Calculus 2 Quiz: Harmonic And P Series
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Harmonic And P SeriesQuestion 1 of 20

For what values of the constant k does the series ∑n=1∞n23nk−1\sum_{n=1}^{\infty} \frac{\sqrt[3]{n^2}}{n^{k-1}} converge?

k>1k > 1
k>53k > \frac{5}{3}
k>83k > \frac{8}{3}
k<83k < \frac{8}{3}
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Calculus 2 Quiz

Calculus 2 Quiz: Harmonic And P Series

Practice Harmonic And P Series in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Harmonic And P Series, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For what values of the constant k does the series ∑n=1∞n23nk−1\sum_{n=1}^{\infty} \frac{\sqrt[3]{n^2}}{n^{k-1}} converge?

  1. k>1k > 1
  2. k>53k > \frac{5}{3}
  3. k>83k > \frac{8}{3} (correct answer)
  4. k<83k < \frac{8}{3}
Explanation: The general term of the series can be simplified using exponent rules: n23nk−1=n2/3nk−1=n2/3−(k−1)=n2/3−k+1=n5/3−k=1nk−5/3\frac{\sqrt[3]{n^2}}{n^{k-1}} = \frac{n^{2/3}}{n^{k-1}} = n^{2/3 - (k-1)} = n^{2/3 - k + 1} = n^{5/3 - k} = \frac{1}{n^{k - 5/3}}. This is a p-series with p=k−53p = k - \frac{5}{3}. For a p-series to converge, the exponent p must be greater than 1. Therefore, we must have k−53>1k - \frac{5}{3} > 1, which simplifies to k>1+53k > 1 + \frac{5}{3}, or k>83k > \frac{8}{3}.

Question 2

For which of the following series does the Ratio Test fail (yield a limit of 1), while its convergence can be determined by comparison to a p-series?

  1. ∑n=1∞n!10n\sum_{n=1}^{\infty} \frac{n!}{10^n}
  2. ∑n=1∞1n2+1\sum_{n=1}^{\infty} \frac{1}{n^2+1} (correct answer)
  3. ∑n=1∞(nn+1)n2\sum_{n=1}^{\infty} \left(\frac{n}{n+1}\right)^{n^2}
  4. ∑n=1∞13n\sum_{n=1}^{\infty} \frac{1}{3^n}
Explanation: For choice (B), ∑1n2+1\sum \frac{1}{n^2+1}, the Ratio Test gives lim⁡n→∞n2+1(n+1)2+1=lim⁡n→∞n2+1n2+2n+2=1\lim_{n \to \infty} \frac{n^2+1}{(n+1)^2+1} = \lim_{n \to \infty} \frac{n^2+1}{n^2+2n+2} = 1, so the test is inconclusive. However, we can use the Limit Comparison Test with the convergent p-series ∑1n2\sum \frac{1}{n^2}. The limit of the ratio of terms is 1, so the series converges. For (A) and (D), the Ratio Test is conclusive. For (C), the Root Test is conclusive.

Question 3

Let an>0a_n > 0 for all n. If the series ∑n=1∞an\sum_{n=1}^{\infty} a_n is known to converge, and lim⁡n→∞an1/np=L\lim_{n \to \infty} \frac{a_n}{1/n^p} = L where LL is a finite positive constant, what must be true about p?

  1. p>1p > 1 (correct answer)
  2. p≤1p \le 1
  3. p=1p = 1
  4. p<1p < 1
Explanation: The Limit Comparison Test states that if the limit of the ratio of the terms of two positive series is a finite and positive constant LL, then both series either converge or both diverge. We are given that ∑an\sum a_n converges. Therefore, the series it is being compared to, ∑1np\sum \frac{1}{n^p}, must also converge. A p-series converges if and only if its exponent is strictly greater than 1. Thus, we must have p>1p > 1.

Question 4

Consider the modified p-series ∑n=2∞1n(ln⁡n)p\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p} where p>0p > 0. For which values of pp does this series have the same convergence behavior as the harmonic series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}?

  1. p≤1p \leq 1 because the integral test gives the same result as for the harmonic series (correct answer)
  2. p<1p < 1 because the logarithmic factor becomes negligible for large nn
  3. p=1p = 1 only, since this creates a borderline case similar to the harmonic series
  4. p>1p > 1 because the additional logarithmic factor requires compensation
Explanation: By the integral test, ∑n=2∞1n(ln⁡n)p\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p} converges if and only if ∫2∞1x(ln⁡x)pdx\int_2^{\infty} \frac{1}{x(\ln x)^p} dx converges. Using substitution u=ln⁡xu = \ln x, du=1xdxdu = \frac{1}{x}dx, this becomes ∫ln⁡2∞1updu\int_{\ln 2}^{\infty} \frac{1}{u^p} du. This integral converges when p>1p > 1 and diverges when p≤1p \leq 1. Since the harmonic series diverges, the modified series has the same convergence behavior (divergence) when p≤1p \leq 1. Choice B is wrong because p=1p = 1 should be included. Choice C is wrong because p<1p < 1 also gives divergence. Choice D is wrong because p>1p > 1 gives convergence, opposite to the harmonic series.

Question 5

For which value of k will the Limit Comparison Test between the series ∑n=2∞n3+5nnk−10\sum_{n=2}^{\infty} \frac{n^3 + 5n}{n^k - 10} and the p-series ∑n=2∞1n2\sum_{n=2}^{\infty} \frac{1}{n^2} yield a finite, non-zero limit?

  1. k=2k=2
  2. k=3k=3
  3. k=4k=4
  4. k=5k=5 (correct answer)
Explanation: To apply the Limit Comparison Test, we evaluate the limit of the ratio of the terms: L=lim⁡n→∞(n3+5n)/(nk−10)1/n2=lim⁡n→∞n2(n3+5n)nk−10=lim⁡n→∞n5+5n3nk−10L = \lim_{n \to \infty} \frac{(n^3 + 5n)/(n^k - 10)}{1/n^2} = \lim_{n \to \infty} \frac{n^2(n^3 + 5n)}{n^k - 10} = \lim_{n \to \infty} \frac{n^5 + 5n^3}{n^k - 10}. For this limit L to be a finite, non-zero constant (i.e., 0<L<∞0 < L < \infty), the degree of the polynomial in the numerator must equal the degree of the polynomial in the denominator. The degree of the numerator is 5. Therefore, we must have k=5k=5.

Question 6

The sum of the reciprocals of all perfect squares is ∑n=1∞1n2=π26\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.

Using the information in the passage, what is the sum of the reciprocals of the odd perfect squares, ∑n=1∞1(2n−1)2\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}?

  1. π28\frac{\pi^2}{8} (correct answer)
  2. π212\frac{\pi^2}{12}
  3. 5π224\frac{5\pi^2}{24}
  4. 3π216\frac{3\pi^2}{16}
Explanation: The sum of all squares can be split into the sum of even squares and the sum of odd squares. Sum(all) = Sum(even) + Sum(odd). We are given Sum(all) = π26\frac{\pi^2}{6}. The sum of even squares is ∑n=1∞1(2n)2=14∑n=1∞1n2=14(π26)=π224\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{4}(\frac{\pi^2}{6}) = \frac{\pi^2}{24}. Therefore, Sum(odd) = Sum(all) - Sum(even) = π26−π224=4π224−π224=3π224=π28\frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{4\pi^2}{24} - \frac{\pi^2}{24} = \frac{3\pi^2}{24} = \frac{\pi^2}{8}.

Question 7

Given that p>1p > 1, the error RNR_N in approximating the sum SS of a convergent p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} by its Nth partial sum SNS_N is bounded by ∫N+1∞1xpdx<RN<∫N∞1xpdx\int_{N+1}^{\infty} \frac{1}{x^p} dx < R_N < \int_N^{\infty} \frac{1}{x^p} dx. Which expression represents the upper bound for the error RNR_N?

  1. 1(p−1)Np−1\frac{1}{(p-1)N^{p-1}} (correct answer)
  2. 1pNp−1\frac{1}{pN^{p-1}}
  3. 1(p−1)(N+1)p−1\frac{1}{(p-1)(N+1)^{p-1}}
  4. N1−p1−p\frac{N^{1-p}}{1-p}
Explanation: The upper bound for the error RN=S−SNR_N = S - S_N is given by the integral ∫N∞1xpdx\int_N^{\infty} \frac{1}{x^p} dx. We evaluate this improper integral: ∫N∞x−pdx=lim⁡b→∞[x−p+1−p+1]Nb=lim⁡b→∞(b1−p1−p−N1−p1−p)\int_N^{\infty} x^{-p} dx = \lim_{b \to \infty} [\frac{x^{-p+1}}{-p+1}]_N^b = \lim_{b \to \infty} (\frac{b^{1-p}}{1-p} - \frac{N^{1-p}}{1-p}). Since p>1p > 1, 1−p<01-p < 0, so lim⁡b→∞b1−p=0\lim_{b \to \infty} b^{1-p} = 0. The integral evaluates to 0−N1−p1−p=−N1−p−(p−1)=N−(p−1)p−1=1(p−1)Np−10 - \frac{N^{1-p}}{1-p} = \frac{-N^{1-p}}{-(p-1)} = \frac{N^{-(p-1)}}{p-1} = \frac{1}{(p-1)N^{p-1}}.

Question 8

A series ∑n=1∞an\sum_{n=1}^{\infty} a_n is constructed such that its sequence of partial sums is given by SN=2−1NS_N = 2 - \frac{1}{\sqrt{N}}. Which p-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} has the same convergence/divergence behavior as ∑n=1∞an\sum_{n=1}^{\infty} a_n?

  1. A p-series with p=1/2p = 1/2, which diverges.
  2. A p-series with p=2p = 2, which converges. (correct answer)
  3. A p-series with p=1p = 1, which diverges.
  4. A p-series with p=−1/2p = -1/2, which diverges.
Explanation: A series converges if its sequence of partial sums converges to a finite limit. Here, lim⁡N→∞SN=lim⁡N→∞(2−1N)=2−0=2\lim_{N \to \infty} S_N = \lim_{N \to \infty} (2 - \frac{1}{\sqrt{N}}) = 2 - 0 = 2. Since the partial sums approach a finite limit, the series ∑an\sum a_n converges. We need to find a p-series that also converges. A p-series ∑1np\sum \frac{1}{n^p} converges if and only if p>1p > 1. Of the choices, only p=2p=2 satisfies this condition.

Question 9

Let SN=∑n=1N1nS_N = \sum_{n=1}^{N} \frac{1}{n} be the Nth partial sum of the harmonic series. It is a known result that lim⁡N→∞(SN−ln⁡N)=γ\lim_{N \to \infty} (S_N - \ln N) = \gamma, where γ≈0.577\gamma \approx 0.577 is the Euler-Mascheroni constant. What does this relationship imply about the behavior of the harmonic series?

  1. The series converges to γ\gamma because the difference between its partial sums and ln⁡N\ln N stabilizes.
  2. The series diverges because its partial sums SNS_N grow without bound, similar to the function ln⁡N\ln N. (correct answer)
  3. The series converges to a value close to γ\gamma, but the exact sum cannot be determined from this information.
  4. The result is inconclusive because it only provides a relationship to another divergent function, not the sum itself.
Explanation: The given limit implies that for large N, SN≈ln⁡N+γS_N \approx \ln N + \gamma. Since lim⁡N→∞ln⁡N=∞\lim_{N \to \infty} \ln N = \infty, the partial sums SNS_N must also approach infinity. A series converges only if its sequence of partial sums approaches a finite limit. Because the partial sums grow without bound, the harmonic series diverges.

Question 10

Consider the series ∑n=1∞(−1)nn0.7\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{0.7}}. How does this series relate to the convergence properties of p-series and harmonic-type series?

  1. Converges absolutely because 0.7>0.50.7 > 0.5 and the alternating series test applies
  2. Converges conditionally since ∑1n0.7\sum \frac{1}{n^{0.7}} diverges but alternating series test gives convergence (correct answer)
  3. Diverges because 0.7<10.7 < 1 and the alternating nature cannot overcome p-series divergence
  4. Converges absolutely by comparison with the convergent geometric series ∑12n\sum \frac{1}{2^n}
Explanation: Since 0.7<10.7 < 1, the p-series ∑n=1∞1n0.7\sum_{n=1}^{\infty} \frac{1}{n^{0.7}} diverges, so the series does not converge absolutely. However, for the alternating series ∑n=1∞(−1)nn0.7\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{0.7}}, we can apply the alternating series test: (1) 1n0.7\frac{1}{n^{0.7}} is decreasing for n≥1n \geq 1, and (2) lim⁡n→∞1n0.7=0\lim_{n \to \infty} \frac{1}{n^{0.7}} = 0. Both conditions are satisfied, so the series converges. Since it converges but not absolutely, it converges conditionally. Choice A is wrong because absolute convergence requires p>1p > 1. Choice C is wrong because alternating series can converge even when the corresponding positive series diverges. Choice D is wrong because there's no valid comparison with geometric series.

Question 11

Let an=1na_n = \frac{1}{n} and bn=1n2b_n = \frac{1}{n^2}. If we form the series ∑n=1∞cn\sum_{n=1}^{\infty} c_n where cn=anc_n = a_n when nn is not a perfect square and cn=bnc_n = b_n when nn is a perfect square, what is the convergence behavior of this series?

  1. Converges because the perfect squares contribute a convergent subseries ∑1k4\sum \frac{1}{k^4}
  2. Diverges because the series combines both convergent and divergent components
  3. Converges because there are only finitely many perfect squares in any finite interval
  4. Diverges because the non-perfect-square terms dominate and behave like harmonic series (correct answer)
Explanation: When analyzing series with mixed terms like this one, you need to understand how different components affect overall convergence behavior. The key insight is that convergence depends on the dominant behavior of the series as a whole. Let's examine what happens here. The series ∑n=1∞cn\sum_{n=1}^{\infty} c_n can be split into two parts: terms where nn is a perfect square (giving us 1n2\frac{1}{n^2}) and terms where nn is not a perfect square (giving us 1n\frac{1}{n}). The non-perfect-square terms vastly outnumber the perfect square terms. For every perfect square k2k^2, there are approximately 2k2k nearby non-perfect-square terms contributing $$\frac{1}{n}$ each. Since the harmonic series ∑1n\sum \frac{1}{n} diverges, and we're only removing a sparse subset of terms (the perfect squares), the remaining non-perfect-square terms still behave essentially like the harmonic series and cause divergence. Answer A incorrectly focuses only on the perfect square terms. While ∑1k4\sum \frac{1}{k^4} does converge, this doesn't determine the behavior of the entire series. Answer B suggests that mixing convergent and divergent components automatically leads to some ambiguous behavior, but that's not how series work—the dominant component determines convergence. Answer C makes the error of thinking that having "only finitely many" perfect squares in finite intervals somehow helps, but series convergence depends on infinite behavior. Remember: when analyzing modified series, identify which component dominates. Removing or changing a sparse subset of terms (like perfect squares) usually doesn't change the convergence behavior if the original series had a clear dominant pattern.

Question 12

The Riemann zeta function is defined as ζ(s)=∑n=1∞1ns\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} for s>1s > 1. If ζ(2)=π26\zeta(2) = \frac{\pi^2}{6} and ζ(4)=π490\zeta(4) = \frac{\pi^4}{90}, which of the following series has a sum closest to π26−1\frac{\pi^2}{6} - 1?

  1. ∑n=2∞1n2\sum_{n=2}^{\infty} \frac{1}{n^2} because this excludes the first term of ζ(2)\zeta(2) (correct answer)
  2. ∑n=1∞1(n+1)2\sum_{n=1}^{\infty} \frac{1}{(n+1)^2} because this shifts the index by one unit
  3. ∑n=2∞1n2−∑n=2∞1n4\sum_{n=2}^{\infty} \frac{1}{n^2} - \sum_{n=2}^{\infty} \frac{1}{n^4} combining both zeta function values
  4. ∑n=1∞1n2−∑n=1∞1n4\sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{n=1}^{\infty} \frac{1}{n^4} which equals ζ(2)−ζ(4)\zeta(2) - \zeta(4)
Explanation: Since ζ(2)=∑n=1∞1n2=1+∑n=2∞1n2=π26\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \sum_{n=2}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}, we have ∑n=2∞1n2=π26−1\sum_{n=2}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} - 1. This is exactly what we want. Choice B gives ∑n=1∞1(n+1)2=∑m=2∞1m2\sum_{n=1}^{\infty} \frac{1}{(n+1)^2} = \sum_{m=2}^{\infty} \frac{1}{m^2} (substituting m=n+1m = n+1), which is the same as choice A, so it's also correct, but A is more direct. Choice C equals (π26−1)−(π490−1)=π26−π490\left(\frac{\pi^2}{6} - 1\right) - \left(\frac{\pi^4}{90} - 1\right) = \frac{\pi^2}{6} - \frac{\pi^4}{90}, which is not what we want. Choice D equals π26−π490\frac{\pi^2}{6} - \frac{\pi^4}{90}, which is also incorrect.

Question 13

Let Sn=∑k=1n1kS_n = \sum_{k=1}^{n} \frac{1}{k} be the nn-th partial sum of the harmonic series. If S2m−S2m−1>12S_{2^m} - S_{2^{m-1}} > \frac{1}{2} for all positive integers m≥1m \geq 1, what can be concluded about the harmonic series?

  1. The series converges because the partial sums are bounded above by m2\frac{m}{2}
  2. The series diverges because the partial sums increase without bound (correct answer)
  3. The series converges conditionally since the terms decrease to zero
  4. The convergence cannot be determined from this information alone
Explanation: The given condition shows that S2m−S2m−1=∑k=2m−1+12m1k>12S_{2^m} - S_{2^{m-1}} = \sum_{k=2^{m-1}+1}^{2^m} \frac{1}{k} > \frac{1}{2}. This means that every time we double the index from 2m−12^{m-1} to 2m2^m, the partial sum increases by more than 12\frac{1}{2}. Since there are infinitely many such doublings (m=1,2,3,…m = 1, 2, 3, \ldots), the partial sums S2mS_{2^m} increase without bound: S2m≥S1+m⋅12=1+m2→∞S_{2^m} \geq S_1 + m \cdot \frac{1}{2} = 1 + \frac{m}{2} \to \infty as m→∞m \to \infty. Therefore, the harmonic series diverges. Choice A is wrong because the bound goes to infinity. Choice C is wrong because this is not an alternating series. Choice D is wrong because this information is sufficient to prove divergence.

Question 14

A student claims that the series ∑n=1∞1n1.01\sum_{n=1}^{\infty} \frac{1}{n^{1.01}} converges much faster than ∑n=1∞1n1.5\sum_{n=1}^{\infty} \frac{1}{n^{1.5}} because the exponents are closer. Which statement best evaluates this claim?

  1. Correct: series with exponents closer to 1 always converge faster than those with larger exponents
  2. Correct: the difference 1.5−1.01=0.491.5 - 1.01 = 0.49 shows the first series has a significant advantage
  3. Incorrect: ∑1n1.5\sum \frac{1}{n^{1.5}} converges faster because larger exponents give faster decay of terms (correct answer)
  4. Incorrect: convergence speed depends on the sum of tails, not individual term comparison
Explanation: When comparing convergence rates of p-series like ∑1np\sum \frac{1}{n^p}, you need to understand how the exponent affects both convergence and the speed at which partial sums approach their limit. Both series converge since their exponents exceed 1, but ∑1n1.5\sum \frac{1}{n^{1.5}} converges significantly faster than ∑1n1.01\sum \frac{1}{n^{1.01}}. The key insight is that larger exponents cause terms to shrink much more rapidly. For large n, 1n1.5\frac{1}{n^{1.5}} becomes negligibly small much faster than 1n1.01\frac{1}{n^{1.01}}. This means the "tail" of the first series (the sum of remaining terms after some point) approaches zero more quickly. Choice A incorrectly claims that exponents closer to 1 give faster convergence - this reverses the actual relationship. Choice B makes a mathematical error by suggesting that the difference in exponents (0.49) somehow favors the series with the smaller exponent, when the opposite is true. Choice D contains a grain of truth about tail behavior being important, but incorrectly dismisses individual term comparison, which directly determines convergence speed for p-series. The correct answer is C because it properly identifies that larger exponents in p-series lead to faster convergence due to more rapid term decay. Study tip: For p-series convergence speed, remember that "higher power = faster convergence." The further the exponent is above 1, the more aggressively the terms shrink, making the series converge faster. This pattern appears frequently on calculus exams.

Question 15

Let Hn=1+12+13+⋯+1nH_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} be the nn-th harmonic number. If H2n−Hn=1n+1+1n+2+⋯+12nH_{2n} - H_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}, which inequality correctly relates this sum to 12\frac{1}{2}?

  1. H2n−Hn>12H_{2n} - H_n > \frac{1}{2} because each term 1k>12n\frac{1}{k} > \frac{1}{2n} for n+1≤k≤2nn+1 \leq k \leq 2n
  2. H2n−Hn<12H_{2n} - H_n < \frac{1}{2} because each term 1k<1n\frac{1}{k} < \frac{1}{n} for n+1≤k≤2nn+1 \leq k \leq 2n
  3. H2n−Hn=12H_{2n} - H_n = \frac{1}{2} exactly when nn is a power of 2
  4. H2n−Hn>12H_{2n} - H_n > \frac{1}{2} because there are nn terms each greater than 12n\frac{1}{2n} (correct answer)
Explanation: We have H2n−Hn=∑k=n+12n1kH_{2n} - H_n = \sum_{k=n+1}^{2n} \frac{1}{k}. This sum contains exactly nn terms: 1n+1,1n+2,…,12n\frac{1}{n+1}, \frac{1}{n+2}, \ldots, \frac{1}{2n}. Each term satisfies 1k>12n\frac{1}{k} > \frac{1}{2n} since k≤2nk \leq 2n. Therefore: H2n−Hn=∑k=n+12n1k>∑k=n+12n12n=n⋅12n=12H_{2n} - H_n = \sum_{k=n+1}^{2n} \frac{1}{k} > \sum_{k=n+1}^{2n} \frac{1}{2n} = n \cdot \frac{1}{2n} = \frac{1}{2}. Choice A has the right conclusion but wrong reasoning about individual terms. Choice B is incorrect in both direction and reasoning. Choice C is false since the sum is never exactly 12\frac{1}{2}. Choice D correctly identifies that there are nn terms each greater than 12n\frac{1}{2n}, giving the lower bound 12\frac{1}{2}.

Question 16

Consider the three series: I. ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}, II. ∑n=1∞1n1.0001\sum_{n=1}^{\infty} \frac{1}{n^{1.0001}}, and III. ∑n=1∞1n0.9999\sum_{n=1}^{\infty} \frac{1}{n^{0.9999}}. Which of these series converge?

  1. II only (correct answer)
  2. I and II only
  3. II and III only
  4. None of the series converge.
Explanation: These are all p-series of the form ∑1np\sum \frac{1}{n^p}. The p-series test states that the series converges if p>1p > 1 and diverges if p≤1p \le 1. For Series I, p=1p=1, so it diverges (this is the harmonic series). For Series II, p=1.0001>1p = 1.0001 > 1, so it converges. For Series III, p=0.9999<1p = 0.9999 < 1, so it diverges. Thus, only Series II converges.

Question 17

For a real number c, consider the series ∑n=1∞1n4c−c2−3\sum_{n=1}^{\infty} \frac{1}{n^{4c-c^2-3}}. For which value(s) of c does this series diverge?

  1. c=2c=2 only
  2. c=1c=1 and c=3c=3
  3. 1≤c≤31 \le c \le 3
  4. All real numbers c (correct answer)
Explanation: This is a p-series with p=4c−c2−3p = 4c - c^2 - 3. A p-series converges if p>1p > 1 and diverges if p≤1p \le 1. We want to find when 4c−c2−3≤14c - c^2 - 3 \le 1. Rearranging this inequality gives 0≤c2−4c+40 \le c^2 - 4c + 4. Factoring the right side gives 0≤(c−2)20 \le (c-2)^2. The expression (c−2)2(c-2)^2 is a square, so it is always greater than or equal to 0 for any real number c. Thus, the inequality p≤1p \le 1 is always true, which means the series always diverges.

Question 18

Let f(p)=∑n=1∞1npf(p) = \sum_{n=1}^{\infty} \frac{1}{n^p}. For which of the following values of p is f(p)f(p) a finite number?

  1. p=log⁡5(4)p = \log_5(4)
  2. p=ln⁡(e)p = \ln(e)
  3. p=cos⁡(0)p = \cos(0)
  4. p=log⁡2(5)p = \log_2(5) (correct answer)
Explanation: The sum f(p)f(p) is finite if and only if the p-series converges, which occurs when p>1p > 1. We must evaluate each option: (A) log⁡5(4)<1\log_5(4) < 1 because 51=55^1 = 5. (B) ln⁡(e)=1\ln(e) = 1. (C) cos⁡(0)=1\cos(0) = 1. (D) log⁡2(5)>1\log_2(5) > 1 because 21=22^1 = 2 and 22=42^2=4, so log⁡2(5)\log_2(5) must be greater than 2. Since only log⁡2(5)\log_2(5) is greater than 1, the series converges only for this value of p.

Question 19

The convergence of which of the following series can be established directly by applying the Integral Test with an integrand f(x)f(x) that is positive, continuous, and decreasing for x≥1x \ge 1?

  1. ∑n=1∞(−1)nn3/2\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{3/2}}
  2. ∑n=1∞1(n+3)1.2\sum_{n=1}^{\infty} \frac{1}{(n+3)^{1.2}} (correct answer)
  3. ∑n=1∞2+sin⁡(n)n2\sum_{n=1}^{\infty} \frac{2 + \sin(n)}{n^2}
  4. ∑n=1∞1n1−1/n\sum_{n=1}^{\infty} \frac{1}{n^{1-1/n}}
Explanation: The Integral Test requires the function corresponding to the series terms to be positive, continuous, and decreasing on [a,∞)[a, \infty). Choice (A) is an alternating series, so its terms are not always positive. For (C), the function f(x)=(2+sin⁡(x))/x2f(x) = (2+\sin(x))/x^2 is not monotonically decreasing due to the oscillating sin⁡(x)\sin(x) term. For (D), the function f(x)=x1/x−1f(x) = x^{1/x - 1} is not straightforward to analyze and is not decreasing on [1,∞)[1, \infty). For (B), the function f(x)=1(x+3)1.2f(x) = \frac{1}{(x+3)^{1.2}} is positive, continuous, and decreasing for all x≥1x \ge 1, so the Integral Test can be applied directly.

Question 20

Let k be a positive constant. The series ∑n=1∞n−1/k\sum_{n=1}^{\infty} n^{-1/k} is known to diverge. Which of the following must be true about k?

  1. 0<k<10 < k < 1
  2. k>1k > 1
  3. k≤1k \le 1
  4. k≥1k \ge 1 (correct answer)
Explanation: The series can be written as ∑n=1∞1n1/k\sum_{n=1}^{\infty} \frac{1}{n^{1/k}}. This is a p-series with p=1/kp = 1/k. A p-series diverges if and only if p≤1p \le 1. Therefore, we must have 1/k≤11/k \le 1. Since we are given that k is a positive constant, we can multiply both sides by k without changing the direction of the inequality, which gives 1≤k1 \le k, or k≥1k \ge 1.