Calculus 2 Quiz: Ratio Test

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Ratio TestQuestion 1 of 20

A series $\sum_{n=1}^\infty a_n$ has the property that $\frac{a_{n+1}}{a_n} = 1 - \frac{1}{n+1}$ for all $n \ge 1$ . What is the conclusion from the Ratio Test?

The series converges conditionally.

The series converges absolutely.

The series diverges.

The test is inconclusive.

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Calculus 2 Quiz

Calculus 2 Quiz: Ratio Test

Practice Ratio Test in Calculus 2 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, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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

A series $\sum_{n=1}^\infty a_n$ has the property that $\frac{a_{n+1}}{a_n} = 1 - \frac{1}{n+1}$ for all $n \ge 1$ . What is the conclusion from the Ratio Test?

The series converges conditionally.
The series converges absolutely.
The series diverges.
The test is inconclusive. (correct answer)

Explanation: When you encounter a series with a given ratio between consecutive terms, the Ratio Test is your go-to tool for determining convergence. The Ratio Test examines

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

and concludes: if this limit is less than 1, the series converges absolutely; if greater than 1, it diverges; if equal to 1, the test is inconclusive. Here, you're given

\frac{a_{n+1}}{a_n} = 1 - \frac{1}{n+1}

. To apply the Ratio Test, you need to find the limit as

n \to \infty

\lim_{n \to \infty} \left|1 - \frac{1}{n+1}\right| = \lim_{n \to \infty} \left|\frac{n+1-1}{n+1}\right| = \lim_{n \to \infty} \left|\frac{n}{n+1}\right| = 1

Since the limit equals 1, the Ratio Test is inconclusive, making D correct. Option A suggests conditional convergence, but the Ratio Test cannot determine this—it only tests absolute convergence or divergence. Option B claims absolute convergence, which would require the limit to be less than 1, not equal to 1. Option C claims divergence, which would need the limit to exceed 1. All three options (A, B, C) incorrectly assume the test provides a definitive conclusion when the limit is 1. Remember: when the Ratio Test limit equals 1, you've hit the "borderline case" where the test fails. You'd need a different convergence test (like the Integral Test or Comparison Test) to determine the series' actual behavior.

Question 2

The Ratio Test is applied to the series $\sum_{n=1}^\infty \frac{n^k}{2^n}$ for a fixed real constant $k$ . What is the value of $L = \lim_{n \to \infty} |a_{n+1}/a_n|$ ?

$0$
$k/2$
$1/2$ (correct answer)
$1$

Explanation: When you encounter a Ratio Test problem, you're examining the behavior of consecutive terms in a series to determine convergence. The Ratio Test finds

L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

, and if

L < 1

, the series converges. For the series

\sum_{n=1}^\infty \frac{n^k}{2^n}

, we have

a_n = \frac{n^k}{2^n}

. Let's calculate the ratio:

\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^k}{2^{n+1}}}{\frac{n^k}{2^n}} = \frac{(n+1)^k}{2^{n+1}} \cdot \frac{2^n}{n^k} = \frac{(n+1)^k}{n^k} \cdot \frac{2^n}{2^{n+1}} = \left(\frac{n+1}{n}\right)^k \cdot \frac{1}{2}

n \to \infty

, we know that

\frac{n+1}{n} = 1 + \frac{1}{n} \to 1

. Therefore,

\left(\frac{n+1}{n}\right)^k \to 1^k = 1

regardless of the value of

k

. This gives us

L = 1 \cdot \frac{1}{2} = \frac{1}{2}

, confirming answer (C). Looking at the wrong answers: (A)

0

would suggest the exponential dominates completely, ignoring the polynomial factor's contribution to the ratio. (B)

k/2

incorrectly assumes the exponent

k

affects the limiting ratio, but polynomial growth rates don't impact the exponential's dominance. (D)

1

forgets about the factor of

\frac{1}{2}

that comes from the exponential terms. Key insight: In ratio tests involving exponentials and polynomials, the exponential base determines the limit—polynomial powers wash out in the ratio calculation.

Question 3

For what values of the positive constant $c$ does the series $\sum_{n=1}^\infty \frac{(cn)^n}{n!}$ converge, according to the Ratio Test?

For all $c > 0$ .
$0 < c < 1$
$c > e$
$0 < c < 1/e$ (correct answer)

Explanation: When you encounter a series with factorials in the denominator and exponential-like terms in the numerator, the Ratio Test is typically your best tool for determining convergence. For the series

\sum_{n=1}^\infty \frac{(cn)^n}{n!}

, apply the Ratio Test by computing

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

. Here,

a_n = \frac{(cn)^n}{n!}

, so:

\frac{a_{n+1}}{a_n} = \frac{(c(n+1))^{n+1}}{(n+1)!} \cdot \frac{n!}{(cn)^n} = \frac{c^{n+1}(n+1)^{n+1}}{(n+1) \cdot n!} \cdot \frac{n!}{c^n n^n} = c \cdot \frac{(n+1)^n}{n^n} = c\left(1 + \frac{1}{n}\right)^n

n \to \infty

, we know that

\left(1 + \frac{1}{n}\right)^n \to e

. Therefore,

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = ce

. The Ratio Test tells us the series converges when this limit is less than 1, so we need

ce < 1

, which gives us

c < \frac{1}{e}

. Since we're told

c

is positive, the answer is

0 < c < \frac{1}{e}

. Choice A is wrong because large values of

c

make the ratio exceed 1. Choice B uses the wrong threshold—it should be

\frac{1}{e} \approx 0.368

, not 1. Choice C completely reverses the inequality direction. Study tip: When applying the Ratio Test to series with factorials, watch for the limit

\left(1 + \frac{1}{n}\right)^n \to e

—it appears frequently and is key to finding the correct convergence threshold.

Question 4

The series $\sum_{n=1}^\infty \frac{x^{3n}}{8^n \sqrt{n}}$ is analyzed for convergence. According to the Ratio Test, for which values of $x$ does the series converge absolutely?

$|x| < 8$
$|x|^3 < 1$
$|x| < 2$ (correct answer)
$|x| < 8/3$

Explanation: When analyzing power series for convergence, the Ratio Test helps you find the radius of convergence by examining the behavior of consecutive terms. For a series

\sum a_n

, you calculate

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

and the series converges absolutely when this limit is less than 1. For the series

\sum_{n=1}^\infty \frac{x^{3n}}{8^n \sqrt{n}}

, let

a_n = \frac{x^{3n}}{8^n \sqrt{n}}

. Then:

a_{n+1} = \frac{x^{3(n+1)}}{8^{n+1} \sqrt{n+1}} = \frac{x^{3n+3}}{8^{n+1} \sqrt{n+1}}

Computing the ratio:

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{3n+3}}{8^{n+1} \sqrt{n+1}} \cdot \frac{8^n \sqrt{n}}{x^{3n}}\right| = \left|\frac{x^3}{8}\right| \sqrt{\frac{n}{n+1}}

n \to \infty

\sqrt{\frac{n}{n+1}} \to 1

, so:

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|^3}{8}

For absolute convergence, we need

\frac{|x|^3}{8} < 1

, which gives us

|x|^3 < 8

, so

|x| < 2

. Option A (

|x| < 8

) is too large—this comes from incorrectly treating the exponent as 1 instead of 3. Option B (

|x|^3 < 1

) stops at an intermediate step without taking the cube root. Option D (

|x| < 8/3

) appears to mix up the algebra somewhere in the calculation. Remember: when applying the Ratio Test to power series with terms like

x^{kn}

, the final condition involves

|x|^k

, so don't forget to take the appropriate root to solve for

|x|

Question 5

A common error in testing a series $\sum a_n$ is to calculate $\lim_{n \to \infty} a_n$ and use it as the result of the Ratio Test. Consider the series $\sum_{n=1}^\infty \frac{n}{n^2+1}$ . Which statement below is true?

The Ratio Test shows divergence, but the common error would incorrectly suggest convergence.
The Ratio Test shows convergence, and the common error also suggests convergence.
The Ratio Test is inconclusive, but the common error would incorrectly suggest convergence. (correct answer)
The Ratio Test is inconclusive, and the common error would incorrectly suggest divergence.

Explanation: When you encounter series convergence problems, it's crucial to distinguish between different tests and apply them correctly. The Ratio Test examines

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

, while the common error mentioned simply looks at

\lim_{n \to \infty} a_n

. For the series

\sum_{n=1}^\infty \frac{n}{n^2+1}

, let's apply the Ratio Test properly. We have

a_n = \frac{n}{n^2+1}

, so:

\frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{(n+1)^2+1}}{\frac{n}{n^2+1}} = \frac{(n+1)(n^2+1)}{n((n+1)^2+1)}

After algebraic simplification, this ratio approaches 1 as

n \to \infty

, making the Ratio Test inconclusive. The "common error" refers to using

\lim_{n \to \infty} a_n

instead. Here,

\lim_{n \to \infty} \frac{n}{n^2+1} = \lim_{n \to \infty} \frac{1/n}{1+1/n^2} = 0

. Someone making this error might think the series converges because the limit is zero. Looking at the answer choices: Choice C correctly identifies that the Ratio Test is inconclusive and the common error suggests convergence. Choice A is wrong because the Ratio Test doesn't show divergence. Choice B is wrong because the Ratio Test doesn't show convergence. Choice D is wrong because the common error (getting limit = 0) would suggest convergence, not divergence. Study tip: Always remember that the Ratio Test uses the ratio of consecutive terms, not the limit of the terms themselves. When the ratio limit equals 1, try other tests like the Limit Comparison Test.

Question 6

A student applies the Ratio Test to $\sum_{n=1}^\infty \frac{(2x-1)^n}{n^2}$ and correctly sets up the condition for convergence as $|2x-1| \lim_{n \to \infty} \frac{n^2}{(n+1)^2} < 1$ . After correctly evaluating the limit, the student deduces the condition $|2x-1| < 1$ . What is the correct interval that follows from this inequality, without considering the endpoints?

$(-1, 1)$
$(0, 1)$ (correct answer)
$(-1/2, 1/2)$
$(-2, 2)$

Explanation: When you encounter power series and the Ratio Test, you're typically finding where the series converges by analyzing the behavior of consecutive terms. The student has correctly applied the Ratio Test and found that convergence requires

|2x-1| < 1

. To solve this absolute value inequality, you need to recognize that

|2x-1| < 1

is equivalent to

-1 < 2x-1 < 1

. Adding 1 to all parts gives

0 < 2x < 2

, and dividing by 2 yields

0 < x < 1

, which is the interval

(0, 1)

. Looking at the wrong answers: Choice A gives

(-1, 1)

, which would result from incorrectly solving

|x| < 1

instead of

|2x-1| < 1

. This represents forgetting about the linear transformation inside the absolute value. Choice C gives

(-1/2, 1/2)

, which you'd get if you mistakenly wrote

|x - 1/2| < 1/2

instead of properly handling the original inequality. Choice D gives

(-2, 2)

, which comes from solving

|x| < 2

, again ignoring the

2x-1

expression entirely. The correct answer is B:

(0, 1)

. Study tip: When solving absolute value inequalities like

|ax + b| < c

, always work with the compound inequality

-c < ax + b < c

and solve step by step. Don't try to "factor out" or simplify the expression inside the absolute value bars first—this leads to common errors.

Question 7

Let $\sum_{n=1}^{\infty} a_n$ be a series of positive terms. Suppose the limit $L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ exists. Which of the following conditions on $L$ is sufficient to conclude that $\lim_{n \to \infty} a_n = 0$ ?

$L < 1$ (correct answer)
$L \le 1$
$L = 1$
$L > 1$

Explanation: If

L < 1

, the Ratio Test implies that the series

\sum a_n

converges absolutely. A necessary condition for any series to converge is that its terms must approach zero. Therefore, if the series converges, we must have

\lim_{n \to \infty} a_n = 0

. The condition

L \le 1

is not sufficient because if

L=1

, the series might diverge in a way that its terms still go to zero (e.g.,

\sum 1/n

) or diverge because its terms do not go to zero (e.g.,

\sum (n+1)/n

). If

L > 1

, the terms do not approach zero, so

\lim a_n \neq 0

Question 8

When applying the Ratio Test to a series $\sum a_n$ , the limit is calculated as $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{k}$ , where $k$ is a positive constant from the series definition. Which condition on $k$ guarantees divergence of the series based on this test?

$0 < k < 1$ (correct answer)
$k > 1$
$k = 1$
$k > 0$

Explanation: The Ratio Test states that a series diverges if the limit

L

is greater than 1. Given

L = 1/k

, the condition for divergence is

1/k > 1

. Since

k

is a positive constant, we can multiply both sides by

k

without changing the inequality direction, which gives

1 > k

. Combining this with the given condition that

k

is positive, the range for

k

that guarantees divergence is

0 < k < 1

Question 9

Two series are given: $S_A = \sum_{n=1}^\infty \frac{n!}{100^n}$ and $S_B = \sum_{n=1}^\infty \frac{100^n}{(2n)!}$ . Which of the following correctly describes the conclusions of the Ratio Test for these series?

Both series diverge.
Both series converge.
$S_A$ diverges and $S_B$ converges. (correct answer)
$S_A$ converges and $S_B$ diverges.

Explanation: When you encounter series with factorials or exponentials, the Ratio Test is typically your most powerful tool. It's designed to handle the rapid growth rates of these expressions by examining how consecutive terms relate to each other. For

S_A = \sum_{n=1}^\infty \frac{n!}{100^n}

, applying the Ratio Test gives us

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(n+1)!}{100^{n+1}} \cdot \frac{100^n}{n!} = \lim_{n \to \infty} \frac{n+1}{100} = \infty

. Since this limit exceeds 1, the series diverges. The factorial in the numerator grows much faster than the exponential in the denominator. For

S_B = \sum_{n=1}^\infty \frac{100^n}{(2n)!}

, we get

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{100^{n+1}}{(2n+2)!} \cdot \frac{(2n)!}{100^n} = \lim_{n \to \infty} \frac{100}{(2n+1)(2n+2)} = 0

. Since this limit is less than 1, the series converges. Here, the factorial in the denominator dominates the exponential in the numerator. Therefore,

S_A

diverges and

S_B

converges, making C correct. Choice A incorrectly claims both diverge, ignoring how rapidly

(2n)!

grows. Choice B incorrectly claims both converge, missing that

n!

eventually outpaces

100^n

. Choice D reverses the conclusions entirely. Remember: factorials grow faster than exponentials, but placement matters—factorials in numerators promote divergence, while factorials in denominators promote convergence.

Question 10

Let $\sum a_n$ and $\sum b_n$ be two series with positive terms. If $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 0.5$ and $\lim_{n \to \infty} \frac{b_{n+1}}{b_n} = 1.5$ , what can be concluded about the series $\sum c_n = \sum (a_n + b_n)$ by applying the Ratio Test?

The test is inconclusive because properties of limits cannot be applied to the ratio.
The limit of the ratio for $\sum c_n$ is 2.0, so the series diverges.
The limit of the ratio for $\sum c_n$ is 1.5, so the series diverges. (correct answer)
The limit of the ratio for $\sum c_n$ is 1.0, so the test is inconclusive.

Explanation: When you encounter a question about the Ratio Test applied to a sum of series, you need to understand how limits behave when series are combined. The Ratio Test examines

\lim_{n \to \infty} \frac{c_{n+1}}{c_n}

where

c_n = a_n + b_n

. To find this limit, we need to analyze

\frac{c_{n+1}}{c_n} = \frac{a_{n+1} + b_{n+1}}{a_n + b_n}

. As

n \to \infty

, we can determine which term dominates. Since

\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 0.5 < 1

, the terms

a_n

decrease exponentially fast. Meanwhile,

\lim_{n \to \infty} \frac{b_{n+1}}{b_n} = 1.5 > 1

means

b_n

grows exponentially. For large

n

, the

b_n

terms completely dominate both the numerator and denominator, so

\frac{a_{n+1} + b_{n+1}}{a_n + b_n} \approx \frac{b_{n+1}}{b_n}

. Therefore,

\lim_{n \to \infty} \frac{c_{n+1}}{c_n} = 1.5

. Since this limit exceeds 1, the Ratio Test tells us

\sum c_n

diverges, making C correct. A is wrong because limit properties do apply here through dominance analysis. B incorrectly adds the two ratios (0.5 + 1.5 = 2.0), but ratios don't add this way. D incorrectly averages the ratios to get 1.0, which is mathematically invalid. Study tip: When applying the Ratio Test to sums of series, identify which series dominates by comparing their ratio limits to 1. The faster-growing series determines the behavior of the combined series.

Question 11

The Ratio Test is applied to a power series $\sum_{n=0}^\infty c_n (x-a)^n$ , and it is found that $\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = K$ , where $K$ is a finite positive number. What is the radius of convergence, $R$ ?

$R = K$
$R = 1/K$ (correct answer)
$R = 1$
$R = \infty$

Explanation: When you encounter a power series with the Ratio Test, you're finding the radius of convergence by determining where the series converges absolutely. The key insight is understanding how the Ratio Test connects to the radius formula. For a power series

\sum_{n=0}^\infty c_n (x-a)^n

, when you apply the Ratio Test, you examine

\lim_{n \to \infty} \left| \frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n} \right| = |x-a| \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|

. This simplifies to

|x-a| \cdot K

, where

K

is your given limit. The Ratio Test tells us the series converges when this expression is less than 1:

|x-a| \cdot K < 1

, which means

|x-a| < \frac{1}{K}

. Since the radius of convergence

R

is the distance from the center

a

where convergence occurs, we have

R = \frac{1}{K}

. This makes choice B correct. Choice A (

R = K

) confuses the coefficient limit with the radius directly. Choice C (

R = 1

) might seem tempting since the Ratio Test uses 1 as the convergence threshold, but this ignores the coefficient behavior entirely. Choice D (

R = \infty

) would only occur if

K = 0

, but the problem states

K

is a finite positive number. Remember this inverse relationship: when the ratio of consecutive coefficients is large, the radius of convergence is small, and vice versa. The formula

R = \frac{1}{\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|}

is fundamental for power series analysis.

Question 12

Consider the series $\sum_{n=1}^\infty \frac{(n!)^k}{(kn)!}$ where $k$ is a positive integer greater than 1. When applying the Ratio Test, the limit $L = \lim_{n\to\infty} |a_{n+1}/a_n|$ is found. What is the value of L?

$L = 1/k!$
$L = 1/k^k$ (correct answer)
$L = 0$
$L = \infty$

Explanation: When you encounter a series with factorials, the Ratio Test is often your best tool because factorials have predictable cancellation patterns that make the limit calculations manageable. To find

L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|

, you need to compute the ratio of consecutive terms. Here,

a_n = \frac{(n!)^k}{(kn)!}

, so:

\frac{a_{n+1}}{a_n} = \frac{((n+1)!)^k}{(k(n+1))!} \cdot \frac{(kn)!}{(n!)^k} = \frac{((n+1)!)^k \cdot (kn)!}{(n!)^k \cdot (k(n+1))!}

Since

(n+1)! = (n+1) \cdot n!

, the numerator becomes

((n+1) \cdot n!)^k \cdot (kn)! = (n+1)^k \cdot (n!)^k \cdot (kn)!

. For the denominator,

(k(n+1))! = (kn+k)! = (kn+k)(kn+k-1)\cdots(kn+1) \cdot (kn)!

. After canceling

(n!)^k

and

(kn)!

, you get:

\frac{a_{n+1}}{a_n} = \frac{(n+1)^k}{(kn+k)(kn+k-1)\cdots(kn+1)}

n \to \infty

, each factor in the denominator behaves like

kn

, giving you

k

factors, so the denominator is approximately

(kn)^k

. The ratio becomes

\frac{(n+1)^k}{(kn)^k} = \frac{1}{k^k}

. Choice A (

1/k!

) confuses factorial with exponential growth rates. Choice C (

0

) would indicate faster factorial decay than actually occurs. Choice D (

\infty

) would suggest divergence, but the factorial in the denominator grows much faster than the numerator. Strategy tip: For factorial ratios, always look for telescoping patterns and remember that

(an)!

grows much faster than

(n!)^a

for any constant

a

Question 13

The Ratio Test is applied to the series $\sum_{n=1}^{\infty} \frac{(x-3)^{2n}}{n \cdot 4^n}$ . Which of the following conditions on $x$ guarantees absolute convergence based on the result of the test?

$|x-3| < 2$ (correct answer)
$|x-3| < 4$
$|x-3| < 1$
$|x-3|^2 < 1/4$

Explanation: To apply the Ratio Test, we compute the limit

L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

. Here,

a_n = \frac{(x-3)^{2n}}{n \cdot 4^n}

. The ratio is

\left| \frac{(x-3)^{2(n+1)}}{(n+1)4^{n+1}} \cdot \frac{n 4^n}{(x-3)^{2n}} \right| = \left| \frac{(x-3)^2}{4} \cdot \frac{n}{n+1} \right|

. As

n \to \infty

, the limit is

L = \frac{|x-3|^2}{4}

. For absolute convergence, we need

L < 1

, which means

\frac{|x-3|^2}{4} < 1

, or

|x-3|^2 < 4

. Taking the square root of both sides gives

|x-3| < 2

Question 14

Consider the series $\sum_{n=1}^{\infty} \frac{2^n \sin(n\pi/3)}{n \cdot 3^n}$ . A student claims the ratio test shows divergence because $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{2}{3} > \frac{1}{2}$ . What is wrong with this reasoning?

The student computed $\sin((n+1)\pi/3) / \sin(n\pi/3)$ incorrectly in the ratio calculation
The student compared the limit to $\frac{1}{2}$ instead of $1$ for the ratio test criterion (correct answer)
The student ignored that the sine function makes some terms zero, invalidating the ratio test
The student should have used absolute convergence testing before applying the ratio test

Explanation: The ratio test states that if

\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L

, then: the series converges if

L < 1

, diverges if

L > 1

, and is inconclusive if

L = 1

. The student incorrectly used

\frac{1}{2}

as the comparison threshold instead of

1

. Since

\frac{2}{3} < 1

, the ratio test actually indicates convergence, not divergence. Choice A addresses a calculation detail that doesn't affect the main error. Choice C is incorrect since

\sin(n\pi/3)

is never zero for positive integers

n

. Choice D is irrelevant to the stated error.

Question 15

The series $\sum_{n=1}^{\infty} \frac{(-2)^n}{n^2 \cdot 3^n}$ is being analyzed. If we apply the ratio test to $\sum |a_n|$ , what limit do we obtain, and what does this tell us about the original series?

The limit is $\frac{2}{3}$ , so the original series converges conditionally but not absolutely
The limit is $\frac{2}{3}$ , so the original series converges absolutely and therefore converges (correct answer)
The limit is $2$ , so we must use the alternating series test instead
The limit is $\frac{4}{9}$ , so the original series converges absolutely and therefore converges

Explanation: When you encounter a series with both positive and negative terms, you need to distinguish between absolute convergence and conditional convergence. The ratio test applied to

\sum |a_n|

determines absolute convergence. For this series,

a_n = \frac{(-2)^n}{n^2 \cdot 3^n}

, so

|a_n| = \frac{2^n}{n^2 \cdot 3^n}

. Applying the ratio test:

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

= \lim_{n \to \infty} \frac{2 \cdot n^2}{3 \cdot (n+1)^2} = \frac{2}{3} \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = \frac{2}{3} \cdot 1 = \frac{2}{3}

Since

\frac{2}{3} < 1

, the series

\sum |a_n|

converges, meaning the original series converges absolutely. Absolute convergence is stronger than regular convergence—if a series converges absolutely, it automatically converges. Choice A incorrectly suggests conditional convergence, but since

\sum |a_n|

converges, we have absolute convergence, not just conditional. Choice C gets the wrong limit value and misses that the ratio test already gives us our answer. Choice D calculates an incorrect limit of

\frac{4}{9}

—this likely comes from incorrectly handling the alternating signs or making an algebraic error in the ratio calculation. Remember: when the ratio test on

\sum |a_n|

gives a limit less than 1, you immediately know the original series converges absolutely, which is the strongest form of convergence.

Question 16

A series has the property that $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$ where $0 < L < 1$ . If we form a new series $\sum_{n=1}^{\infty} c_n$ where $c_n = a_{2n}$ (taking only even-indexed terms), what can we conclude about the convergence of $\sum c_n$ ?

The ratio test is inconclusive for $\sum c_n$ since we changed the indexing structure
$\sum c_n$ converges, and the ratio test gives $\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = L$
$\sum c_n$ may diverge because we removed half the terms from a convergent series
$\sum c_n$ converges, and the ratio test gives $\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = L^2$ (correct answer)

Explanation: When you encounter problems involving subsequences and convergence tests, the key insight is understanding how ratio limits behave when you extract every other term from a sequence. Since the original series has

\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L < 1

, the ratio test tells us

\sum a_n

converges. For the new series

\sum c_n

where

c_n = a_{2n}

, we need to find

\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|

. This becomes

\lim_{n \to \infty} \left| \frac{a_{2(n+1)}}{a_{2n}} \right| = \lim_{n \to \infty} \left| \frac{a_{2n+2}}{a_{2n}} \right|

. We can rewrite this as:

\lim_{n \to \infty} \left| \frac{a_{2n+2}}{a_{2n+1}} \cdot \frac{a_{2n+1}}{a_{2n}} \right| = \lim_{n \to \infty} \left| \frac{a_{2n+2}}{a_{2n+1}} \right| \cdot \lim_{n \to \infty} \left| \frac{a_{2n+1}}{a_{2n}} \right|

Since both limits equal

L

(they're just subsequences of the original ratio sequence), we get

L \cdot L = L^2 < 1

. Therefore,

\sum c_n

converges by the ratio test. Option A is wrong because the ratio test works fine on subsequences. Option B incorrectly claims the limit remains

L

instead of

L^2

. Option C misunderstands that removing terms from a convergent series with positive terms cannot make the remaining terms diverge—if anything, it makes convergence more likely. Strategy tip: When working with subsequences and ratio tests, remember that skipping terms effectively "jumps" the ratio by the number of terms skipped, leading to powers in the limit.

Question 17

For the series $\sum_{n=1}^{\infty} \frac{n! \cdot a^n}{n^n}$ where $a > 0$ , the ratio test gives $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{a}{e}$ . For which values of $a$ does this series converge?

$0 < a < e$ only, since the ratio test requires the limit to be less than 1 (correct answer)
$0 < a \leq e$ , since the test is inconclusive when $a = e$
$a > e$ only, since Stirling's approximation reverses the inequality
All $a > 0$ , since the factorial and exponential terms balance exactly

Explanation: Using Stirling's approximation or careful ratio analysis:

\frac{a_{n+1}}{a_n} = \frac{(n+1)! \cdot a^{n+1}}{(n+1)^{n+1}} \cdot \frac{n^n}{n! \cdot a^n} = a \cdot \frac{n+1}{n+1} \cdot \left(\frac{n}{n+1}\right)^n = a \cdot \left(1 + \frac{1}{n}\right)^{-n}

. As

n \to \infty

\left(1 + \frac{1}{n}\right)^{-n} \to \frac{1}{e}

, so the limit is

\frac{a}{e}

. For convergence by ratio test, we need

\frac{a}{e} < 1

, so

a < e

. Choice B incorrectly includes the boundary case. Choice C reverses the inequality. Choice D ignores the ratio test conclusion.

Question 18

The series $\sum_{n=1}^{\infty} a_n$ satisfies $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{3}{4}$ . If we define $b_n = (a_n)^2$ , what can we conclude about $\sum_{n=1}^{\infty} b_n$ using the ratio test?

The series $\sum b_n$ converges since $\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = \frac{9}{16} < 1$ (correct answer)
The series $\sum b_n$ converges since $\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = \frac{3}{4} < 1$
The ratio test is inconclusive since $\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = 1$
The series $\sum b_n$ diverges since $\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = \frac{4}{3} > 1$

Explanation: Since

b_n = (a_n)^2

, we have

\frac{b_{n+1}}{b_n} = \frac{(a_{n+1})^2}{(a_n)^2} = \left(\frac{a_{n+1}}{a_n}\right)^2

. Therefore,

\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} < 1

, so

\sum b_n

converges. Choice B uses the original ratio instead of squaring it. Choice C incorrectly suggests the limit is 1. Choice D inverts and then squares incorrectly.

Question 19

Let $\sum a_n$ be a series where $a_n > 0$ for all $n$ . If $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$ , which statement guarantees that the series $\sum n \cdot a_n$ converges?

$L < 1$ (correct answer)
$L = 0$
$L < 1/n$
The new series $\sum n \cdot a_n$ is guaranteed to diverge.

Explanation: To test the convergence of

\sum n \cdot a_n

, we apply the Ratio Test to its terms. Let

b_n = n \cdot a_n

. The ratio is

\frac{b_{n+1}}{b_n} = \frac{(n+1)a_{n+1}}{n a_n} = \left(\frac{n+1}{n}\right) \left(\frac{a_{n+1}}{a_n}\right)

. The limit of this new ratio is

\lim_{n \to \infty} \left(\frac{n+1}{n}\right) \cdot \lim_{n \to \infty} \left(\frac{a_{n+1}}{a_n}\right) = 1 \cdot L = L

. Therefore, the series

\sum n \cdot a_n

converges if its ratio limit

L

is less than 1, which is the same condition for the convergence of

\sum a_n

Question 20

For the series $\sum_{n=2}^{\infty} \frac{(-1)^n \cdot 2^n \cdot n^2}{3^n + n}$ , the ratio test gives a limit of $L$ . Which statement is correct?

$L = \frac{2}{3}$ and the series converges absolutely by the ratio test (correct answer)
$L = \frac{2}{3}$ and the series converges conditionally by the ratio test
$L = \frac{4}{9}$ and the series converges absolutely by the ratio test
$L = 2$ and the series diverges by the ratio test

Explanation: Since we're testing absolute convergence with the ratio test, we consider

|a_n| = \frac{2^n \cdot n^2}{3^n + n}

. Then

\frac{|a_{n+1}|}{|a_n|} = \frac{2^{n+1}(n+1)^2}{3^{n+1} + (n+1)} \cdot \frac{3^n + n}{2^n \cdot n^2} = 2 \cdot \frac{(n+1)^2}{n^2} \cdot \frac{3^n + n}{3 \cdot 3^n + (n+1)}

. As

n \to \infty

, this approaches

2 \cdot 1 \cdot \frac{1}{3} = \frac{2}{3} < 1

, so the series converges absolutely. Choice B incorrectly suggests conditional convergence. Choice C miscalculates the limit. Choice D uses

2

without the factor of

\frac{1}{3}