nth Term Test for Divergence - Calculus 2

1. For the series $\sum_{n=1}^\infty \left(\frac{n+k}{n}\right)^n$, where $k$ is a real constant, the nth Term Test is used. Which statement is correct?
   Answer: The test proves divergence for all real numbers $k$. We evaluate the limit of the terms: $\lim_{n \to \infty} \left(\frac{n+k}{n}\right)^n = \lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n$. This is a standard limit form equal to $e^k$. The nth Term Test states that if this limit is not zero, the series diverges. The value of $e^k$ is never zero for any real number $k$. Even for $k=0$, the limit is $e^0=1$, which is not zero. Therefore, the test proves that the series diverges for all real values of $k$.
2. For a series $\sum_{n=1}^\infty a_n$, it is found that $\lim_{n \to \infty} a_n = L$. The nth Term Test is then applied. For which value of $L$ is the test's conclusion that the series must diverge?
   Answer: Any $L \neq 0$. 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 $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum_{n=1}^\infty a_n$ must diverge. This makes intuitive sense—if the terms aren't approaching zero, they can't possibly sum to a finite value. However, if $\lim_{n \to \infty} a_n = 0$, the test is inconclusive; the series might converge or diverge. The correct answer is A because when $L \neq 0$ (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 $L = 0$, 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 $L = 0$, 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.
3. Let $S_n$ be the nth partial sum of the series $\sum_{k=1}^\infty a_k$. If the sequence of partial sums $\{S_n\}$ converges to 10, what can be concluded about the terms $\{a_n\}$ in the context of the nth Term Test?
   Answer: $\lim_{n \to \infty} a_n = 0$, which makes the nth Term Test inconclusive for this series. The convergence of the series $\sum a_k$ to a sum of 10 means that $\lim_{n \to \infty} S_n = 10$. The term $a_n$ can be expressed as $a_n = S_n - S_{n-1}$ for $n > 1$. Taking the limit as $n \to \infty$, we get $\lim_{n \to \infty} a_n = \lim_{n \to \infty} (S_n - S_{n-1}) = \lim_{n \to \infty} S_n - \lim_{n \to \infty} S_{n-1} = 10 - 10 = 0$. 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.
4. A student correctly determines that for a series $\sum a_n$, the sequence of terms $\{a_n\}$ converges to $L = 0.01$. Based on the nth Term Test, what can be concluded about the series $\sum a_n$?
   Answer: The series must diverge. The nth Term Test for Divergence states that if $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum a_n$ diverges. In this case, the limit of the terms is $L = 0.01$, 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.
5. Let $p$ be a fixed real number. What conclusion does the nth Term Test provide for the series $\sum_{n=1}^\infty n^p e^{-n}$?
   Answer: The test is inconclusive for all real values of $p$. To apply the nth Term Test, we evaluate $\lim_{n \to \infty} n^p e^{-n} = \lim_{n \to \infty} \frac{n^p}{e^n}$. The exponential function $e^n$ in the denominator grows faster than any power function $n^p$ in the numerator, regardless of the value of $p$. Therefore, for any fixed real number $p$, the limit is 0. Since $\lim_{n \to \infty} a_n = 0$, the nth Term Test is inconclusive for all real values of $p$.
6. For which of the following series would the nth Term Test for Divergence provide a definitive conclusion about the behavior of the series?
   Answer: $\sum_{n=1}^{\infty} \frac{4n^3 - 2n + 1}{3n^2 + 7n - 2}$ because the limit of terms grows without bound as $n \to \infty$ The nth Term Test gives a definitive conclusion only when $\lim_{n \to \infty} a_n \neq 0$, which proves divergence. For choice B: $\lim_{n \to \infty} \frac{4n^3 - 2n + 1}{3n^2 + 7n - 2} = \lim_{n \to \infty} \frac{4n - \frac{2}{n} + \frac{1}{n^2}}{3 + \frac{7}{n} - \frac{2}{n^2}} = +\infty \neq 0$, so the test proves divergence. Choice A has limit 0 (inconclusive). Choice C has limit 0 (inconclusive). Choice D has limit 0 since $|\cos(n)| \leq 1$ (inconclusive).
7. Let the sequence $\{a_n\}$ be defined as $a_n = \frac{(-1)^n n}{n+1}$ if $n$ is a multiple of 3, and $a_n = \frac{1}{n^2}$ otherwise. What is the conclusion of the nth Term Test for the series $\sum_{n=1}^\infty a_n$?
   Answer: The series diverges because the limit of the terms does not exist. To determine the limit of the sequence $\{a_n\}$, we examine its subsequences. For $n$ that are not multiples of 3, $a_n = 1/n^2$, and this subsequence converges to 0. For $n$ that are multiples of 3, the subsequence is $a_{3k} = \frac{(-1)^{3k} 3k}{3k+1}$. As $k \to \infty$, the term $\frac{3k}{3k+1} \to 1$, but $(-1)^{3k}$ 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 $\lim_{n \to \infty} a_n$ does not exist. By the nth Term Test, the series diverges.
8. The condition $\lim_{n \to \infty} a_n = 0$ is a necessary, but not sufficient, condition for the convergence of $\sum a_n$. How does the nth Term Test relate to this principle?
   Answer: The test is the contrapositive statement: if the necessary condition is not met, then convergence is not possible. The statement "If $\sum a_n$ converges, then $\lim_{n \to \infty} a_n = 0$" means that having the limit be zero is a necessary requirement for convergence. The nth Term Test is the logical contrapositive of this statement. The contrapositive of "If P, then Q" is "If not Q, then not P". So, the contrapositive is: "If $\lim_{n \to \infty} a_n \neq 0$, then $\sum a_n$ does not converge (i.e., it diverges)." This is precisely the statement of the nth Term Test.
9. For the series $\sum_{n=1}^{\infty} \left(1 - \frac{3}{n}\right)^n$, what conclusion can be drawn from the nth Term Test for Divergence?
   Answer: The series diverges because the limit of the terms is $e^{-3}$. To apply the nth Term Test, we must evaluate $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(1 - \frac{3}{n}\right)^n$. This limit is of the indeterminate form $1^\infty$ and corresponds to the definition of the exponential function, $\lim_{n \to \infty} (1 + x/n)^n = e^x$. In this case, $x = -3$. Therefore, the limit is $e^{-3}$. Since $e^{-3} \approx 0.0498$, the limit is not zero. By the nth Term Test, the series diverges.
10. Let $\sum a_n$ be a series of positive terms. If the sequence of terms $\{a_n\}$ converges, what can be concluded about the series $\sum a_n$ using only the nth Term Test?
    Answer: The series $\sum a_n$ diverges if $\lim_{n \to \infty} a_n \neq 0$, and the test is inconclusive if $\lim_{n \to \infty} a_n = 0$. If the sequence $\{a_n\}$ converges, its limit $L$ is a finite number. The nth Term Test has two possible outcomes based on this limit. Case 1: $L \neq 0$. In this case, the test concludes that the series $\sum a_n$ diverges. Case 2: $L = 0$. 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.
11. For what real values of the constant $k$ does the nth Term Test conclusively prove that the series $\sum_{n=1}^\infty \frac{kn^2 + n}{3n^2 - 1}$ diverges?
    Answer: For all $k \neq 0$. The nth Term Test proves divergence if the limit of the terms is non-zero. Let's find the limit: $$\lim_{n \to \infty} \frac{kn^2 + n}{3n^2 - 1} = \lim_{n \to \infty} \frac{k + 1/n}{3 - 1/n^2} = \frac{k}{3}$$ The test is conclusive and proves divergence if this limit is not equal to zero. The equation $\frac{k}{3} \neq 0$ is true for all values of $k$ except $k=0$. If $k=0$, the limit is 0, and the test is inconclusive.
12. Consider the series $\sum_{n=1}^{\infty} \frac{\ln(n^2 + 1)}{n^2}$. When applying the nth Term Test for Divergence, what can be concluded?
    Answer: The test is inconclusive because $\lim_{n \to \infty} \frac{\ln(n^2 + 1)}{n^2} = 0$, requiring other convergence tests for determination When you encounter a series convergence problem, the nth Term Test for Divergence is often your first step. This test states that if $\lim_{n \to \infty} a_n \neq 0$, then $\sum a_n$ diverges. However, if $\lim_{n \to \infty} a_n = 0$, the test tells us nothing about convergence. Let's find $\lim_{n \to \infty} \frac{\ln(n^2 + 1)}{n^2}$. This is an $\frac{\infty}{\infty}$ indeterminate form, so we can apply L'Hôpital's rule: $\lim_{n \to \infty} \frac{\ln(n^2 + 1)}{n^2} = \lim_{n \to \infty} \frac{\frac{2n}{n^2 + 1}}{2n} = \lim_{n \to \infty} \frac{2n}{2n(n^2 + 1)} = \lim_{n \to \infty} \frac{1}{n^2 + 1} = 0$ 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 $\lim_{n \to \infty} a_n = 0$ 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.
13. The nth Term Test for Divergence is inconclusive for three of the following series. For which series does it conclusively prove divergence?
    Answer: $\sum_{n=1}^{\infty} n \sin(1/n)$ We check the limit of the terms for each series. The test is conclusive only if the limit is non-zero. A) $\lim_{n \to \infty} \frac{\ln n}{n^2} = 0$ by L'Hôpital's Rule. Inconclusive. B) As $n \to \infty$, $1/n \to 0$, so $\lim_{n \to \infty} \arctan(1/n) = \arctan(0) = 0$. Inconclusive. C) For $n>1$, $0 < \frac{n!}{n^n} = \frac{1 \cdot 2 \cdots n}{n \cdot n \cdots n} \le \frac{1}{n}$. By Squeeze Theorem, the limit is 0. Inconclusive. D) We evaluate $\lim_{n \to \infty} n \sin(1/n) = \lim_{n \to \infty} \frac{\sin(1/n)}{1/n}$. Let $x=1/n$. As $n \to \infty$, $x \to 0$. The limit becomes $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. Since the limit is 1 (not 0), the test conclusively proves divergence.
14. For the series $\sum_{n=1}^{\infty} \frac{2^n}{n \cdot 3^n}$, a student incorrectly concludes that the nth Term Test proves divergence. What error did the student most likely make?
    Answer: The student incorrectly calculated $\lim_{n \to \infty} \frac{2^n}{n \cdot 3^n} = +\infty$ by focusing only on exponential growth in the numerator The correct limit is $\lim_{n \to \infty} \frac{2^n}{n \cdot 3^n} = \lim_{n \to \infty} \frac{1}{n} \cdot \left(\frac{2}{3}\right)^n = 0$ since $\left(\frac{2}{3}\right)^n \to 0$ exponentially fast, dominating the $\frac{1}{n}$ factor. This makes the nth Term Test inconclusive, not proving divergence. The most common error would be focusing on the $2^n$ term and incorrectly concluding the limit is infinite, ignoring that $3^n$ grows faster than $2^n$. 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.
15. A student argues: "For the series $\sum_{n=1}^{\infty} \frac{1}{n}$, the limit of the terms is $\lim_{n \to \infty} \frac{1}{n} = 0$. 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?
    Answer: The student has misunderstood the conclusion of the nth Term Test when the limit is zero. 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 $\lim_{n \to \infty} a_n = 0$, 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.
16. Let $\sum a_n$ and $\sum b_n$ be two series. It is known that $\lim_{n \to \infty} a_n = 2$ and the series $\sum b_n$ converges. What can be concluded about the series $\sum (a_n - b_n)$ using the nth Term Test?
    Answer: The series $\sum (a_n - b_n)$ diverges because the limit of its terms is 2. Let $c_n = a_n - b_n$. We need to evaluate $\lim_{n \to \infty} c_n$ to apply the nth Term Test. We are given $\lim_{n \to \infty} a_n = 2$. Since the series $\sum b_n$ converges, it is a necessary condition that the limit of its terms is zero, so $\lim_{n \to \infty} b_n = 0$. Using the limit properties, $\lim_{n \to \infty} c_n = \lim_{n \to \infty} (a_n - b_n) = \lim_{n \to \infty} a_n - \lim_{n \to \infty} b_n = 2 - 0 = 2$. Since the limit of the terms is 2, which is not zero, the nth Term Test concludes that the series $\sum (a_n - b_n)$ diverges.
17. Consider the series $\sum (a_n + c)$, where $c$ is a non-zero constant and $\lim_{n \to \infty} a_n = 0$. What can be concluded about this series using the nth Term Test?
    Answer: The series diverges for any non-zero value of $c$. Let the terms of the new series be $b_n = a_n + c$. To apply the nth Term Test, we find the limit of $b_n$: $$\lim_{n \to \infty} b_n = \lim_{n \to \infty} (a_n + c) = (\lim_{n \to \infty} a_n) + (\lim_{n \to \infty} c) = 0 + c = c$$ Since $c$ is a non-zero constant, the limit of the terms is not zero. According to the nth Term Test, the series $\sum (a_n + c)$ must diverge.
18. Consider the series $\sum_{n=1}^\infty \cos(n\pi) \arctan(n)$. What is the conclusion of the nth Term Test for Divergence?
    Answer: The series diverges because the limit of the terms does not exist. Let $a_n = \cos(n\pi) \arctan(n)$. We know that $\cos(n\pi) = (-1)^n$ and $\lim_{n \to \infty} \arctan(n) = \pi/2$. The sequence of terms $a_n$ becomes approximately $(-1)^n(\pi/2)$ for large $n$. This means the terms oscillate between values close to $\pi/2$ and $-\pi/2$. Since the sequence of terms does not approach a single value, the limit $\lim_{n \to \infty} a_n$ does not exist. According to the nth Term Test, if the limit of the terms does not exist, the series diverges.
19. Consider the series $\sum_{n=1}^\infty \frac{(n+1)!}{3^n n!}$. What is the conclusion of the nth Term Test?
    Answer: The test is inconclusive because the limit of the terms is 0. First, we simplify the general term $a_n = \frac{(n+1)!}{3^n n!}$. We can write $(n+1)! = (n+1) \cdot n!$. So, $a_n = \frac{(n+1)n!}{3^n n!} = \frac{n+1}{3^n}$. Now we apply the nth Term Test by finding the limit: $\lim_{n \to \infty} \frac{n+1}{3^n}$. Since exponential functions grow much faster than linear functions, this limit is 0. This can be formally shown using L'Hôpital's Rule on $f(x) = (x+1)/3^x$. Because the limit of the terms is 0, the nth Term Test is inconclusive.
20. Consider the series $\sum_{n=1}^{\infty} a_n$ where $a_n = \frac{2^n + (-1)^n}{3^n + 1}$. A student applies the nth Term Test for Divergence and concludes the series diverges. Which analysis best explains whether this conclusion is justified?
    Answer: The conclusion is unjustified because $\lim_{n \to \infty} a_n = 0$ since $3^n$ grows faster than $2^n$, making the test inconclusive We have $a_n = \frac{2^n + (-1)^n}{3^n + 1}$. Dividing numerator and denominator by $3^n$: $a_n = \frac{\left(\frac{2}{3}\right)^n + \frac{(-1)^n}{3^n}}{1 + \frac{1}{3^n}}$. As $n \to \infty$, $\left(\frac{2}{3}\right)^n \to 0$ and $\frac{(-1)^n}{3^n} \to 0$, so $\lim_{n \to \infty} a_n = 0$. 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.

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