Calculus 2 Quiz: Nth Term Test For Divergence
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Nth Term Test For DivergenceQuestion 1 of 20

For what real values of the constant kk does the nth Term Test conclusively prove that the series ∑n=1∞kn2+n3n2−1\sum_{n=1}^\infty \frac{kn^2 + n}{3n^2 - 1} diverges?

For all real values of kk.
For k=0k=0 only.
For all k≠0k \neq 0.
For all k>0k > 0.
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Calculus 2 Quiz

Calculus 2 Quiz: Nth Term Test For Divergence

Practice Nth Term Test For Divergence in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Nth Term Test For Divergence, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

For what real values of the constant kk does the nth Term Test conclusively prove that the series ∑n=1∞kn2+n3n2−1\sum_{n=1}^\infty \frac{kn^2 + n}{3n^2 - 1} diverges?

  1. For all real values of kk.
  2. For k=0k=0 only.
  3. For all k≠0k \neq 0. (correct answer)
  4. For all k>0k > 0.
Explanation: The nth Term Test proves divergence if the limit of the terms is non-zero. Let's find the limit: \nlim⁡n→∞kn2+n3n2−1=lim⁡n→∞k+1/n3−1/n2=k3\lim_{n \to \infty} \frac{kn^2 + n}{3n^2 - 1} = \lim_{n \to \infty} \frac{k + 1/n}{3 - 1/n^2} = \frac{k}{3} \nThe test is conclusive and proves divergence if this limit is not equal to zero. The equation k3≠0\frac{k}{3} \neq 0 is true for all values of kk except k=0k=0. If k=0k=0, the limit is 0, and the test is inconclusive.

Question 2

A student correctly determines that for a series ∑an\sum a_n, the sequence of terms {an}\{a_n\} converges to L=0.01L = 0.01. Based on the nth Term Test, what can be concluded about the series ∑an\sum a_n?

  1. The series must converge to a value close to 0.01.
  2. The series must diverge. (correct answer)
  3. The series must converge, but its sum cannot be determined from this information.
  4. The nth Term Test is inconclusive for this series.
Explanation: The nth Term Test for Divergence states that if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0, then the series ∑an\sum a_n diverges. In this case, the limit of the terms is L=0.01L = 0.01, which is not zero. Therefore, the test is conclusive and proves that the series diverges. The series adds terms that are close to 0.01 infinitely many times, so the sum must grow infinitely large.

Question 3

Let SnS_n be the nth partial sum of the series ∑k=1∞ak\sum_{k=1}^\infty a_k. If the sequence of partial sums {Sn}\{S_n\} converges to 10, what can be concluded about the terms {an}\{a_n\} in the context of the nth Term Test?

  1. lim⁡n→∞an=10\lim_{n \to \infty} a_n = 10, so the nth Term Test proves divergence.
  2. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, so the nth Term Test proves convergence.
  3. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, which makes the nth Term Test inconclusive for this series. (correct answer)
  4. The value of lim⁡n→∞an\lim_{n \to \infty} a_n cannot be determined from the given information.
Explanation: The convergence of the series ∑ak\sum a_k to a sum of 10 means that lim⁡n→∞Sn=10\lim_{n \to \infty} S_n = 10. The term ana_n can be expressed as an=Sn−Sn−1a_n = S_n - S_{n-1} for n>1n > 1. Taking the limit as n→∞n \to \infty, we get lim⁡n→∞an=lim⁡n→∞(Sn−Sn−1)=lim⁡n→∞Sn−lim⁡n→∞Sn−1=10−10=0\lim_{n \to \infty} a_n = \lim_{n \to \infty} (S_n - S_{n-1}) = \lim_{n \to \infty} S_n - \lim_{n \to \infty} S_{n-1} = 10 - 10 = 0. Since the limit of the terms is 0, the nth Term Test for Divergence is inconclusive. This is consistent with the fact that the series converges.

Question 4

For a series ∑n=1∞an\sum_{n=1}^\infty a_n, it is found that lim⁡n→∞an=L\lim_{n \to \infty} a_n = L. The nth Term Test is then applied. For which value of LL is the test's conclusion that the series must diverge?

  1. Any L≠0L \neq 0. (correct answer)
  2. L=0L=0 only.
  3. LL can be any real number.
  4. Any L>0L > 0.
Explanation: When you encounter questions about the nth Term Test (also called the Divergence Test), remember that this test has a very specific and limited purpose: it can only definitively prove divergence in certain cases. The nth Term Test states that if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0, then the series ∑n=1∞an\sum_{n=1}^\infty a_n must diverge. This makes intuitive sense—if the terms aren't approaching zero, they can't possibly sum to a finite value. However, if lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the test is inconclusive; the series might converge or diverge. The correct answer is A because when L≠0L \neq 0 (whether positive or negative), the nth Term Test definitively concludes the series must diverge. This is the only scenario where the test gives a definitive "must diverge" conclusion. Choice B is wrong because when L=0L = 0, the nth Term Test cannot conclude anything—it's inconclusive. Choice C is incorrect because the test doesn't conclude "must diverge" for all real numbers; when L=0L = 0, it's inconclusive. Choice D is too restrictive because the test concludes divergence for any nonzero limit, not just positive values—negative limits also guarantee divergence. Study tip: Remember the nth Term Test's asymmetry. It can prove divergence (when the limit isn't zero) but never prove convergence. When you see "must diverge" in nth Term Test questions, look for any nonzero limit value.

Question 5

A common mistake is to assume that if lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, then ∑an\sum a_n converges. Which pair of series best illustrates that this assumption is false and that the nth Term Test is inconclusive in this case?

  1. ∑1n2\sum \frac{1}{n^2} which converges and ∑1n3\sum \frac{1}{n^3} which also converges.
  2. ∑(1.1)n\sum (1.1)^n which diverges and ∑(0.9)n\sum (0.9)^n which converges.
  3. ∑1n\sum \frac{1}{n} which diverges and ∑nn+1\sum \frac{n}{n+1} which diverges.
  4. ∑1n\sum \frac{1}{n} which diverges and ∑1n2\sum \frac{1}{n^2} which converges. (correct answer)
Explanation: To illustrate the inconclusiveness of the nth Term Test when lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, we need a pair of series that both satisfy this limit condition, yet one converges and the other diverges. \nA) Both series converge, so this pair doesn't show a differing outcome. \nB) For ∑(1.1)n\sum (1.1)^n, the limit of terms is ∞\infty, not 0. \nC) For ∑nn+1\sum \frac{n}{n+1}, the limit of terms is 1, not 0. \nD) For both ∑1n\sum \frac{1}{n} and ∑1n2\sum \frac{1}{n^2}, the limit of the terms is 0. However, the harmonic series ∑1n\sum \frac{1}{n} diverges, while the p-series ∑1n2\sum \frac{1}{n^2} converges. This pair perfectly demonstrates that when the limit of terms is 0, the series might do either, making the test inconclusive.

Question 6

A student argues: "For the series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}, the limit of the terms is lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0. However, this is the harmonic series, which diverges. Therefore, the nth Term Test for Divergence is contradicted by this example." What is the primary flaw in this reasoning?

  1. The student's calculation of the limit is incorrect.
  2. The harmonic series is actually a convergent series.
  3. The student has misunderstood the conclusion of the nth Term Test when the limit is zero. (correct answer)
  4. The nth Term Test is not applicable to p-series like the harmonic series.
Explanation: The student's reasoning is flawed because they misinterpret the nth Term Test. The test states that if the limit is not zero, the series diverges. It does not state that if the limit is zero, the series converges. When lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the test is inconclusive. The harmonic series is a classic example illustrating that even when the terms go to zero, the series can still diverge. This does not contradict the test; it merely shows its limitation.

Question 7

Let ∑an\sum a_n and ∑bn\sum b_n be two series. It is known that lim⁡n→∞an=2\lim_{n \to \infty} a_n = 2 and the series ∑bn\sum b_n converges. What can be concluded about the series ∑(an−bn)\sum (a_n - b_n) using the nth Term Test?

  1. The series ∑(an−bn)\sum (a_n - b_n) diverges because the limit of its terms is 2. (correct answer)
  2. The series ∑(an−bn)\sum (a_n - b_n) converges if ∑an\sum a_n also converges.
  3. The test is inconclusive because the behavior of ∑an\sum a_n is not fully known.
  4. The behavior of ∑(an−bn)\sum (a_n - b_n) cannot be determined from the given information.
Explanation: Let cn=an−bnc_n = a_n - b_n. We need to evaluate lim⁡n→∞cn\lim_{n \to \infty} c_n to apply the nth Term Test. We are given lim⁡n→∞an=2\lim_{n \to \infty} a_n = 2. Since the series ∑bn\sum b_n converges, it is a necessary condition that the limit of its terms is zero, so lim⁡n→∞bn=0\lim_{n \to \infty} b_n = 0. Using the limit properties, lim⁡n→∞cn=lim⁡n→∞(an−bn)=lim⁡n→∞an−lim⁡n→∞bn=2−0=2\lim_{n \to \infty} c_n = \lim_{n \to \infty} (a_n - b_n) = \lim_{n \to \infty} a_n - \lim_{n \to \infty} b_n = 2 - 0 = 2. Since the limit of the terms is 2, which is not zero, the nth Term Test concludes that the series ∑(an−bn)\sum (a_n - b_n) diverges.

Question 8

Let the sequence {an}\{a_n\} be defined as an=(−1)nnn+1a_n = \frac{(-1)^n n}{n+1} if nn is a multiple of 3, and an=1n2a_n = \frac{1}{n^2} otherwise. What is the conclusion of the nth Term Test for the series ∑n=1∞an\sum_{n=1}^\infty a_n?

  1. The series converges because the terms an=1/n2a_n = 1/n^2 form a convergent p-series.
  2. The test is inconclusive because some terms approach 0 while others do not.
  3. The series diverges because the limit of the terms does not exist. (correct answer)
  4. The series diverges because the limit of the terms is -1.
Explanation: To determine the limit of the sequence {an}\{a_n\}, we examine its subsequences. For nn that are not multiples of 3, an=1/n2a_n = 1/n^2, and this subsequence converges to 0. For nn that are multiples of 3, the subsequence is a3k=(−1)3k3k3k+1a_{3k} = \frac{(-1)^{3k} 3k}{3k+1}. As k→∞k \to \infty, the term 3k3k+1→1\frac{3k}{3k+1} \to 1, but (−1)3k(-1)^{3k} alternates between -1 (for odd k) and +1 (for even k). Thus, this subsequence has its own subsequences converging to -1 and 1. Since the overall sequence has subsequences converging to -1, 0, and 1, the limit lim⁡n→∞an\lim_{n \to \infty} a_n does not exist. By the nth Term Test, the series diverges.

Question 9

Consider the series ∑n=1∞3n2+5n−12n3−n2+4\sum_{n=1}^{\infty} \frac{3n^2 + 5n - 1}{2n^3 - n^2 + 4}. Which statement about applying the nth Term Test for Divergence is correct?

  1. The test is inconclusive because lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, so we cannot determine convergence or divergence (correct answer)
  2. The test proves divergence because lim⁡n→∞an=32≠0\lim_{n \to \infty} a_n = \frac{3}{2} \neq 0, violating the necessary condition for convergence
  3. The test proves convergence because lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, satisfying the sufficient condition for convergence
  4. The test is inconclusive because the limit oscillates between positive and negative values as n→∞n \to \infty
Explanation: To apply the nth Term Test, we find lim⁡n→∞3n2+5n−12n3−n2+4\lim_{n \to \infty} \frac{3n^2 + 5n - 1}{2n^3 - n^2 + 4}. Dividing numerator and denominator by n3n^3: lim⁡n→∞3n+5n2−1n32−1n+4n3=02=0\lim_{n \to \infty} \frac{\frac{3}{n} + \frac{5}{n^2} - \frac{1}{n^3}}{2 - \frac{1}{n} + \frac{4}{n^3}} = \frac{0}{2} = 0. Since the limit is 0, the nth Term Test is inconclusive - we cannot determine convergence or divergence from this test alone. Choice B incorrectly calculates the limit. Choice C incorrectly states that lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0 is sufficient for convergence (it's only necessary). Choice D incorrectly claims the limit oscillates.

Question 10

Consider the series ∑n=1∞an\sum_{n=1}^{\infty} a_n where an=2n+(−1)n3n+1a_n = \frac{2^n + (-1)^n}{3^n + 1}. A student applies the nth Term Test for Divergence and concludes the series diverges. Which analysis best explains whether this conclusion is justified?

  1. The conclusion is unjustified because lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0 since 3n3^n grows faster than 2n2^n, making the test inconclusive (correct answer)
  2. The conclusion is justified because the (−1)n(-1)^n term causes oscillation, preventing lim⁡n→∞an\lim_{n \to \infty} a_n from existing
  3. The conclusion is unjustified because lim⁡n→∞an=23\lim_{n \to \infty} a_n = \frac{2}{3}, but this limit being nonzero doesn't guarantee divergence
  4. The conclusion is justified because lim⁡n→∞an\lim_{n \to \infty} a_n does not exist due to the competing exponential terms
Explanation: We have an=2n+(−1)n3n+1a_n = \frac{2^n + (-1)^n}{3^n + 1}. Dividing numerator and denominator by 3n3^n: an=(23)n+(−1)n3n1+13na_n = \frac{\left(\frac{2}{3}\right)^n + \frac{(-1)^n}{3^n}}{1 + \frac{1}{3^n}}. As n→∞n \to \infty, (23)n→0\left(\frac{2}{3}\right)^n \to 0 and (−1)n3n→0\frac{(-1)^n}{3^n} \to 0, so lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0. Since the limit is 0, the nth Term Test is inconclusive. Choice B incorrectly focuses on oscillation without recognizing the limit exists. Choice C miscalculates the limit. Choice D incorrectly claims the limit doesn't exist.

Question 11

Two students are debating about ∑n=1∞n!(n+2)!\sum_{n=1}^{\infty} \frac{n!}{(n+2)!}. Student A claims the nth Term Test proves convergence because the terms approach zero. Student B claims the test only shows the series might converge. Who is correct and why?

  1. Student A is correct because when lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the nth Term Test guarantees convergence of the series
  2. Student B is correct because lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0 is necessary but not sufficient for convergence; the test cannot prove convergence (correct answer)
  3. Student A is correct because lim⁡n→∞n!(n+2)!=lim⁡n→∞1(n+1)(n+2)=0\lim_{n \to \infty} \frac{n!}{(n+2)!} = \lim_{n \to \infty} \frac{1}{(n+1)(n+2)} = 0 definitively proves convergence
  4. Student B is correct because factorial functions require special convergence tests; the nth Term Test cannot be applied to series with factorials
Explanation: Student B is correct. We have an=n!(n+2)!=n!(n+2)(n+1)n!=1(n+1)(n+2)a_n = \frac{n!}{(n+2)!} = \frac{n!}{(n+2)(n+1)n!} = \frac{1}{(n+1)(n+2)}, so lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0. However, the nth Term Test for Divergence only provides a sufficient condition for divergence (if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0), not for convergence. When the limit equals zero, the test is inconclusive. Choice A incorrectly states the test proves convergence when the limit is zero. Choice C makes the same error. Choice D incorrectly claims the test cannot be applied to factorials.

Question 12

Consider the series ∑n=1∞ln⁡(n2+1)n2\sum_{n=1}^{\infty} \frac{\ln(n^2 + 1)}{n^2}. When applying the nth Term Test for Divergence, what can be concluded?

  1. The series diverges because lim⁡n→∞ln⁡(n2+1)n2=+∞\lim_{n \to \infty} \frac{\ln(n^2 + 1)}{n^2} = +\infty due to the logarithmic growth in the numerator
  2. The test cannot be applied because the logarithmic function creates undefined behavior in the limit calculation process
  3. The series converges because lim⁡n→∞ln⁡(n2+1)n2=0\lim_{n \to \infty} \frac{\ln(n^2 + 1)}{n^2} = 0, satisfying the necessary and sufficient condition
  4. The test is inconclusive because lim⁡n→∞ln⁡(n2+1)n2=0\lim_{n \to \infty} \frac{\ln(n^2 + 1)}{n^2} = 0, requiring other convergence tests for determination (correct answer)
Explanation: When you encounter a series convergence problem, the nth Term Test for Divergence is often your first step. This test states that if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0, then ∑an\sum a_n diverges. However, if lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the test tells us nothing about convergence. Let's find lim⁡n→∞ln⁡(n2+1)n2\lim_{n \to \infty} \frac{\ln(n^2 + 1)}{n^2}. This is an ∞∞\frac{\infty}{\infty} indeterminate form, so we can apply L'Hôpital's rule: lim⁡n→∞ln⁡(n2+1)n2=lim⁡n→∞2nn2+12n=lim⁡n→∞2n2n(n2+1)=lim⁡n→∞1n2+1=0\lim_{n \to \infty} \frac{\ln(n^2 + 1)}{n^2} = \lim_{n \to \infty} \frac{\frac{2n}{n^2 + 1}}{2n} = \lim_{n \to \infty} \frac{2n}{2n(n^2 + 1)} = \lim_{n \to \infty} \frac{1}{n^2 + 1} = 0 Since the limit equals 0, the nth Term Test is inconclusive—it cannot determine whether the series converges or diverges. Choice A is wrong because the limit is 0, not +∞+\infty. Logarithmic functions grow slower than polynomial functions. Choice B is incorrect because logarithmic functions are well-defined for positive arguments, and the limit calculation is straightforward using L'Hôpital's rule. Choice C makes a critical error by claiming that lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0 is a sufficient condition for convergence—it's only necessary. The nth Term Test cannot prove convergence. Remember: The nth Term Test can only prove divergence, never convergence. When the limit of terms equals zero, you must use other tests like the integral test, comparison tests, or ratio test to determine convergence.

Question 13

A student claims that for any series ∑an\sum a_n where lim⁡n→∞an=L≠0\lim_{n \to \infty} a_n = L \neq 0, the nth Term Test for Divergence proves the series diverges, but if L=0L = 0, the test proves convergence. What is wrong with this reasoning?

  1. The reasoning is correct for L=0L = 0 but wrong for L≠0L \neq 0; the test requires ∣L∣>1|L| > 1 to prove divergence, not just L≠0L \neq 0
  2. The reasoning is wrong in both cases; the nth Term Test can only prove convergence, never divergence, regardless of the limit value
  3. The reasoning is correct for L≠0L \neq 0 but wrong for L=0L = 0; when the limit is zero, the test is inconclusive, not proof of convergence (correct answer)
  4. The reasoning is completely wrong; the nth Term Test applies only to alternating series and cannot be used for general series
Explanation: When you encounter questions about convergence tests, it's crucial to understand exactly what each test can and cannot prove. The nth Term Test for Divergence is one of the most basic but frequently misunderstood tests. The nth Term Test states: If lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0, then ∑an\sum a_n diverges. However, if lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the test tells us nothing—the series might converge or diverge. This is why answer C is correct: the student's reasoning is right when L≠0L \neq 0 (the series does diverge), but wrong when L=0L = 0 because the test becomes inconclusive, not proof of convergence. Answer A incorrectly suggests you need ∣L∣>1|L| > 1 for divergence, but any nonzero limit causes divergence. Answer B completely reverses the test's purpose—it can prove divergence but never convergence. Answer D is nonsensical since the nth Term Test applies to all series, not just alternating ones. The classic example illustrating the L=0L = 0 case is comparing ∑1n\sum \frac{1}{n} (diverges) with ∑1n2\sum \frac{1}{n^2} (converges). Both have lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, yet they behave differently. Study tip: Remember the nth Term Test as a "divergence detector only." If the limit isn't zero, you're done—it diverges. If the limit is zero, you've learned nothing and must try other tests. Never let a zero limit fool you into thinking the series converges.

Question 14

For the series ∑n=1∞2nn⋅3n\sum_{n=1}^{\infty} \frac{2^n}{n \cdot 3^n}, a student incorrectly concludes that the nth Term Test proves divergence. What error did the student most likely make?

  1. The student incorrectly calculated lim⁡n→∞2nn⋅3n=+∞\lim_{n \to \infty} \frac{2^n}{n \cdot 3^n} = +\infty by focusing only on exponential growth in the numerator (correct answer)
  2. The student incorrectly applied the test to the partial sums instead of the individual terms of the series
  3. The student correctly found lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0 but incorrectly concluded this proves divergence rather than being inconclusive
  4. The student incorrectly assumed that series with exponential terms automatically diverge by the nth Term Test
Explanation: The correct limit is lim⁡n→∞2nn⋅3n=lim⁡n→∞1n⋅(23)n=0\lim_{n \to \infty} \frac{2^n}{n \cdot 3^n} = \lim_{n \to \infty} \frac{1}{n} \cdot \left(\frac{2}{3}\right)^n = 0 since (23)n→0\left(\frac{2}{3}\right)^n \to 0 exponentially fast, dominating the 1n\frac{1}{n} factor. This makes the nth Term Test inconclusive, not proving divergence. The most common error would be focusing on the 2n2^n term and incorrectly concluding the limit is infinite, ignoring that 3n3^n grows faster than 2n2^n. Choice B describes an unusual procedural error. Choice C describes the opposite error (correct limit, wrong conclusion). Choice D describes a conceptual misunderstanding about exponential terms.

Question 15

Let ∑an\sum a_n be a series of positive terms. If the sequence of terms {an}\{a_n\} converges, what can be concluded about the series ∑an\sum a_n using only the nth Term Test?

  1. The series ∑an\sum a_n must diverge, because the test is always conclusive for a convergent sequence of terms.
  2. The series ∑an\sum a_n must converge, because its terms approach a single finite value.
  3. The series ∑an\sum a_n diverges if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0, and the test is inconclusive if lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0. (correct answer)
  4. The series ∑an\sum a_n must have the same convergence status as the sequence {an}\{a_n\}.
Explanation: If the sequence {an}\{a_n\} converges, its limit LL is a finite number. The nth Term Test has two possible outcomes based on this limit. Case 1: L≠0L \neq 0. In this case, the test concludes that the series ∑an\sum a_n diverges. Case 2: L=0L = 0. In this case, the test is inconclusive, and another test must be used to determine the series' behavior. Therefore, the outcome depends on whether the limit is zero or not.

Question 16

Consider the series ∑n=2∞ln⁡(n3)n\sum_{n=2}^\infty \frac{\ln(n^3)}{\sqrt{n}}. What is the result of applying the nth Term Test for Divergence?

  1. The test proves the series diverges because ln⁡(n3)\ln(n^3) approaches infinity.
  2. The test proves the series converges because the limit of the terms is 0.
  3. The test is inconclusive because the limit of the terms is 0. (correct answer)
  4. The test proves the series diverges because the limit of the terms is infinite.
Explanation: We evaluate lim⁡n→∞an=lim⁡n→∞ln⁡(n3)n=lim⁡n→∞3ln⁡(n)n1/2\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\ln(n^3)}{\sqrt{n}} = \lim_{n \to \infty} \frac{3\ln(n)}{n^{1/2}}. This is an indeterminate form ∞∞\frac{\infty}{\infty}, so we can use L'Hôpital's Rule. \nlim⁡n→∞3ln⁡(n)n1/2=lim⁡n→∞ddn(3ln⁡(n))ddn(n1/2)=lim⁡n→∞3/n12n−1/2=lim⁡n→∞6n=0\lim_{n \to \infty} \frac{3\ln(n)}{n^{1/2}} = \lim_{n \to \infty} \frac{\frac{d}{dn}(3\ln(n))}{\frac{d}{dn}(n^{1/2})} = \lim_{n \to \infty} \frac{3/n}{\frac{1}{2}n^{-1/2}} = \lim_{n \to \infty} \frac{6}{\sqrt{n}} = 0 \nSince the limit of the terms is 0, the nth Term Test is inconclusive. It provides no information about whether the series converges or diverges.

Question 17

For which of the following series is the nth Term Test for Divergence conclusive?

  1. ∑n=1∞sin⁡2(n)n2\sum_{n=1}^\infty \frac{\sin^2(n)}{n^2}
  2. ∑n=1∞ne−n2\sum_{n=1}^\infty n e^{-n^2}
  3. ∑n=1∞1n\sum_{n=1}^\infty \frac{1}{\sqrt{n}}
  4. ∑n=1∞n2−5n1+n+2n2\sum_{n=1}^\infty \frac{n^2 - 5n}{1 + n + 2n^2} (correct answer)
Explanation: The nth Term Test is conclusive if the limit of the terms is not zero. Let's check the limits for each choice:\nA) 0≤sin⁡2(n)≤10 \le \sin^2(n) \le 1, so 0≤sin⁡2(n)n2≤1n20 \le \frac{\sin^2(n)}{n^2} \le \frac{1}{n^2}. By the Squeeze Theorem, lim⁡n→∞sin⁡2(n)n2=0\lim_{n \to \infty} \frac{\sin^2(n)}{n^2} = 0. Inconclusive.\nB) Using L'Hôpital's Rule on nen2\frac{n}{e^{n^2}}, the limit is 0. Inconclusive.\nC) lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0. Inconclusive.\nD) lim⁡n→∞n2−5n1+n+2n2=12\lim_{n \to \infty} \frac{n^2 - 5n}{1 + n + 2n^2} = \frac{1}{2}. Since the limit is not 0, the test is conclusive and proves divergence.

Question 18

Consider the series ∑(an+c)\sum (a_n + c), where cc is a non-zero constant and lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0. What can be concluded about this series using the nth Term Test?

  1. The test is inconclusive, as its result depends on the convergence of ∑an\sum a_n.
  2. The series converges if c<0c < 0 and diverges if c>0c > 0.
  3. The series diverges for any non-zero value of cc. (correct answer)
  4. The series converges to the same sum as ∑an\sum a_n, shifted by the constant cc.
Explanation: Let the terms of the new series be bn=an+cb_n = a_n + c. To apply the nth Term Test, we find the limit of bnb_n: \nlim⁡n→∞bn=lim⁡n→∞(an+c)=(lim⁡n→∞an)+(lim⁡n→∞c)=0+c=c\lim_{n \to \infty} b_n = \lim_{n \to \infty} (a_n + c) = (\lim_{n \to \infty} a_n) + (\lim_{n \to \infty} c) = 0 + c = c \nSince cc is a non-zero constant, the limit of the terms is not zero. According to the nth Term Test, the series ∑(an+c)\sum (a_n + c) must diverge.

Question 19

The nth Term Test for Divergence is inconclusive for three of the following series. For which series does it conclusively prove divergence?

  1. ∑n=1∞ln⁡nn2\sum_{n=1}^{\infty} \frac{\ln n}{n^2}
  2. ∑n=1∞arctan⁡(1/n)\sum_{n=1}^{\infty} \arctan(1/n)
  3. ∑n=1∞n!nn\sum_{n=1}^{\infty} \frac{n!}{n^n}
  4. ∑n=1∞nsin⁡(1/n)\sum_{n=1}^{\infty} n \sin(1/n) (correct answer)
Explanation: We check the limit of the terms for each series. The test is conclusive only if the limit is non-zero.\nA) lim⁡n→∞ln⁡nn2=0\lim_{n \to \infty} \frac{\ln n}{n^2} = 0 by L'Hôpital's Rule. Inconclusive.\nB) As n→∞n \to \infty, 1/n→01/n \to 0, so lim⁡n→∞arctan⁡(1/n)=arctan⁡(0)=0\lim_{n \to \infty} \arctan(1/n) = \arctan(0) = 0. Inconclusive.\nC) For n>1n>1, 0<n!nn=1⋅2⋯nn⋅n⋯n≤1n0 < \frac{n!}{n^n} = \frac{1 \cdot 2 \cdots n}{n \cdot n \cdots n} \le \frac{1}{n}. By Squeeze Theorem, the limit is 0. Inconclusive.\nD) We evaluate lim⁡n→∞nsin⁡(1/n)=lim⁡n→∞sin⁡(1/n)1/n\lim_{n \to \infty} n \sin(1/n) = \lim_{n \to \infty} \frac{\sin(1/n)}{1/n}. Let x=1/nx=1/n. As n→∞n \to \infty, x→0x \to 0. The limit becomes lim⁡x→0sin⁡(x)x=1\lim_{x \to 0} \frac{\sin(x)}{x} = 1. Since the limit is 1 (not 0), the test conclusively proves divergence.

Question 20

Let pp be a fixed real number. What conclusion does the nth Term Test provide for the series ∑n=1∞npe−n\sum_{n=1}^\infty n^p e^{-n}?

  1. The test proves divergence for p>0p > 0 and is inconclusive otherwise.
  2. The test is inconclusive for all real values of pp. (correct answer)
  3. The test proves divergence for all real values of pp.
  4. The test proves convergence for p<0p < 0 and is inconclusive otherwise.
Explanation: To apply the nth Term Test, we evaluate lim⁡n→∞npe−n=lim⁡n→∞npen\lim_{n \to \infty} n^p e^{-n} = \lim_{n \to \infty} \frac{n^p}{e^n}. The exponential function ene^n in the denominator grows faster than any power function npn^p in the numerator, regardless of the value of pp. Therefore, for any fixed real number pp, the limit is 0. Since lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the nth Term Test is inconclusive for all real values of pp.