The nth Term Test for Divergence
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AP Calculus BC › The nth Term Test for Divergence
A series is $\sum_{n=1}^{\infty} \left(\frac{n}{n+1}\right)^n$. What does the nth-term test conclude?
No conclusion can be drawn because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$ exists.
The series converges because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n=0$.
The series diverges because the base is less than $1$.
The series diverges because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\neq 0$.
No conclusion can be drawn because $\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\neq 0$.
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.
A sum is $\sum_{n=1}^{\infty} \frac{1}{n^3}\sin\left(\frac{1}{n}\right)$. What does the nth-term test conclude?
The series diverges because $\sin(1/n)$ is approximately $1/n$.
The series converges because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$.
The series converges because the terms are products.
The series diverges because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$.
No conclusion can be drawn because $\lim_{n\to\infty}\frac{1}{n^3}\sin\left(\frac{1}{n}\right)=0$.
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.
For $\sum_{n=1}^{\infty} \frac{n^5}{n^5+2}$, what does the nth-term test imply about divergence?
The series diverges because $\lim_{n\to\infty}\frac{n^5}{n^5+2}\neq 0$.
The series diverges because the limit exists.
The series converges because the terms are less than $1$.
No conclusion can be drawn because the limit equals $1$.
The series converges because $\lim_{n\to\infty}\frac{n^5}{n^5+2}=1$.
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 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.
For $\sum_{n=1}^{\infty} \frac{2n+3}{n^2}$, what does the nth-term test conclude?
The series diverges because $\lim_{n\to\infty}\frac{2n+3}{n^2}=2$.
No conclusion can be drawn because $\lim_{n\to\infty}\frac{2n+3}{n^2}=0$.
The series converges because $\lim_{n\to\infty}\frac{2n+3}{n^2}=0$.
The series converges because the denominator is quadratic.
The series diverges because the numerator is linear.
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.
A series is $\sum_{n=1}^{\infty} \frac{\cos n}{\sqrt{n}}$. What does the nth-term test conclude?
The series diverges because $\lim_{n\to\infty}\frac{\cos n}{\sqrt{n}}\neq 0$.
The series converges because $\lim_{n\to\infty}\frac{\cos n}{\sqrt{n}}=0$.
The series converges because $|\cos n|\le 1$.
No conclusion can be drawn because $\lim_{n\to\infty}\frac{\cos n}{\sqrt{n}}=0$.
The series diverges because $\cos n$ does not have a limit.
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.
For $\sum_{n=1}^{\infty} \frac{n^2}{n+1}$, what does the nth-term test imply about divergence?
No conclusion can be drawn because the limit is infinite.
The series diverges because the terms are rational.
The series converges because polynomials cancel.
The series diverges because $\lim_{n\to\infty}\frac{n^2}{n+1}\neq 0$.
The series converges because $\lim_{n\to\infty}\frac{n^2}{n+1}=0$.
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.
For $\sum_{n=1}^{\infty} \frac{n^2+4}{n^2}$, what does the nth-term test imply about divergence?
No conclusion can be drawn because the limit is $1$.
The series converges because $\lim_{n\to\infty}\frac{n^2+4}{n^2}=1$.
The series diverges because the terms are greater than $1$.
The series diverges because $\lim_{n\to\infty}\frac{n^2+4}{n^2}\neq 0$.
The series converges because $\frac{n^2+4}{n^2}$ is decreasing.
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.
Consider $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{\sqrt{n}+1}$. What does the nth-term test imply?
The series converges because $\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n}+1}=1$.
The series diverges because $\sqrt{n}$ increases.
The series converges because the terms are less than $1$.
No conclusion can be drawn because the limit equals $1$.
The series diverges because $\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n}+1}\neq 0$.
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.
Consider $\sum_{n=1}^{\infty} \frac{2n}{n^2+1}$. What does the nth-term test imply?
The series converges because $\lim_{n\to\infty}\frac{2n}{n^2+1}=0$.
The series converges because the terms are rational.
The series diverges because the numerator is linear.
The series diverges because $\lim_{n\to\infty}\frac{2n}{n^2+1}=0$.
No conclusion can be drawn because $\lim_{n\to\infty}\frac{2n}{n^2+1}=0$.
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.
A series is $\sum_{n=1}^{\infty} \frac{1}{n}\cos\left(\frac{\pi}{n}\right)$. What does the nth-term test conclude?
The series diverges because $\lim_{n\to\infty}\frac{1}{n}\cos\left(\frac{\pi}{n}\right)\neq 0$.
The series converges because $\lim_{n\to\infty}\frac{1}{n}\cos\left(\frac{\pi}{n}\right)=0$.
The series diverges because $\cos(\pi/n)\to 1$.
No conclusion can be drawn because $\lim_{n\to\infty}\frac{1}{n}\cos\left(\frac{\pi}{n}\right)=0$.
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.