Defining Convergent and Divergent Infinite Series - AP Calculus BC

1. A sensor correction uses the infinite series $\sum_{n=1}^{\infty} \frac{3}{n^2+1}$. Does the series converge or diverge?
   Answer: Converges by comparison with a $p$-series This problem tests your ability to classify infinite series as convergent or divergent. The series $\sum_{n=1}^{\infty} \frac{3}{n^2+1}$ has terms that behave like $\frac{3}{n^2}$ for large $n$, since the '+1' becomes negligible compared to $n^2$. We can use the comparison test with the convergent p-series $\sum \frac{3}{n^2}$ (where $p=2>1$). Since $\frac{3}{n^2+1} < \frac{3}{n^2}$ for all $n \geq 1$, and the comparison series converges, our original series also converges. The integral test (option E) would also work but requires more computation than the direct comparison. When terms resemble $\frac{1}{n^p}$, always check if you can compare with a p-series first.
2. A sequence of updates is $\sum_{n=1}^{\infty} \frac{n!}{n^n}$. Does the series converge or diverge?
   Answer: Converges by the ratio test This problem involves the series $\sum_{n=1}^{\infty} \frac{n!}{n^n}$, which requires the ratio test. Computing $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} = \lim_{n\to\infty} \frac{(n+1) \cdot n^n}{(n+1)^{n+1}} = \lim_{n\to\infty} \frac{n^n}{(n+1)^n}$. This simplifies to $\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n = \lim_{n\to\infty} \left(1 - \frac{1}{n+1}\right)^n = \frac{1}{e} < 1$. Since the ratio test limit is less than 1, the series converges. The nth-term test (option A) would only check if terms go to zero, not prove convergence. For series involving factorials and exponentials, the ratio test is typically the most effective approach.
3. A model predicts total growth $\sum_{n=2}^{\infty} \frac{1}{n\ln n}$. Does this infinite series converge or diverge?
   Answer: Diverges by the integral test This question tests the skill of classifying infinite series as convergent or divergent. To assess ∑ 1/(n ln n) from n=2 to ∞, apply the integral test by considering ∫_2^∞ dx/(x ln x), which evaluates to ln(ln x) from 2 to ∞, diverging to infinity. Since the integral diverges and the function is positive, continuous, and decreasing for n≥2, the series diverges. This test is appropriate for series with logarithmic terms that decrease slowly. A tempting distractor is choice D, which claims convergence because the terms approach 0, but this fails as the nth-term test is necessary but not sufficient for convergence. A transferable strategy for classifying series is to use the integral test for monotonically decreasing positive terms, comparing the improper integral's behavior to determine convergence.
4. A power-series tail appears in a model: $\sum_{n=1}^{\infty}\frac{(-2)^n}{n^3}$; does it converge or diverge?
   Answer: Diverges by the Ratio Test This problem tests classification of series with alternating signs and exponential growth. For the series $\sum_{n=1}^{\infty}\frac{(-2)^n}{n^3}$, the Ratio Test is most appropriate given the exponential term. Applying the Ratio Test: $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{(-2)^{n+1}/(n+1)^3}{(-2)^n/n^3}\right| = \lim_{n\to\infty}\frac{2n^3}{(n+1)^3} = 2\cdot\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^3 = 2 > 1$. Since the limit exceeds 1, the series diverges by the Ratio Test. Choice E correctly identifies that $|-2| > 1$ matters, but doesn't specify the proper test; the Ratio Test provides the rigorous justification. For series with exponential terms $r^n$ where $|r| > 1$, the Ratio Test typically confirms divergence most directly.
5. A process alternates adjustments totaling $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$. Does the series converge or diverge?
   Answer: Converges by the alternating series test This problem involves determining convergence of an alternating series. The series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ is the alternating harmonic series, where terms alternate in sign and have magnitude $1/n$. To apply the alternating series test, we check: (1) terms decrease in magnitude: $\frac{1}{n+1} < \frac{1}{n}$, and (2) $\lim_{n \to \infty} \frac{1}{n} = 0$. Both conditions are satisfied, so the series converges by the alternating series test. Choice A incorrectly claims divergence—the alternating series test never proves divergence, only convergence. For alternating series $\sum (-1)^n a_n$, check if $a_n$ decreases to 0 for convergence.
6. A signal alternates with terms $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{\sqrt{n}}$. Does the series converge or diverge?
   Answer: Converges conditionally by the alternating series test This problem tests convergence of the alternating series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{\sqrt{n}}$. For alternating series, we check two conditions: (1) the terms $a_n = \frac{1}{\sqrt{n}}$ decrease to zero, and (2) $\lim_{n\to\infty} a_n = 0$. Both conditions are satisfied, so the series converges by the alternating series test. However, to check absolute convergence, we examine $\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}$, which is a divergent p-series with $p = \frac{1}{2} < 1$. Since the series converges but not absolutely, it converges conditionally. Option B incorrectly claims absolute convergence. When an alternating series converges but its absolute value series diverges, always classify it as conditionally convergent.
7. A stacked error term is $\sum_{n=1}^{\infty} \frac{4n+1}{n^2}$. Does this infinite series converge or diverge?
   Answer: Diverges by comparison to $\sum \frac{1}{n}$ This question tests the skill of classifying infinite series as convergent or divergent. Rewrite ∑ (4n+1)/n² as ∑ (4/n + 1/n²), where the dominant term 4/n suggests comparison to the divergent harmonic series 4 ∑ 1/n. Since (4n+1)/n² > 4/n for all n≥1 and ∑ 4/n diverges, the series diverges by the comparison test. This approach works for rational functions by analyzing asymptotic behavior. A tempting distractor is choice A, which claims convergence by comparison to ∑ 1/n², but this fails because the terms are larger than those of a divergent series, not smaller than a convergent one. A transferable strategy for classifying series is to simplify terms asymptotically and compare with p-series, ensuring the inequality direction matches the known series' behavior.
8. An error estimate uses $\sum_{n=1}^{\infty} \frac{n}{n^2+4}$. Determine whether the series converges or diverges.
   Answer: Diverges by comparison with the harmonic series This problem requires analyzing the series $\sum_{n=1}^{\infty} \frac{n}{n^2+4}$. For large $n$, the terms behave like $\frac{n}{n^2} = \frac{1}{n}$, suggesting comparison with the harmonic series. We can verify that $\lim_{n\to\infty} \frac{n/(n^2+4)}{1/n} = \lim_{n\to\infty} \frac{n^2}{n^2+4} = 1$. Since this limit is positive and finite, and the harmonic series $\sum \frac{1}{n}$ diverges, our series also diverges by the limit comparison test. The comparison with a p-series (option A) would be incorrect because the series behaves like $p=1$, not $p>1$. When the highest powers of $n$ in numerator and denominator differ by exactly 1, the series typically behaves like the harmonic series.
9. A bouncing ball travels distances forming $\sum_{n=0}^{\infty}5\left(\frac{2}{3}\right)^n$ meters; does the total distance converge?
   Answer: Converges as a geometric series This problem asks you to determine if an infinite series converges or diverges. The series $\sum_{n=0}^{\infty}5\left(\frac{2}{3}\right)^n$ is a geometric series with first term $a = 5$ and common ratio $r = \frac{2}{3}$. Since $|r| = \frac{2}{3} < 1$, the geometric series converges. The sum equals $\frac{a}{1-r} = \frac{5}{1-\frac{2}{3}} = \frac{5}{\frac{1}{3}} = 15$ meters. Choice A might seem plausible, but the Divergence Test only applies when terms don't approach zero, which isn't the case here since $\lim_{n\to\infty}5\left(\frac{2}{3}\right)^n = 0$. When you recognize a series as geometric with $|r| < 1$, you can immediately conclude it converges without needing more sophisticated tests.
10. A physics model uses $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$. Does this series converge or diverge?
    Answer: Diverges by the integral test This question tests your ability to classify infinite series as convergent or divergent. The series $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ can be analyzed using the integral test since $f(x) = \frac{1}{x\ln x}$ is positive, continuous, and decreasing for $x \geq 2$. The integral $\int_2^{\infty} \frac{1}{x\ln x}dx$ can be evaluated by substitution: let $u = \ln x$, then $du = \frac{1}{x}dx$, giving $\int \frac{1}{u}du = \ln|u| = \ln(\ln x)$. Since $\lim_{x\to\infty} \ln(\ln x) = \infty$, the integral diverges, so the series diverges. Choice E incorrectly assumes that terms approaching zero guarantees convergence. Remember: the integral test is powerful for series involving logarithms—if $\int f(x)dx$ diverges, then $\sum f(n)$ diverges.
11. A decaying input produces terms $a_n=\frac{n}{n+1}$. Does $\sum_{n=1}^{\infty} a_n$ converge or diverge?
    Answer: Diverges because $\lim_{n\to\infty} a_n\neq 0$ This question tests your ability to classify infinite series as convergent or divergent. For the series $\sum_{n=1}^{\infty} \frac{n}{n+1}$, we first check if the terms approach zero: $\lim_{n\to\infty} \frac{n}{n+1} = \lim_{n\to\infty} \frac{1}{1+\frac{1}{n}} = 1 \neq 0$. By the divergence test (nth term test), if $\lim_{n\to\infty} a_n \neq 0$, then $\sum a_n$ must diverge. The series diverges because we're essentially adding terms that approach 1, giving an infinite sum. Choice D incorrectly suggests that having $\lim a_n = 1$ would make the series converge. Remember: if terms don't approach zero, the series must diverge—this is the first test to try.
12. To estimate error, analyze $\sum_{n=1}^{\infty}\frac{n}{n^2+1}$; does this series converge or diverge?
    Answer: Diverges by comparison to $\sum\frac{1}{n}$ This problem requires determining convergence or divergence of an infinite series. For the series $\sum_{n=1}^{\infty}\frac{n}{n^2+1}$, we can use the Limit Comparison Test with the harmonic series $\sum\frac{1}{n}$. Computing the limit: $\lim_{n\to\infty}\frac{n/(n^2+1)}{1/n} = \lim_{n\to\infty}\frac{n^2}{n^2+1} = \lim_{n\to\infty}\frac{1}{1+1/n^2} = 1$. Since this limit is positive and finite, and the harmonic series diverges, our series also diverges by the Limit Comparison Test. Choice C might seem reasonable since the denominator has degree 2, but the numerator's degree 1 prevents the terms from decreasing fast enough for convergence. When a rational function's numerator and denominator have degrees differing by exactly 1, compare to the appropriate p-series for a definitive answer.
13. A sensor's error is modeled by $\sum_{n=1}^{\infty} \frac{3}{n^2+1}$. Does the series converge or diverge?
    Answer: Converges by comparison to $\sum \frac{1}{n^2}$ This question tests the skill of classifying infinite series as convergent or divergent. To determine convergence, compare the given series ∑ 3/(n² + 1) to the known convergent p-series ∑ 1/n², since for n ≥ 1, 3/(n² + 1) < 3/n² and ∑ 3/n² converges as p=2>1. The comparison test confirms that if a smaller positive series converges, so does the original. This direct comparison works because the terms are positive and the inequality holds for all n. A tempting distractor is choice E, which claims divergence by comparison to the harmonic series ∑ 1/n, but this fails because 3/(n² + 1) is actually much smaller than 1/n for large n, not larger. A transferable strategy for classifying series is to bound the terms with known convergent or divergent series like p-series or geometrics, ensuring the comparison direction supports the conclusion.
14. A filter output includes $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+n}}$. Determine whether the series converges or diverges.
    Answer: Diverges by comparison with the harmonic series This problem requires analyzing $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+n}}$. For large $n$, we can factor: $\sqrt{n^2+n} = \sqrt{n^2(1+1/n)} = n\sqrt{1+1/n} \approx n$ as $n \to \infty$. Therefore, the terms behave like $\frac{1}{n}$, suggesting comparison with the harmonic series. Using limit comparison: $\lim_{n\to\infty} \frac{1/\sqrt{n^2+n}}{1/n} = \lim_{n\to\infty} \frac{n}{\sqrt{n^2+n}} = \lim_{n\to\infty} \frac{1}{\sqrt{1+1/n}} = 1$. Since this limit is positive and finite, and the harmonic series diverges, our series also diverges. The comparison with a p-series (option A) would require $p > 1$, but here we have $p = 1$. When square roots contain polynomials, factor out the highest power to determine the series' behavior.
15. A population model uses $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}+1}$ as a correction; does the series converge or diverge?
    Answer: Diverges by comparison to a $p$-series with $p=\tfrac12$ This question examines ∑1/(√n+1), testing comparison with p-series for convergence analysis. For large n, √n+1 behaves like √n, so terms behave like 1/√n = 1/$n^{1/2}$. Using limit comparison with the divergent p-series ∑1/$n^{1/2}$ (p = 1/2 < 1): lim[1/(√n+1)]/(1/√n) = lim[√n/(√n+1)] = lim[1/(1+1/√n)] = 1. Since this limit is positive and finite, and ∑1/$n^{1/2}$ diverges, our series also diverges. Students might think the +1 in the denominator helps convergence, but it doesn't change the growth rate enough. When square roots appear with added constants, use limit comparison with the corresponding p-series to determine convergence.
16. An algorithm's error terms are $\sum_{n=1}^{\infty}\frac{5}{2^n}$. Does this infinite series converge or diverge?
    Answer: Converges because it is a geometric series with $|r|<1$ This question tests your ability to classify infinite series as convergent or divergent. The series $\sum_{n=1}^{\infty}\frac{5}{2^n} = 5\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n$ is a geometric series with first term $a = \frac{5}{2}$ and common ratio $r = \frac{1}{2}$. Since $|r| = \frac{1}{2} < 1$, the geometric series converges to $5 \cdot \frac{\frac{1}{2}}{1-\frac{1}{2}} = 5$. Choice A incorrectly claims $|r| > 1$, perhaps confusing the base 2 in the denominator with the ratio. Remember: rewrite series in the form $\sum ar^n$ to identify geometric series and check if $|r| < 1$.
17. A feedback process accumulates $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$. Does the series converge or diverge?
    Answer: Converges absolutely by comparison to $\sum \frac{1}{n^2}$ This question tests the skill of classifying infinite series as convergent or divergent. The series ∑ $(-1)^n$ / n² has absolute series ∑ 1/n², which converges as a p-series with p=2>1. Since the absolute series converges, the original converges absolutely by the absolute convergence theorem. This implies conditional convergence as well, but absolute is stronger. A tempting distractor is choice D, which claims conditional convergence by the alternating series test, but this fails because absolute convergence is proven, making it absolutely convergent. A transferable strategy for classifying series is to check absolute convergence first for alternating series using tests like comparison or p-series, as it provides more information than conditional alone.
18. In a fractal perimeter model, consider $\sum_{n=1}^{\infty}\frac{\ln n}{n^{3/2}}$; does this series converge or diverge?
    Answer: Converges by comparison to $\sum\frac{1}{n^{4/3}}$ This problem tests convergence classification for series combining logarithmic and power functions. For the series $\sum_{n=1}^{\infty}\frac{\ln n}{n^{3/2}}$, we can use the Limit Comparison Test with $\sum\frac{1}{n^{4/3}}$. We need to verify that $\lim_{n\to\infty}\frac{\ln n/n^{3/2}}{1/n^{4/3}} = \lim_{n\to\infty}\frac{\ln n \cdot n^{4/3}}{n^{3/2}} = \lim_{n\to\infty}\frac{\ln n}{n^{1/6}}$. By L'Hôpital's rule: $\lim_{n\to\infty}\frac{\ln n}{n^{1/6}} = \lim_{n\to\infty}\frac{1/n}{(1/6)n^{-5/6}} = \lim_{n\to\infty}\frac{6}{n^{1/6}} = 0$. Since this limit is 0 and $\sum\frac{1}{n^{4/3}}$ converges (p-series with $p = 4/3 > 1$), our series also converges. Choice B incorrectly suggests divergence; while $\ln n$ grows, it grows slowly enough that $n^{3/2}$ in the denominator ensures convergence. For series with $\ln n$ in the numerator, remember that logarithms grow more slowly than any positive power of $n$.
19. A dataset's error is modeled by $\sum_{n=1}^{\infty} \frac{2^n}{n!}$. Does the series converge or diverge?
    Answer: Converges by the ratio test This question involves determining convergence using the ratio test. For the series $\sum_{n=1}^{\infty} \frac{2^n}{n!}$, apply the ratio test: $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{2^{n+1}/(n+1)!}{2^n/n!} = \lim_{n \to \infty} \frac{2}{n+1} = 0 < 1$. Since the limit is less than 1, the series converges by the ratio test. The geometric series test (choice D) doesn't apply because this isn't a geometric series. When factorials appear in series, the ratio test is often the most effective approach.
20. A chemical concentration is approximated by $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$. Does the series converge or diverge?
    Answer: Converges by comparison to a $p$-series with $p=2$ This problem tests series convergence through algebraic manipulation. The series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ can be rewritten using partial fractions: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. This creates a telescoping series where most terms cancel, leaving a finite sum that converges. Alternatively, since $\frac{1}{n(n+1)} < \frac{1}{n^2}$ for all n ≥ 1, and $\sum \frac{1}{n^2}$ converges (p-series with p = 2), the comparison test shows convergence. Choice C incorrectly suggests divergence by comparison. When you see products in denominators, consider partial fractions or comparison to simpler series.

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