The nth Term Test for Divergence - AP Calculus BC

1. For $\sum_{n=1}^{\infty} \frac{n^2}{n+1}$, what does the nth-term test imply about divergence?
   Answer: The series diverges because $\lim_{n\to\infty}\frac{n^2}{n+1}\neq 0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice A, which claims convergence because the limit is zero, but the limit is actually infinity, proving divergence. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
2. A series is $\sum_{n=1}^{\infty} \frac{(-2)^n}{n!}$. What does the nth-term test conclude?
   Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{(-2)^n}{n!}=0$. In AP Calculus BC, the nth-term test for divergence helps classify series. It only proves divergence for nonzero limits; zero is inconclusive. Factorials often lead to zero limits. $\lim_{n\to\infty} \frac{(-2)^n}{n!} = 0$, no conclusion. A tempting mistake is convergence due to factorials, but the test does not confirm it. Apply the nth-term test as a starting point, using ratio test for zero-limit cases like this.
3. A student defines $\sum_{n=1}^{\infty} \frac{n}{n+5}$. What does the nth-term test imply about this series?
   Answer: The series diverges because $\lim_{n\to\infty} \frac{n}{n+5} \neq 0$. The nth-term test for divergence is a key skill in AP Calculus BC for assessing series behavior. This test asserts that if the limit of the nth term as n approaches infinity is not zero, the series diverges because the terms do not become sufficiently small. For convergence, terms must approach zero to allow partial sums to stabilize, but a nonzero limit means additions remain significant indefinitely. Thus, since lim (n/(n+5)) = 1 ≠ 0, the series diverges. A tempting distractor might claim convergence because the limit is 1, but the test requires the limit to be exactly zero for any chance of convergence. Always apply the nth-term test first to potentially rule out convergence quickly when the limit is nonzero.
4. A series is $\sum_{n=1}^{\infty} \left(\frac{n}{n+1}\right)^n$. What does the nth-term test conclude?
   Answer: The series diverges because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\neq 0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice B, which claims convergence because the limit is zero, but the limit is actually 1/e, not zero, leading to divergence. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
5. A series is $\sum_{n=1}^{\infty} \frac{n}{\ln(n+2)}$. What does the nth-term test imply?
   Answer: The series diverges because $\lim_{n\to\infty}\frac{n}{\ln(n+2)}\neq 0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice A, which claims convergence because the limit is zero, but the limit is actually infinity, not zero, proving divergence. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
6. For $\sum_{n=1}^{\infty} \frac{2n+3}{n^2}$, what does the nth-term test conclude?
   Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{2n+3}{n^2}=0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice B, which claims convergence because the limit is zero, but the test does not confirm convergence here. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
7. Consider $\sum_{n=1}^{\infty} \left(\frac{n+1}{n}\right)$. What does the nth-term test imply?
   Answer: The series diverges because $\lim_{n\to\infty}\frac{n+1}{n}\neq 0$. The nth-term test for divergence is a key AP Calculus BC concept for series. It asserts divergence if lim $a_n$ ≠ 0, since terms must vanish for convergence. Nonzero limits cause endless significant additions. Lim ((n+1)/n) = 1 ≠ 0, so diverges. Temptingly, one might claim convergence due to decreasing terms, but decreasing to 1 ≠ 0 still diverges. Strategically, evaluate lim $a_n$ first to rule out convergence swiftly when nonzero.
8. A series is $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n}$. What does the nth-term test conclude?
   Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{\sqrt{n}}{n}=0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice A, which claims convergence because the limit is zero, but the test is inconclusive when the limit is zero. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
9. A series is $\sum_{n=1}^{\infty} \frac{n}{n^2+n+1}$. What does the nth-term test conclude?
   Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{n}{n^2+n+1}=0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice A, which claims convergence because the limit is zero, but the test is inconclusive when the limit is zero. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
10. A sum is $\sum_{n=1}^{\infty} \frac{n}{n^2-1}$. What does the nth-term test conclude?
    Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{n}{n^2-1}=0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice A, which claims convergence because the limit is zero, but the test is inconclusive when the limit is zero. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
11. Consider $\sum_{n=1}^{\infty} \frac{2n}{n^2+1}$. What does the nth-term test imply?
    Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{2n}{n^2+1}=0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice A, which claims convergence because the limit is zero, but the test does not confirm convergence in this case. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
12. A sum is defined as $\sum_{n=1}^{\infty} \frac{1}{1+n^4}$. What can the nth-term test conclude?
    Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{1}{1+n^4}=0$. AP Calculus BC includes the nth-term test for divergence as a core skill. The test spots divergence via nonzero limits but is inconclusive for zero. Further tests are required then. Lim (1/(1+n⁴)) = 0, no conclusion. A distractor suggests divergence because n⁴ grows quickly, but quick growth ensures zero limit, not divergence. Use the nth-term test first: nonzero means diverge; zero means continue with p-series or comparison.
13. Consider $\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$. What does the nth-term test conclude?
    Answer: No conclusion can be drawn because $\lim_{n\to\infty}\left(\frac12\right)^n=0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as $n \to \infty$ does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice B, which claims convergence because the limit is zero, but the test is inconclusive when the limit is zero. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
14. Consider $\sum_{n=1}^{\infty} \frac{3n+1}{n}$. What does the nth-term test indicate?
    Answer: The series diverges because $\lim_{n\to\infty}\frac{3n+1}{n}\neq 0$. The nth-term test for divergence is a fundamental skill in series analysis within AP Calculus BC. It states that a series diverges if the limit of its terms does not equal zero, as persistent nonzero additions prevent the sum from converging. The partial sums would keep growing without bound if terms approach a value like 3. Here, lim ((3n+1)/n) = 3 ≠ 0, confirming divergence. One might be tempted to say it converges because the terms are positive, but positivity alone does not imply convergence without the limit being zero. Remember to use the nth-term test as an initial check to identify divergence when limits are nonzero, though zero limits require further tests.
15. For $\sum_{n=1}^{\infty} \frac{n^2+4}{n^2}$, what does the nth-term test imply about divergence?
    Answer: The series diverges because $\lim_{n\to\infty}\frac{n^2+4}{n^2}\neq 0$. The nth-term test for divergence is essential in AP Calculus BC series topics. It declares divergence when lim $a_n$ ≠ 0, preventing sum finiteness. Nonzero limits imply unbounded growth. Lim ((n²+4)/n²) = 1 ≠ 0, so diverges. Temptingly, one might say convergence because decreasing, but decreasing to 1 ≠ 0 diverges. Always begin with the nth-term test for quick divergence detection, then apply alternatives if needed.
16. Consider $\sum_{n=1}^{\infty} \frac{1}{n+\cos n}$. What can the nth-term test conclude?
    Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{1}{n+\cos n}=0$. The nth-term test for divergence is a vital skill for series in AP Calculus BC. It concludes divergence solely when lim $a_n$ ≠ 0, as non-vanishing terms prevent sum stabilization. If terms approach zero, additional analysis is required. For this series, lim (1/(n + cos n)) = 0, yielding no conclusion. Temptingly, one might say it diverges due to oscillating cos n, but the limit is zero regardless. Use the nth-term test as a first step: nonzero limit means diverge; zero means investigate further with tools like p-series or ratio tests.
17. For $\sum_{n=1}^{\infty} \frac{n+\sin n}{n}$, what does the nth-term test imply?
    Answer: The series diverges because $\lim_{n\to\infty}\frac{n+\sin n}{n}\neq 0$. The nth-term test for divergence is fundamental in AP Calculus BC. It asserts divergence if lim $a_n$ ≠ 0, with terms needing to vanish. Bounded oscillations do not prevent nonzero limits here. Lim ((n + sin n)/n) = 1 ≠ 0, so diverges. One distractor says no conclusion due to oscillation, but the limit exists and is 1. Always compute lim $a_n$ carefully, concluding divergence if nonzero, and investigate further if zero.
18. For $\sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)$, what does the nth-term test say about the series?
    Answer: The series diverges because $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\neq 0$. The nth-term test for divergence is a core concept in AP Calculus BC series topics. It declares a series divergent if lim $a_n$ ≠ 0, since terms must vanish for the sum to possibly converge. Nonzero limiting terms continually add a fixed amount, causing divergence. In this case, lim (1 + 1/n) = 1 ≠ 0, so it diverges. One might incorrectly choose convergence because terms decrease, but decreasing to a nonzero value still leads to divergence. As a strategy, prioritize the nth-term test to swiftly detect divergence via nonzero limits, reserving other methods for zero-limit cases.
19. For $\sum_{n=1}^{\infty} \frac{\arctan(n)}{n}$, what does the nth-term test conclude?
    Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{\arctan(n)}{n}=0$. The skill being tested is the nth-term test for divergence. This test states that if the limit of the sequence terms as n approaches infinity does not equal zero, then the infinite series must diverge. The reasoning is that for the partial sums to approach a finite value, the added terms must become negligible, which fails if they approach a nonzero number. If the terms do not shrink to zero, their accumulation cannot settle to a limit. A tempting distractor is choice B, which claims convergence because the limit is zero, but while the limit is zero, the test is inconclusive. A transferable strategy for the nth-term test is to always compute the limit of $a_n$ first; if it is not zero, conclude divergence, otherwise apply other convergence tests.
20. A series is $\sum_{n=1}^{\infty} \frac{\cos n}{\sqrt{n}}$. What does the nth-term test conclude?
    Answer: No conclusion can be drawn because $\lim_{n\to\infty}\frac{\cos n}{\sqrt{n}}=0$. In AP Calculus BC, the nth-term test for divergence is used to assess series. It concludes divergence only for nonzero limits; zero limits leave options open. Oscillation with diminishing amplitude still hits zero. Lim (cos n / √n) = 0, so inconclusive. A common error is divergence due to no limit for cos n, but the overall limit is zero by squeeze theorem. Strategically, compute lim $a_n$; if not zero, diverge; if zero, use absolute convergence or other tests.

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