=1}^{\infty} \frac{n}{2^n}$ converges or diverges.
Answer: Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1$ . 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.
Using the ratio test, determine whether $\sum_{n=1}^{\infty} \frac{3^n}{n!}$ converges or diverges.
Answer: Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=0<1$ . 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.
Using the ratio test, determine if $\sum_{n=1}^{\infty} \frac{n^2}{4^n}$ converges.
Answer: Converges because $\lim \left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{4}<1$ . This problem involves applying the ratio test to a series with polynomial and exponential terms. For $\sum_{n=1}^{\infty} \frac{n^2}{4^n}$ , we find $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^2/4^{n+1}}{n^2/4^n} = \frac{(n+1)^2}{4n^2} = \frac{1}{4} \cdot \left(\frac{n+1}{n}\right)^2$ . As $n \to \infty$ , this becomes $\frac{1}{4} \cdot 1^2 = \frac{1}{4} < 1$ , so the series converges. Choice A incorrectly states the limit is 4, likely by inverting the base of the exponential. When exponential terms appear in denominators with base greater than 1, they typically lead to convergence.
A probability computation uses $\sum_{n=1}^{\infty} \frac{n^5}{n!}$ ; apply the ratio test to determine convergence.
Answer: Converges because $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1$ . The ratio test is a key method for determining the convergence of infinite series, particularly those involving high-degree polynomials over factorials. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for $a_n$ = $n^5$ / n!, the ratio | $a_{n+1}$ / $a_n$ | = ( $(n+1)^5$ / $n^5$ ) / (n+1). As n approaches infinity, this simplifies to 0, which is less than 1. Therefore, the ratio test indicates that the series converges. A tempting distractor might be choice A, which claims the test is inconclusive with limit 1, but this underestimates the factorial's dominance. Always remember that in the ratio test, polynomials of any degree over factorials yield a limit of 0, guaranteeing convergence.

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