Alternating Series Test for Convergence
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AP Calculus BC › Alternating Series Test for Convergence
A damping term is $\sum_{n=1}^{\infty}(-1)^n\frac{n}{n^2+1}$. Does the series converge?
Converges by the Alternating Series Test because $\frac{n}{n^2+1}$ decreases and approaches $0$.
Converges because any alternating rational function converges.
Diverges because $\frac{n}{n^2+1}$ is not decreasing.
Diverges because $\lim_{n\to\infty}\frac{n}{n^2+1}=1$.
Converges absolutely because $\sum \frac{n}{n^2+1}$ converges.
Explanation
The skill being tested is the Alternating Series Test for convergence. The test applies when b_n > 0, is eventually decreasing, and limits to 0. For b_n = $n/(n^2$+1), it approaches 0 like 1/n, and is decreasing for n ≥ 1 as its derivative is negative. Thus, the series converges by the test. Choice D is tempting but wrong because lim $n/(n^2$+1) = 0, not 1; miscalculating the limit leads to errors. A key strategy is to analyze the asymptotic behavior of b_n to confirm it decreases and vanishes at infinity.
A series used in simulation is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\sin(1/n)}{n}$. Does it converge?
Converges because $\sin(1/n)$ is negative for large $n$.
Converges absolutely because $\sum \frac{\sin(1/n)}{n}$ converges.
Converges by the Alternating Series Test because $\frac{\sin(1/n)}{n}$ decreases for large $n$ and approaches $0$.
Diverges because $\lim_{n\to\infty}\frac{\sin(1/n)}{n}\neq 0$.
Diverges because $\sin(1/n)$ oscillates.
Explanation
The skill being tested here is the Alternating Series Test for convergence. The conditions are b_n positive, decreasing (eventually), and limiting to 0 for convergence. Eventual monotonicity suffices for large n. For ∑ $(-1)^{n+1}$ rac{sin(1/n)}{n}, b_n ≈ $1/n^2$ for large n, decreasing to 0, so it converges. A tempting distractor is choice B, claiming divergence due to oscillation in sin, but sin(1/n) is positive and the whole b_n decreases. A transferable strategy for alternating series is to use approximations like Taylor expansions for complex b_n.
A series is defined by $\sum_{n=1}^{\infty}(-1)^n\frac{1}{n+(-1)^n}$. Does it converge?
Diverges because $\lim_{n\to\infty}\frac{1}{n+(-1)^n}\neq 0$.
Diverges because $\frac{1}{n+(-1)^n}$ is not monotone decreasing.
Converges absolutely because $\sum \frac{1}{n+(-1)^n}$ converges.
Converges by the Alternating Series Test because $\frac{1}{n+(-1)^n}$ decreases and approaches $0$.
Converges because the denominator alternates.
Explanation
The skill being tested here is the Alternating Series Test for convergence. For ∑ $(-1)^n$ b_n to converge, b_n must be positive, the sequence must be monotonically decreasing, and lim b_n = 0. Monotonicity is crucial to prevent oscillations that could cause divergence. In this case, b_n = rac{1}{n + $(-1)^n$} oscillates and is not monotonically decreasing, so the series diverges. A tempting distractor is choice A, which assumes it decreases to 0, but the oscillation violates monotonicity. A transferable strategy for alternating series is to plot or compute a few terms to check for strict monotonicity in b_n.
A series appears as $\sum_{n=1}^{\infty}(-1)^n\frac{1}{1+\sin^2 n}$. Does it converge?
Converges because $\sin^2 n$ is periodic.
Converges because $\frac{1}{1+\sin^2 n}$ is bounded.
Converges absolutely because $\sum \frac{1}{1+\sin^2 n}$ converges.
Converges by the Alternating Series Test because $\frac{1}{1+\sin^2 n}$ decreases and approaches $0$.
Diverges because $\lim_{n\to\infty}\frac{1}{1+\sin^2 n}\neq 0$.
Explanation
The skill being tested is the Alternating Series Test for convergence. Test requires lim b_n=0 among others. Here, b_n = $1/(1+sin^2$ n) doesn't →0 as it oscillates between 1/2 and 1 densely. Thus, diverges by nth-term test. Choice A tempts but fails on limit condition due to oscillation. Be cautious with oscillatory terms; ensure the limit exists and is zero.
A model uses $\sum_{n=2}^{\infty}(-1)^n\frac{1}{\ln n}$. Does the series converge?
Diverges because $\sum \frac{1}{\ln n}$ diverges, so the alternating series diverges.
Diverges because $\frac{1}{\ln n}$ is increasing.
Converges absolutely because $\sum \frac{1}{\ln n}$ converges.
Converges because $\lim_{n\to\infty}\frac{1}{\ln n}=1$.
Converges by the Alternating Series Test because $\frac{1}{\ln n}$ decreases and approaches $0$.
Explanation
The skill being tested is the Alternating Series Test for convergence. This test requires an alternating series ∑ $(-1)^n$ b_n with b_n positive, decreasing to 0 in the limit. Here, b_n = 1/ln n for n ≥ 2 is positive, decreases because ln n increases slower than any positive power, and lim (1/ln n) = 0. Therefore, the series converges by the test. Choice D is tempting but fails because divergence of the absolute series does not imply divergence of the alternating one; the test specifically allows conditional convergence. Remember to verify the decreasing condition for n beyond a certain point, as initial terms may not strictly decrease but the tail determines convergence.
A series used in estimation is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n^2+5n}$. Does it converge?
Converges because $\frac{1}{n^2+5n}$ is increasing.
Diverges because $\frac{1}{n^2+5n}$ is not positive.
Diverges because $\lim_{n\to\infty}\frac{1}{n^2+5n}=1$.
Converges by the Alternating Series Test because $\frac{1}{n^2+5n}$ decreases and approaches $0$.
Diverges because $\sum \frac{1}{n^2+5n}$ diverges.
Explanation
The skill being tested here is the Alternating Series Test for convergence. The test needs b_n positive, decreasing monotonically, and limiting to 0. Partial fraction decomposition can help verify. For ∑ $(-1)^{n+1}$ $rac{1}{n^2$ + 5n}, b_n = 1/(n(n+5)) decreases to 0, converges. A tempting distractor is choice D, saying divergence because absolute diverges, but AST allows that. A transferable strategy for alternating series is to decompose rational b_n to assess decrease.
An algorithm sums $\sum_{n=1}^{\infty}(-1)^n\frac{\sqrt{n}}{n+1}$. Does the series converge?
Diverges because $\frac{\sqrt{n}}{n+1}$ is increasing for all $n$.
Converges by the Alternating Series Test because $\frac{\sqrt{n}}{n+1}$ decreases and approaches $0$.
Converges absolutely because $\sum \frac{\sqrt{n}}{n+1}$ converges.
Diverges because $\frac{\sqrt{n}}{n+1}$ approaches $1$.
Converges because $\sum \frac{\sqrt{n}}{n+1}$ diverges.
Explanation
The skill being tested is the Alternating Series Test for convergence. The test requires eventual decrease of b_n to 0. For b_n = √n/(n+1) ≈ 1/√n, it limits to 0 and decreases for n ≥ 1 as per derivative. Hence, converges conditionally. Choice B is tempting but false since it approaches 0, not 1; simplify expressions asymptotically. For rational b_n, divide numerator and denominator by highest power to check limits and behavior.
A series for a measurement error is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n^3+1}$. Does it converge?
Diverges because $\frac{n}{n^3+1}$ is increasing.
Converges by the Alternating Series Test because $\frac{n}{n^3+1}$ decreases and approaches $0$.
Diverges because alternating series require $b_n$ to increase.
Converges because $\sum \frac{n}{n^3+1}$ diverges.
Diverges because $\frac{n}{n^3+1}$ does not approach $0$.
Explanation
The skill being tested is the Alternating Series Test for convergence. The conditions are b_n > 0, eventually decreasing, and lim b_n = 0. For b_n = $n/(n^3$+1) ≈ $1/n^2$, it limits to 0 and decreases for n ≥ 1 based on derivative analysis. Thus, the series converges. Choice A distracts by claiming it doesn't approach 0, but it does; always compute limits carefully. Use approximation for large n to quickly check both conditions in alternating series.
A series for corrections is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n(\ln n)^2}$ for $n\ge2$. Does it converge?
Diverges because $\frac{1}{n(\ln n)^2}$ increases.
Diverges because alternating series require $b_n$ to be constant.
Converges because $\sum \frac{1}{n(\ln n)^2}$ diverges.
Diverges because $\lim_{n\to\infty}\frac{1}{n(\ln n)^2}=1$.
Converges by the Alternating Series Test because $\frac{1}{n(\ln n)^2}$ decreases and approaches $0$.
Explanation
The skill being tested here is the Alternating Series Test for convergence. The test checks if b_n is positive, decreasing, and approaches 0, ensuring bounded partial sums. This is useful for series with slow-decaying terms. For ∑ $(-1)^{n+1}$ rac{1}{n (ln $n)^2$} (n ≥ 2), b_n decreases to 0, so it converges. A tempting distractor is choice D, claiming convergence because the absolute diverges, but that's irrelevant to AST. A transferable strategy for alternating series is to use integral tests on |b_n| for additional insights, but not as a requirement.
A computation uses $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\ln n}{n}$. Does the series converge?
Converges by the Alternating Series Test because $\frac{\ln n}{n}$ decreases for large $n$ and approaches $0$.
Diverges because $\frac{\ln n}{n}$ increases without bound.
Diverges because $\sum \frac{\ln n}{n}$ diverges, so the alternating series diverges.
Converges absolutely because $\sum \frac{\ln n}{n}$ converges.
Diverges because $\lim_{n\to\infty}\frac{\ln n}{n}\neq 0$.
Explanation
The skill being tested is the Alternating Series Test for convergence. This involves checking if b_n is positive, eventually decreasing, and lim b_n = 0. For b_n = (ln n)/n, it limits to 0 by L'Hôpital's rule and decreases for n ≥ 3 after an initial increase. Therefore, the series converges conditionally since the absolute version diverges by integral test. Choice E distracts by claiming lim (ln n)/n ≠ 0, but it does equal 0, emphasizing careful limit evaluation. Always use calculus tools like derivatives to verify the decreasing condition for non-obvious b_n.