AP Calculus BC Quiz: The Nth Term Test For Divergence

What this quiz covers

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

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.

Question 1

1. The series diverges because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\neq 0$. (correct answer)
2. The series converges because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$ exists.
4. No conclusion can be drawn because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\neq 0$.
5. The series diverges because the base is less than $1$.

Explanation: 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.

Question 2

1. The series diverges because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$.
2. The series converges because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$. (correct answer)
4. The series diverges because $\sin(1/n)$ is approximately $1/n$.
5. The series converges because the terms are products.

Explanation: AP Calculus BC teaches the nth-term test for divergence for series evaluation. The test identifies divergence only when lim a_n ≠ 0; zero limits are neutral. Approximations like sin(1/n) ≈ 1/n help confirm zero. Lim ((1/n³) sin(1/n)) = 0, inconclusive. A common distractor is divergence because sin(1/n) ≈ 1/n, but overall it's 1/n⁴ → 0. Use the nth-term test first, then p-series or integral for zero-limit series.

Question 3

1. The series diverges because $\lim_{n\to\infty}\frac{n^5}{n^5+2}\neq 0$. (correct answer)
2. The series converges because $\lim_{n\to\infty}\frac{n^5}{n^5+2}=1$.
3. No conclusion can be drawn because the limit equals $1$.
4. The series converges because the terms are less than $1$.
5. The series diverges because the limit exists.

Explanation: 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 1, but the test indicates divergence when the limit is not 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.

Question 4

1. The series diverges because $\lim_{n\to\infty}\frac{2n+3}{n^2}=2$.
2. The series converges because $\lim_{n\to\infty}\frac{2n+3}{n^2}=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{2n+3}{n^2}=0$. (correct answer)
4. The series diverges because the numerator is linear.
5. The series converges because the denominator is quadratic.

Explanation: 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.

Question 5

1. The series converges because $\lim_{n\to\infty}\frac{\cos n}{\sqrt{n}}=0$.
2. The series diverges because $\cos n$ does not have a limit.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{\cos n}{\sqrt{n}}=0$. (correct answer)
4. The series diverges because $\lim_{n\to\infty}\frac{\cos n}{\sqrt{n}}\neq 0$.
5. The series converges because $|\cos n|\le 1$.

Explanation: 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.

Question 6

1. The series converges because $\lim_{n\to\infty}\frac{n^2}{n+1}=0$.
2. No conclusion can be drawn because the limit is infinite.
3. The series diverges because $\lim_{n\to\infty}\frac{n^2}{n+1}\neq 0$. (correct answer)
4. The series converges because polynomials cancel.
5. The series diverges because the terms are rational.

Explanation: 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.

Question 7

1. No conclusion can be drawn because the limit is $1$.
2. The series converges because $\lim_{n\to\infty}\frac{n^2+4}{n^2}=1$.
3. The series diverges because $\lim_{n\to\infty}\frac{n^2+4}{n^2}\neq 0$. (correct answer)
4. The series converges because $\frac{n^2+4}{n^2}$ is decreasing.
5. The series diverges because the terms are greater than $1$.

Explanation: 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.

Question 8

1. The series converges because $\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n}+1}=1$.
2. No conclusion can be drawn because the limit equals $1$.
3. The series diverges because $\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n}+1}\neq 0$. (correct answer)
4. The series converges because the terms are less than $1$.
5. The series diverges because $\sqrt{n}$ increases.

Explanation: The nth-term test for divergence is a core skill in AP Calculus BC. It concludes divergence for nonzero term limits, requiring zero for potential convergence. Limits approaching $1$ indicate divergence. $\lim \left( \frac{\sqrt{n}}{\sqrt{n} + 1} \right) = 1 \neq 0$, so diverges. Temptingly, one might say convergence because $<1$, but approaching $1 \neq 0$ diverges. Strategically, check $\lim a_n$ upfront to detect divergence quickly, saving effort for ambiguous cases.

Question 9

1. The series converges because $\lim_{n\to\infty}\frac{2n}{n^2+1}=0$.
2. The series diverges because $\lim_{n\to\infty}\frac{2n}{n^2+1}=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{2n}{n^2+1}=0$. (correct answer)
4. The series diverges because the numerator is linear.
5. The series converges because the terms are rational.

Explanation: 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.

Question 10

1. The series converges because $\lim_{n\to\infty}\frac{1}{n}\cos\left(\frac{\pi}{n}\right)=0$.
2. No conclusion can be drawn because $\lim_{n\to\infty}\frac{1}{n}\cos\left(\frac{\pi}{n}\right)=0$. (correct answer)
3. The series diverges because $\cos(\pi/n)\to 1$.
4. The series diverges because $\lim_{n\to\infty}\frac{1}{n}\cos\left(\frac{\pi}{n}\right)\neq 0$.
5. The series converges because cosine is bounded.

Explanation: 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 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.

Question 11

1. The series diverges because $\lim_{n\to\infty}\left(\frac12\right)^n=0$.
2. The series converges because $\lim_{n\to\infty}\left(\frac12\right)^n=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\left(\frac12\right)^n=0$. (correct answer)
4. The series diverges because the ratio is $\frac12$.
5. The series converges because it is geometric, so nth-term test is unnecessary.

Explanation: 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.

Question 12

1. The series diverges because $\lim_{n\to\infty}\frac{1}{1+n^4}=0$.
2. No conclusion can be drawn because $\lim_{n\to\infty}\frac{1}{1+n^4}=0$. (correct answer)
3. The series converges because $\lim_{n\to\infty}\frac{1}{1+n^4}=0$.
4. The series diverges because $n^4$ grows quickly.
5. The series converges because the terms are positive.

Explanation: 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.

Question 13

1. The series converges because $\lim_{n\to\infty}\frac{\sqrt{n^2+1}}{n}=1$.
2. The series diverges because $\lim_{n\to\infty}\frac{\sqrt{n^2+1}}{n}\neq 0$. (correct answer)
3. No conclusion can be drawn because the limit equals $1$.
4. The series converges because $\frac{\sqrt{n^2+1}}{n}$ is decreasing.
5. The series diverges because the terms are irrational.

Explanation: 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 1, but the test proves divergence when the limit is not 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.

Question 14

1. The series converges because $\lim_{n\to\infty}\frac{\sqrt{n}}{n}=0$.
2. The series diverges because $\lim_{n\to\infty}\frac{\sqrt{n}}{n}=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{\sqrt{n}}{n}=0$. (correct answer)
4. The series converges because $\sqrt{n}$ grows.
5. The series diverges because the terms are decreasing.

Explanation: 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.

Question 15

1. The series converges because $\lim_{n\to\infty}\frac{n}{n^2-1}=0$.
2. The series diverges because $\lim_{n\to\infty}\frac{n}{n^2-1}=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{n}{n^2-1}=0$. (correct answer)
4. The series diverges because $n^2-1$ factors.
5. The series converges because the denominator is quadratic.

Explanation: 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.

Question 16

1. The series diverges because $\lim_{n\to\infty}\frac{\arctan(n)}{n}=\frac{\pi}{2}$.
2. The series converges because $\lim_{n\to\infty}\frac{\arctan(n)}{n}=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{\arctan(n)}{n}=0$. (correct answer)
4. The series diverges because $\arctan(n)$ has a limit.
5. The series converges because $\arctan(n)\le \frac{\pi}{2}$.

Explanation: 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.

Question 17

1. The series diverges because $\lim_{n\to\infty}\frac{2n-1}{2n}\neq 0$. (correct answer)
2. The series converges because $\lim_{n\to\infty}\frac{2n-1}{2n}=1$.
3. No conclusion can be drawn because the limit equals $1$.
4. The series converges because the terms are less than $1$.
5. The series diverges because the terms are decreasing.

Explanation: The nth-term test for divergence is a staple in AP Calculus BC for series. It confirms divergence if lim a_n ≠ 0, due to non-vanishing contributions. Terms approaching zero are essential for convergence. Lim ((2n-1)/(2n)) = 1 ≠ 0, hence diverges. One distractor claims convergence because terms <1, but approaching 1 ≠ 0 causes divergence. Prioritize the nth-term test to identify divergence efficiently before deeper analysis.

Question 18

1. The series diverges because $\lim_{n\to\infty}\frac{5}{\sqrt{n}}=0$.
2. The series converges because $\lim_{n\to\infty}\frac{5}{\sqrt{n}}=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{5}{\sqrt{n}}=0$. (correct answer)
4. The series diverges because the limit is $5$.
5. The series converges because $\sqrt{n}$ grows.

Explanation: In AP Calculus BC, the nth-term test for divergence aids in series classification. The test identifies divergence when the term limit is not zero, but is silent otherwise. Zero limits require further investigation. Lim (5/√n) = 0, yielding no conclusion. A distractor might say convergence because √n grows, but growth ensuring zero limit does not prove convergence. Always apply the nth-term test as a preliminary step, proceeding to other tests if the limit is zero.

Question 19

1. The series diverges because $\lim_{n\to\infty}\frac{n!}{(n+1)!}=1$.
2. The series converges because $\lim_{n\to\infty}\frac{n!}{(n+1)!}=0$.
3. No conclusion can be drawn because $\lim_{n\to\infty}\frac{n!}{(n+1)!}=0$. (correct answer)
4. The series diverges because factorials grow fast.
5. The series converges because the terms are fractions.

Explanation: 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 indeed zero, the test remains 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.

Question 20

1. The series converges because $\lim_{n\to\infty}\frac{3n+1}{n}=3$.
2. The series diverges because $\lim_{n\to\infty}\frac{3n+1}{n}\neq 0$. (correct answer)
3. No conclusion can be drawn because the terms are not alternating.
4. The series converges because $\frac{3n+1}{n}>0$.
5. The series diverges because $\frac{3n+1}{n}$ is decreasing.

Explanation: 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.