Calculus 2 Quiz: Absolute Vs Conditional Convergence

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Absolute Vs Conditional ConvergenceQuestion 1 of 20

Let $a_n = \frac{(-1)^n}{n^p}$ . The series $\sum a_n$ converges conditionally if and only if:

p > 1

p > 0

0 < p < 1

0 < p \le 1

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Calculus 2 Quiz

Calculus 2 Quiz: Absolute Vs Conditional Convergence

Practice Absolute Vs Conditional Convergence 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 Absolute Vs Conditional Convergence, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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

Let $a_n = \frac{(-1)^n}{n^p}$ . The series $\sum a_n$ converges conditionally if and only if:

$p > 1$
$p > 0$
$0 < p < 1$
$0 < p \le 1$ (correct answer)

Explanation: This is an alternating p-series. The series of absolute values is

\sum |a_n| = \sum \frac{1}{n^p}

, which is a standard p-series. It converges if

p > 1

and diverges if

p \le 1

. For the original series

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

to converge, we can use the Alternating Series Test. The conditions are that

b_n = \frac{1}{n^p}

must be positive, decreasing, and approach 0. All these conditions are met if

p > 0

. Conditional convergence occurs when the series converges (

p > 0

) but does not converge absolutely (

p \le 1

). The intersection of these two conditions is

0 < p \le 1

Question 2

For the series $\sum_{n=1}^{\infty} \frac{(-1)^n \sin(1/n)}{\sqrt{n}}$ , which analysis correctly determines the convergence behavior?

Converges absolutely since $|\sin(1/n)| \leq 1$ and $\sum \frac{1}{\sqrt{n}}$ provides a convergent comparison
Converges conditionally since $\sin(1/n) \sim \frac{1}{n}$ gives $\frac{\sin(1/n)}{\sqrt{n}} \sim \frac{1}{n\sqrt{n}} = \frac{1}{n^{3/2}}$ , making absolute convergence hold
Converges conditionally since $\sin(1/n) \sim \frac{1}{n}$ gives $\frac{\sin(1/n)}{\sqrt{n}} \sim \frac{1}{n^{3/2}}$ , but the alternating series test fails
Converges absolutely since $\sin(1/n) \sim \frac{1}{n}$ and $\sum \frac{1}{n^{3/2}}$ converges by comparison (correct answer)

Explanation: As

n \to \infty

\sin(1/n) \sim \frac{1}{n}

, so

\frac{\sin(1/n)}{\sqrt{n}} \sim \frac{1/n}{\sqrt{n}} = \frac{1}{n^{3/2}}

. Since

\sum \frac{1}{n^{3/2}}

converges (p-series with

p = 3/2 > 1

), by limit comparison test,

\sum \frac{|\sin(1/n)|}{\sqrt{n}}

converges. Therefore the original series converges absolutely. Choice A incorrectly states that

\sum \frac{1}{\sqrt{n}}

converges. Choice B has the right asymptotic analysis but wrong conclusion about absolute convergence. Choice C incorrectly claims the alternating series test fails.

Question 3

If $\sum_{n=1}^{\infty} |a_n|$ converges and $\sum_{n=1}^{\infty} a_n$ also converges, which statement about the series $\sum_{n=1}^{\infty} (-1)^n a_n$ is necessarily true?

The series $\sum_{n=1}^{\infty} (-1)^n a_n$ converges absolutely since $\sum |(-1)^n a_n| = \sum |a_n|$ converges (correct answer)
The series $\sum_{n=1}^{\infty} (-1)^n a_n$ converges conditionally since it's an alternating series with convergent terms
The series $\sum_{n=1}^{\infty} (-1)^n a_n$ may diverge even though both given series converge
The convergence type of $\sum_{n=1}^{\infty} (-1)^n a_n$ depends on whether $a_n$ changes sign

Explanation: Since

\sum |a_n|

converges by hypothesis, and

|(-1)^n a_n| = |(-1)^n| \cdot |a_n| = |a_n|

, we have

\sum |(-1)^n a_n| = \sum |a_n|

converges. This means

\sum (-1)^n a_n

converges absolutely. Choice B incorrectly concludes conditional convergence when we have absolute convergence. Choice C is wrong because absolute convergence of a series guarantees convergence of the series itself. Choice D is irrelevant since we've established absolute convergence regardless of sign changes in

a_n

Question 4

The series $\sum_{n=2}^{\infty} \frac{(-1)^n}{n(\ln n)^p}$ converges conditionally when $p$ satisfies which condition?

$p > 1$ only, since this ensures both absolute and conditional convergence simultaneously
$0 < p \leq 1$ , since the alternating series converges but absolute convergence fails (correct answer)
$p \leq 0$ , since only then does the alternating series test apply correctly
$p = 1$ exactly, since this is the boundary case between convergence and divergence

Explanation: For conditional convergence, we need the alternating series to converge but not absolutely. The alternating series test requires

\frac{1}{n(\ln n)^p}

to be decreasing and approach 0, which holds for

p > 0

. For absolute convergence, we examine

\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}

. By the integral test, this converges when

p > 1

and diverges when

p \leq 1

. Therefore, conditional convergence occurs when

0 < p \leq 1

. Choice A gives absolute convergence. Choice C fails the alternating series test. Choice D is too restrictive.

Question 5

For what values of the real number $k$ does the series $\sum_{n=1}^{\infty} (-1)^n \left(\frac{k^2-3}{5}\right)^n$ converge absolutely?

$-\sqrt{8} < k < \sqrt{8}$ (correct answer)
$\sqrt{3} < |k| < \sqrt{8}$
$-\sqrt{3} < k < \sqrt{3}$
The series never converges absolutely.

Explanation: This is a geometric series with ratio

r = -\left(\frac{k^2-3}{5}\right)

. A geometric series converges absolutely if and only if

|r| < 1

\left| -\frac{k^2-3}{5} \right| < 1

\frac{|k^2-3|}{5} < 1

|k^2-3| < 5

This inequality is equivalent to

-5 < k^2-3 < 5

. Adding 3 to all parts gives

-2 < k^2 < 8

. Since

k^2

is always non-negative, the left part

k^2 > -2

is always true for any real

k

. We only need to solve

k^2 < 8

, which gives

-\sqrt{8} < k < \sqrt{8}

, or

-2\sqrt{2} < k < 2\sqrt{2}

Question 6

Let $\sum a_n$ be a series of non-zero terms. Which of the following conditions is sufficient to guarantee that $\sum a_n$ converges conditionally?

The sequence $\{a_n\}$ is alternating, and $\lim_{n\to\infty} |a_n| = 0$ .
The series $\sum a_n$ converges, and the series $\sum |a_n|$ diverges. (correct answer)
$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = 1$ and $\sum a_n$ converges.
The series $\sum a_n$ satisfies the conditions of the Alternating Series Test.

Explanation: The definition of conditional convergence for a series

\sum a_n

is that the series itself converges, while the series of its absolute values,

\sum |a_n|

, diverges. Choice B is a direct statement of this definition. Choice A is not sufficient because it omits the condition that the terms

|a_n|

must be decreasing. Choice C is not sufficient; while an inconclusive Ratio Test is common for conditionally convergent series, it doesn't guarantee the divergence of

\sum |a_n|

. Choice D guarantees convergence, but it could be absolute convergence (e.g.,

\sum (-1)^n/n^2

), so it does not guarantee conditional convergence.

Question 7

How does the series $\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{\sqrt{n^2+1}}$ behave?

It converges absolutely.
It converges conditionally. (correct answer)
It diverges by the nth Term Test.
It diverges by Limit Comparison Test with $\sum \frac{1}{n}$ .

Explanation: First, recognize that

\cos(n\pi) = (-1)^n

. The series is

\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n^2+1}}

. To check for absolute convergence, consider

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

. Using the Limit Comparison Test with the divergent harmonic series

\sum \frac{1}{n}

, we find

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

. Since the limit is a finite positive number, and

\sum \frac{1}{n}

diverges, the series of absolute values also diverges. Thus, the original series does not converge absolutely. To check for conditional convergence, we apply the Alternating Series Test. Let

b_n = \frac{1}{\sqrt{n^2+1}}

. The terms are positive,

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

, and the terms are decreasing. Therefore, the series converges. Since it converges but not absolutely, it converges conditionally.

Question 8

For which values of the real number $p$ does the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^p + 3}$ converge conditionally?

$p > 1$
$p > 2$
$1 < p \le 2$ (correct answer)
$0 < p \le 1$

Explanation: First, test for absolute convergence using the Limit Comparison Test on

\sum \frac{n}{n^p + 3}

with the p-series

\sum \frac{n}{n^p} = \sum \frac{1}{n^{p-1}}

. The limit of the ratio of terms is 1. The p-series

\sum \frac{1}{n^{p-1}}

converges if

p-1 > 1

, which means

p > 2

. So, the original series converges absolutely for

p > 2

. Next, test for convergence using the Alternating Series Test. The terms

b_n = \frac{n}{n^p + 3}

must approach 0, which requires

p > 1

. The terms must also be decreasing for large

n

, which is also true for

p > 1

. Thus, the series converges for all

p > 1

. Conditional convergence occurs when the series converges but not absolutely. This corresponds to the interval

p > 1

(for convergence) but not

p > 2

(for absolute convergence), which is the interval

1 < p \le 2

Question 9

Determine the convergence behavior of the series $\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{n!}$ .

The series converges conditionally.
The Ratio Test is inconclusive for this series.
The series diverges.
The series converges absolutely. (correct answer)

Explanation: When you encounter a series with alternating signs and factorials, you need to test for both absolute and conditional convergence. This series has the form

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

, so start by checking absolute convergence using

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

. The Ratio Test works perfectly here since factorials grow extremely rapidly. For

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

, compute:

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

Since this limit is less than 1, the series of absolute values converges, meaning the original alternating series converges absolutely. Choice A is wrong because conditional convergence occurs when a series converges but its absolute values diverge—that's not the case here. Choice B is incorrect because the Ratio Test gives a definitive result of 0, which is conclusive. Choice C is wrong because we've shown the series does converge. The key insight is recognizing that factorials dominate polynomial growth so dramatically that even with the

n^2

term, absolute convergence occurs. Remember: when you see factorials in denominators, the Ratio Test is usually your best friend, and absolute convergence is often the result due to factorial's explosive growth rate.

Question 10

If $\sum_{n=1}^\infty a_n$ converges conditionally, which statement about $\sum_{n=1}^\infty |a_n|$ must be true?

$\sum_{n=1}^\infty |a_n|$ converges to 0.
$\sum_{n=1}^\infty |a_n|$ converges to a positive number.
$\sum_{n=1}^\infty |a_n|$ diverges to infinity. (correct answer)
$\sum_{n=1}^\infty |a_n|$ could either converge or diverge.

Explanation: The definition of conditional convergence is that the series

\sum a_n

converges, but the series of absolute values

\sum |a_n|

diverges. Since the terms

|a_n|

are all non-negative, if the series

\sum |a_n|

diverges, its partial sums must increase without bound. Therefore,

\sum |a_n|

must diverge to infinity.

Question 11

Let $\sum a_n$ and $\sum b_n$ be two series. If $\sum a_n$ converges absolutely and $\sum b_n$ converges conditionally, what can be concluded about the series $\sum (a_n + b_n)$ ?

It must converge absolutely.
It may converge or diverge.
It must diverge.
It must converge conditionally. (correct answer)

Explanation: When you encounter problems about combining convergent series, you need to carefully consider how different types of convergence interact with each other. Let's think about what happens when we add

a_n + b_n

. Since

\sum a_n

converges absolutely, we know

\sum |a_n|

converges, which means

\sum a_n

also converges (absolute convergence implies convergence). Since

\sum b_n

converges conditionally, it converges but

\sum |b_n|

diverges. By the basic properties of convergent series, when you add two convergent series term by term, the resulting series

\sum (a_n + b_n)

must converge. This rules out options B and C immediately. Now, does this sum converge absolutely or conditionally? For absolute convergence, we'd need

\sum |a_n + b_n|

to converge. However, since

\sum |b_n|

diverges and

|a_n + b_n| \geq ||b_n| - |a_n||

by the reverse triangle inequality, the series

\sum |a_n + b_n|

will generally diverge because the

|b_n|

terms dominate. Therefore,

\sum (a_n + b_n)

converges but not absolutely—it converges conditionally. Option A is wrong because absolute convergence is prevented by the conditionally convergent series

\sum b_n

. Options B and C are incorrect because the sum of two convergent series always converges. Option D correctly identifies that we get conditional convergence. Study tip: Remember that conditional convergence is "weaker" than absolute convergence, so when combining series, the weaker type of convergence typically determines the behavior of the sum.

Question 12

The series $\sum_{n=1}^\infty a_n$ is conditionally convergent. What can be concluded about the convergence of the series $\sum_{n=1}^\infty a_n^2$ ?

It must converge.
It must diverge.
It may converge or it may diverge. (correct answer)
It must converge conditionally.

Explanation: If

\sum a_n

is conditionally convergent, we know

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

. This implies that for large

n

|a_n| < 1

, and thus

a_n^2 < |a_n|

. However, this comparison is not sufficient to guarantee convergence of

\sum a_n^2

. Let's consider two examples.

Let $a_n = \frac{(-1)^n}{\sqrt{n}}$ . $\sum a_n$ converges conditionally. Then $\sum a_n^2 = \sum \frac{1}{n}$ , which is the divergent harmonic series.
Let $a_n = \frac{(-1)^n}{n}$ . $\sum a_n$ converges conditionally. Then $\sum a_n^2 = \sum \frac{1}{n^2}$ , which is a convergent p-series. Since we have found examples where $\sum a_n^2$ converges and examples where it diverges, we must conclude that it may converge or it may diverge.

Question 13

The series $\sum_{n=1}^\infty \frac{(-1)^n (n+1)}{n^2+1}$ is best described as:

Conditionally convergent by the Alternating Series Test and Limit Comparison Test. (correct answer)
Absolutely convergent by the Ratio Test.
Divergent by the nth Term Test for Divergence.
Absolutely convergent by comparison with $\sum \frac{1}{n^2}$ .

Explanation: When you encounter an alternating series like this one, you need to test for both conditional and absolute convergence. An alternating series converges conditionally if the series itself converges but the series of absolute values diverges. For the series

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

, first apply the Alternating Series Test. You need: (1) the terms

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

must decrease, and (2)

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

. The limit equals zero since the denominator grows faster than the numerator. For large n, the terms do decrease, so the series converges by the Alternating Series Test. Next, test for absolute convergence by examining

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

. Using the Limit Comparison Test with

\sum \frac{1}{n}

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

. Since this limit is positive and finite, and

\sum \frac{1}{n}

diverges, the series of absolute values diverges. Therefore, the series is conditionally convergent. Option B is wrong because the Ratio Test would show convergence to 1, indicating inconclusive results, not absolute convergence. Option C fails because

\lim_{n \to \infty} \frac{(-1)^n (n+1)}{n^2+1} = 0

. Option D is incorrect because comparison with

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

doesn't work—the terms behave like

\frac{1}{n}

, not

\frac{1}{n^2}

. Study tip: Always check both conditional and absolute convergence for alternating series—use limit comparison to find the right comparison series based on the dominant terms.

Question 14

Suppose $a_n > 0$ for all $n$ , and the series $\sum (-1)^n a_n$ converges. What additional condition would guarantee that this convergence is conditional and not absolute?

$\lim_{n\to\infty} a_n = 0$ .
The sequence $\{a_n\}$ is decreasing.
$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 1$ .
The series $\sum a_n$ diverges. (correct answer)

Explanation: Conditional convergence means the series converges, but not absolutely. We are given that

\sum (-1)^n a_n

converges. For the convergence to be conditional, the series of absolute values,

\sum |(-1)^n a_n| = \sum a_n

, must diverge. Therefore, the additional condition needed is that

\sum a_n

diverges. Choices A and B are conditions of the AST, which are necessary for the convergence of

\sum (-1)^n a_n

but don't say anything about whether the convergence is absolute or conditional. Choice C represents an inconclusive Ratio Test, which is common for such series but doesn't guarantee divergence of

\sum a_n

(e.g., for

\sum 1/n^2

, the ratio limit is 1 but the series converges).

Question 15

Consider the series $S = \sum_{n=1}^\infty (-1)^n a_n$ , where $a_n = \int_n^{n+1} \frac{1}{x} dx$ . Which statement is correct?

The series S converges absolutely.
The series S converges conditionally. (correct answer)
The series S diverges because $\lim_{n\to\infty} a_n \ne 0$ .
The series S diverges for another reason.

Explanation: First, evaluate the term

a_n

a_n = \int_n^{n+1} \frac{1}{x} dx = [\ln|x|]_n^{n+1} = \ln(n+1) - \ln(n) = \ln\left(\frac{n+1}{n}\right)

. The series is

\sum_{n=1}^\infty (-1)^n \ln(1 + \frac{1}{n})

. Let's check for absolute convergence:

\sum \ln(1 + \frac{1}{n})

. Using the Limit Comparison Test with

\sum \frac{1}{n}

, we have

\lim_{n\to\infty} \frac{\ln(1+1/n)}{1/n} = 1

(a standard limit, provable with L'Hopital's rule on

\frac{\ln(1+x)}{x}

x\to 0

). Since

\sum \frac{1}{n}

diverges, the series of absolute values diverges. Now, apply the AST to the original series. Let

b_n = \ln(1+1/n)

. The terms are positive.

\lim_{n\to\infty} \ln(1+1/n) = \ln(1) = 0

. The terms are decreasing since

1+1/n

is a decreasing function and

\ln(x)

is an increasing function. Thus, the series converges by AST. Since it converges but not absolutely, it is conditionally convergent.

Question 16

Which of the following series converges conditionally?

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n\sqrt{n}}$
$\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2+1}$
$\sum_{n=1}^{\infty} (-1)^n \frac{n^2+1}{n^3}$ (correct answer)
$\sum_{n=1}^{\infty} (-1)^n \frac{n}{n+5}$

Explanation: A series converges conditionally if it converges, but its series of absolute values diverges. A)

\sum|a_n| = \sum \frac{1}{n^{3/2}}

is a convergent p-series (p=3/2 > 1), so the original series converges absolutely. B)

\sum|a_n| = \sum \frac{|\sin(n)|}{n^2+1} \le \sum \frac{1}{n^2+1}

, which converges by comparison to

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

. The original series converges absolutely. D) The limit of the terms is

\lim_{n\to\infty} (-1)^n \frac{n}{n+5}

, which does not exist (it oscillates between -1 and 1). Since the limit is not 0, the series diverges by the nth Term Test. C) The series of absolute values is

\sum \frac{n^2+1}{n^3}

. By Limit Comparison Test with

\sum \frac{1}{n}

, this series diverges. However, the original alternating series converges by the Alternating Series Test, as

b_n = \frac{n^2+1}{n^3}

is positive, decreasing for

n>1

, and its limit is 0. Thus, the series converges conditionally.

Question 17

Consider the series $\sum_{n=2}^{\infty} \frac{(-1)^n}{n \ln n}$ Which of the following statements accurately describes its convergence?

The series converges absolutely.
The series converges conditionally. (correct answer)
The series diverges.
The Ratio Test is inconclusive, so convergence cannot be determined.

Explanation: To test for absolute convergence, we examine the series

\sum_{n=2}^{\infty} |\frac{(-1)^n}{n \ln n}| = \sum_{n=2}^{\infty} \frac{1}{n \ln n}

. We can use the Integral Test. The integral

\int_2^\infty \frac{1}{x \ln x} dx

with substitution

u = \ln x

becomes

\int_{\ln 2}^\infty \frac{1}{u} du = [\ln|u|]_{\ln 2}^\infty

, which diverges. Thus, the series does not converge absolutely. To test for conditional convergence, we use the Alternating Series Test (AST). Let

b_n = \frac{1}{n \ln n}

. The terms are positive,

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

, and the terms are decreasing since

f(x) = x \ln x

is an increasing function for

x \ge 2

. Since the conditions for AST are met, the series converges. Because the series converges but does not converge absolutely, it converges conditionally.

Question 18

Given that $\sum_{n=1}^\infty a_n$ converges, but $\sum_{n=1}^\infty (-1)^n a_n$ diverges. What can be concluded about the series $\sum a_n$ ?

It converges absolutely.
It converges conditionally. (correct answer)
It diverges.
No conclusion can be drawn.

Explanation: Let

\sum a_n

be the series. We are given that it converges. So it must be either absolutely or conditionally convergent. If

\sum a_n

were absolutely convergent, then

\sum |a_n|

would converge. This would imply that

\sum |(-1)^n a_n| = \sum |a_n|

also converges, meaning

\sum (-1)^n a_n

must be absolutely convergent, and therefore convergent. However, we are told that

\sum (-1)^n a_n

diverges. This creates a contradiction. Therefore, the initial assumption that

\sum a_n

converges absolutely must be false. Since

\sum a_n

converges, but not absolutely, it must converge conditionally.

Question 19

Let $\sum a_n$ be a series that converges to a sum $L$ . If the terms of the series are rearranged to form a new series $\sum b_n$ , under which condition must $\sum b_n$ also converge to $L$ ?

The series $\sum a_n$ is an alternating series.
The series $\sum a_n$ converges absolutely. (correct answer)
The series $\sum a_n$ converges conditionally.
The sequence of partial sums of $\sum a_n$ is bounded.

Explanation: This question refers to the Riemann Rearrangement Theorem. The theorem states that if a series is absolutely convergent, then any rearrangement of its terms results in a series that converges to the same sum. If a series is conditionally convergent, its terms can be rearranged to converge to any real number, or to diverge. Therefore, absolute convergence is the condition required to guarantee that the sum remains unchanged after rearrangement.

Question 20

Let $p_n$ denote the $n$ -th prime number. The series $\sum_{n=1}^\infty \frac{(-1)^n}{p_n}$ converges. What is the nature of its convergence?

It converges conditionally, as $\sum \frac{1}{p_n}$ is a known divergent series. (correct answer)
It converges absolutely because prime numbers grow quickly.
It diverges, because the spacing between primes is irregular.
It converges absolutely, by comparison with $\sum \frac{1}{n^2}$ .

Explanation: When you encounter an alternating series with terms involving primes, you need to determine whether it converges absolutely or conditionally. This requires checking both the original alternating series and the series of absolute values. The series

\sum_{n=1}^\infty \frac{(-1)^n}{p_n}

is alternating with terms

\frac{1}{p_n}

. Since prime numbers increase without bound,

\frac{1}{p_n} \to 0

, and the terms decrease monotonically. By the Alternating Series Test, this series converges. However, for absolute convergence, we must examine

\sum_{n=1}^\infty \left|\frac{(-1)^n}{p_n}\right| = \sum_{n=1}^\infty \frac{1}{p_n}

. This is the famous series of reciprocals of primes, which Euler proved diverges. Since the alternating series converges but the series of absolute values diverges, the convergence is conditional. Choice A correctly identifies this reasoning. Choice B is wrong because even though primes grow quickly, they don't grow fast enough to make

\sum \frac{1}{p_n}

converge—the growth rate of primes is similar to

n \ln n

, which still leads to divergence. Choice C incorrectly claims the original series diverges, but the Alternating Series Test guarantees convergence. Choice D falsely suggests absolute convergence by comparison with

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

, but since

\sum \frac{1}{p_n}

diverges, no such comparison works. Remember: for alternating series, always check both conditional and absolute convergence separately. Many series converge conditionally but not absolutely, especially those involving primes or harmonic-like terms.