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