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 because any alternating rational function converges.
Converges absolutely because $\sum \frac{n}{n^2+1}$ converges.
Diverges because $\frac{n}{n^2+1}$ is not decreasing.
Diverges because $\lim_{n\to\infty}\frac{n}{n^2+1}=1$.
Converges by the Alternating Series Test because $\frac{n}{n^2+1}$ decreases and approaches $0$.
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 $n$.
Diverges because $\sin(1/n)$ oscillates.
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 $\lim_{n\to\infty}\frac{\sin(1/n)}{n}\neq 0$.
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?
Converges because the denominator alternates.
Diverges because $\lim_{n\to\infty}\frac{1}{n+(-1)^n}\neq 0$.
Converges absolutely because $\sum \frac{1}{n+(-1)^n}$ converges.
Diverges because $\frac{1}{n+(-1)^n}$ is not monotone decreasing.
Converges by the Alternating Series Test because $\frac{1}{n+(-1)^n}$ decreases and approaches $0$.
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 $\frac{1}{1+\sin^2 n}$ is bounded.
Converges by the Alternating Series Test because $\frac{1}{1+\sin^2 n}$ decreases and approaches $0$.
Converges absolutely because $\sum \frac{1}{1+\sin^2 n}$ converges.
Diverges because $\lim_{n\to\infty}\frac{1}{1+\sin^2 n}\neq 0$.
Converges because $\sin^2 n$ is periodic.
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?
Converges by the Alternating Series Test because $\frac{1}{\ln n}$ decreases and approaches $0$.
Converges because $\lim_{n\to\infty}\frac{1}{\ln n}=1$.
Diverges because $\sum \frac{1}{\ln n}$ diverges, so the alternating series diverges.
Converges absolutely because $\sum \frac{1}{\ln n}$ converges.
Diverges because $\frac{1}{\ln n}$ is increasing.
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 for a measurement error is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n^3+1}$. Does it converge?
Converges because $\sum \frac{n}{n^3+1}$ diverges.
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.
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 $\lim_{n\to\infty}\frac{1}{n(\ln n)^2}=1$.
Converges by the Alternating Series Test because $\frac{1}{n(\ln n)^2}$ decreases and approaches $0$.
Diverges because alternating series require $b_n$ to be constant.
Diverges because $\frac{1}{n(\ln n)^2}$ increases.
Converges because $\sum \frac{1}{n(\ln n)^2}$ diverges.
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 series for alternating work is $\sum_{n=1}^{\infty}(-1)^n\frac{\arctan(n)}{n}$. Does it converge?
Diverges because $\frac{\arctan(n)}{n}$ is increasing.
Diverges because $\arctan(n)$ does not approach $0$.
Diverges because alternating series require $b_n$ to be constant.
Converges absolutely because $\sum \frac{\arctan(n)}{n}$ converges.
Converges by the Alternating Series Test because $\frac{\arctan(n)}{n}$ decreases for large $n$ and approaches $0$.
Explanation
The skill being tested is the Alternating Series Test for convergence. Test needs b_n eventually decreasing to 0. b_n = arctan(n)/n → 0 since arctan bounds to π/2, and decreases after initial terms. Thus, converges. Choice B is tempting but wrong as arctan(n) → π/2, but divided by n →0. Examine bounded numerators over growing denominators for limit zero in such series.
A Fourier-like term is $\sum_{n=1}^{\infty}(-1)^{n}\frac{2n+1}{n}$. Does it converge?
Diverges because $\frac{2n+1}{n}$ is decreasing.
Converges because the terms alternate in sign.
Converges by the Alternating Series Test because $\frac{2n+1}{n}$ decreases to $0$.
Converges absolutely because $\sum \frac{2n+1}{n}$ converges.
Diverges because $\lim_{n\to\infty}\frac{2n+1}{n}\neq 0$.
Explanation
The skill being tested is the Alternating Series Test for convergence. The test requires $b_n > 0$, decreasing, and approaching 0. Here, $b_n = \frac{2n+1}{n} = 2 + \frac{1}{n}$ approaches $2 \neq 0$, so terms do not go to zero, and the series diverges. The alternating nature cannot overcome non-vanishing terms. Choice A is a distractor that falsely claims it decreases to 0, but it decreases to 2, violating the limit condition. Prioritize checking the limit of $b_n$ before assessing monotonicity in alternating series analysis.
A series is $\sum_{n=1}^{\infty}(-1)^n\frac{n+2}{n^2}$. Does it converge?
Diverges because $\frac{n+2}{n^2}$ is increasing.
Converges by the Alternating Series Test because $\frac{n+2}{n^2}$ decreases and approaches $0$.
Converges absolutely because $\sum \frac{n+2}{n^2}$ diverges.
Diverges because alternating series require $b_n$ to be increasing.
Diverges because $\lim_{n\to\infty}\frac{n+2}{n^2}=1$.
Explanation
The skill being tested here is the Alternating Series Test for convergence. To apply it, confirm b_n > 0, b_n is decreasing (at least eventually), and lim b_n = 0. These ensure the remainders decrease appropriately. For ∑ $(-1)^n$ $rac{n+2}{n^2$}, b_n approaches 0 and decreases for n ≥ 1, satisfying the test. A tempting distractor is choice A, saying it diverges because b_n increases, but actually it decreases like 1/n. A transferable strategy for alternating series is to simplify b_n asymptotically to check the limit and monotonicity.