Harmonic & p-Series - Calculus 2

1. Let $a_n = \frac{1}{n}$ and $b_n = \frac{1}{n^2}$. If we form the series $\sum_{n=1}^{\infty} c_n$ where $c_n = a_n$ when $n$ is not a perfect square and $c_n = b_n$ when $n$ is a perfect square, what is the convergence behavior of this series?
   Answer: Diverges because the non-perfect-square terms dominate and behave like harmonic series 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 $\sum_{n=1}^{\infty} c_n$ can be split into two parts: terms where $n$ is a perfect square (giving us $\frac{1}{n^2}$) and terms where $n$ is not a perfect square (giving us $\frac{1}{n}$). The non-perfect-square terms vastly outnumber the perfect square terms. For every perfect square $k^2$, there are approximately $2k$ nearby non-perfect-square terms contributing $\frac{1}{n}\$ each. Since the harmonic series$\sum \frac{1}{n}$diverges, and we're only removing a sparse subset of terms (the perfect squares), the remaining non-perfect-square terms still behave essentially like the harmonic series and cause divergence. Answer A incorrectly focuses only on the perfect square terms. While$\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.
2. Let $f(p) = \sum_{n=1}^{\infty} \frac{1}{n^p}$. For which of the following values of p is $f(p)$ a finite number?
   Answer: $p = \log_2(5)$ The sum $f(p)$ is finite if and only if the p-series converges, which occurs when $p > 1$. We must evaluate each option: (A) $\log_5(4) < 1$ because $5^1 = 5$. (B) $\ln(e) = 1$. (C) $\cos(0) = 1$. (D) $\log_2(5) > 1$ because $2^1 = 2$ and $2^2=4$, so $\log_2(5)$ must be greater than 2. Since only $\log_2(5)$ is greater than 1, the series converges only for this value of p.
3. Let $S_p = \sum_{n=1}^{\infty} \frac{1}{n^p}$. Which of the following inequalities is true?
   Answer: $S_3 < S_2$ The sum $S_p = \sum_{n=1}^{\infty} \frac{1}{n^p}$ is a function of p for $p > 1$. For any given $n > 1$, if $p_1 < p_2$, then $n^{p_1} < n^{p_2}$, which implies $\frac{1}{n^{p_1}} > \frac{1}{n^{p_2}}$. Since every term (after n=1) in the series with the smaller p is larger than the corresponding term in the series with the larger p, its sum will be larger. Therefore, if $p_1 < p_2$, then $S_{p_1} > S_{p_2}$. Comparing the options: (A) False, since $2 < 3$, $S_2 > S_3$. (B) False, since $1.1 > 1.01$, $S_{1.1} < S_{1.01}$. (C) False, since $4 > e$, $S_4 < S_e$. (D) True, since $3 > 2$, $S_3 < S_2$.
4. A series $\sum_{n=1}^{\infty} a_n$ is constructed such that its sequence of partial sums is given by $S_N = 2 - \frac{1}{\sqrt{N}}$. Which p-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ has the same convergence/divergence behavior as $\sum_{n=1}^{\infty} a_n$?
   Answer: A p-series with $p = 2$, which converges. A series converges if its sequence of partial sums converges to a finite limit. Here, $\lim_{N \to \infty} S_N = \lim_{N \to \infty} (2 - \frac{1}{\sqrt{N}}) = 2 - 0 = 2$. Since the partial sums approach a finite limit, the series $\sum a_n$ converges. We need to find a p-series that also converges. A p-series $\sum \frac{1}{n^p}$ converges if and only if $p > 1$. Of the choices, only $p=2$ satisfies this condition.
5. The Riemann zeta function is defined as $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$ for $s > 1$. If $\zeta(2) = \frac{\pi^2}{6}$ and $\zeta(4) = \frac{\pi^4}{90}$, which of the following series has a sum closest to $\frac{\pi^2}{6} - 1$?
   Answer: $\sum_{n=2}^{\infty} \frac{1}{n^2}$ because this excludes the first term of $\zeta(2)$ Since $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \sum_{n=2}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$, we have $\sum_{n=2}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} - 1$. This is exactly what we want. Choice B gives $\sum_{n=1}^{\infty} \frac{1}{(n+1)^2} = \sum_{m=2}^{\infty} \frac{1}{m^2}$ (substituting $m = n+1$), which is the same as choice A, so it's also correct, but A is more direct. Choice C equals $\left(\frac{\pi^2}{6} - 1\right) - \left(\frac{\pi^4}{90} - 1\right) = \frac{\pi^2}{6} - \frac{\pi^4}{90}$, which is not what we want. Choice D equals $\frac{\pi^2}{6} - \frac{\pi^4}{90}$, which is also incorrect.
6. Consider two series: Series I is $\sum_{n=1}^{\infty} \frac{1}{n^{p/2}}$ and Series II is $\sum_{n=1}^{\infty} \frac{1}{(3n)^p}$. For which of the following ranges of p do both series converge?
   Answer: $p > 2$ Series I, $\sum_{n=1}^{\infty} \frac{1}{n^{p/2}}$, is a p-series with exponent $p/2$. It converges if $p/2 > 1$, which means $p > 2$. Series II, $\sum_{n=1}^{\infty} \frac{1}{(3n)^p} = \sum_{n=1}^{\infty} \frac{1}{3^p n^p} = \frac{1}{3^p} \sum_{n=1}^{\infty} \frac{1}{n^p}$, is a constant multiple of a p-series with exponent p. It converges if $p > 1$. For both series to converge, both conditions must be met. The intersection of $p > 2$ and $p > 1$ is $p > 2$.
7. Let $S_N = \sum_{n=1}^{N} \frac{1}{n}$ be the Nth partial sum of the harmonic series. It is a known result that $\lim_{N \to \infty} (S_N - \ln N) = \gamma$, where $\gamma \approx 0.577$ is the Euler-Mascheroni constant. What does this relationship imply about the behavior of the harmonic series?
   Answer: The series diverges because its partial sums $S_N$ grow without bound, similar to the function $\ln N$. The given limit implies that for large N, $S_N \approx \ln N + \gamma$. Since $\lim_{N \to \infty} \ln N = \infty$, the partial sums $S_N$ 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.
8. A student claims that the series $\sum_{n=1}^{\infty} \frac{1}{n^{1.01}}$ converges much faster than $\sum_{n=1}^{\infty} \frac{1}{n^{1.5}}$ because the exponents are closer. Which statement best evaluates this claim?
   Answer: Incorrect: $\sum \frac{1}{n^{1.5}}$ converges faster because larger exponents give faster decay of terms When comparing convergence rates of p-series like $\sum \frac{1}{n^p}$, 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 $\sum \frac{1}{n^{1.5}}$ converges significantly faster than $\sum \frac{1}{n^{1.01}}$. The key insight is that larger exponents cause terms to shrink much more rapidly. For large n, $\frac{1}{n^{1.5}}$ becomes negligibly small much faster than $\frac{1}{n^{1.01}}$. 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.
9. If the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges to S, what is the value of $\sum_{n=3}^{\infty} \frac{1}{n^p}$?
   Answer: $S - 1 - \frac{1}{2^p}$ The sum $S = \sum_{n=1}^{\infty} \frac{1}{n^p}$ can be written as the sum of its first two terms plus the rest of the series. $S = \frac{1}{1^p} + \frac{1}{2^p} + \sum_{n=3}^{\infty} \frac{1}{n^p}$. The desired sum is $\sum_{n=3}^{\infty} \frac{1}{n^p}$. By rearranging the equation, we get $\sum_{n=3}^{\infty} \frac{1}{n^p} = S - \frac{1}{1^p} - \frac{1}{2^p} = S - 1 - \frac{1}{2^p}$.
10. It is a famous result that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$. Using this fact, what is the sum of the series $\sum_{n=1}^{\infty} \frac{1}{(2n)^2}$?
    Answer: $\frac{\pi^2}{24}$ We can rewrite the terms of the given series and factor out a constant: $\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \sum_{n=1}^{\infty} \frac{1}{4n^2} = \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$. Now, we can substitute the given sum: $\frac{1}{4} \left( \frac{\pi^2}{6} \right) = \frac{\pi^2}{24}$.
11. Let k be a positive constant. The series $\sum_{n=1}^{\infty} n^{-1/k}$ is known to diverge. Which of the following must be true about k?
    Answer: $k \ge 1$ The series can be written as $\sum_{n=1}^{\infty} \frac{1}{n^{1/k}}$. This is a p-series with $p = 1/k$. A p-series diverges if and only if $p \le 1$. Therefore, we must have $1/k \le 1$. 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 $1 \le k$, or $k \ge 1$.
12. For a real number c, consider the series $\sum_{n=1}^{\infty} \frac{1}{n^{4c-c^2-3}}$. For which value(s) of c does this series diverge?
    Answer: All real numbers c This is a p-series with $p = 4c - c^2 - 3$. A p-series converges if $p > 1$ and diverges if $p \le 1$. We want to find when $4c - c^2 - 3 \le 1$. Rearranging this inequality gives $0 \le c^2 - 4c + 4$. Factoring the right side gives $0 \le (c-2)^2$. The expression $(c-2)^2$ is a square, so it is always greater than or equal to 0 for any real number c. Thus, the inequality $p \le 1$ is always true, which means the series always diverges.
13. For what values of the constant k does the series $\sum_{n=1}^{\infty} \frac{\sqrt[3]{n^2}}{n^{k-1}}$ converge?
    Answer: $k > \frac{8}{3}$ The general term of the series can be simplified using exponent rules: $\frac{\sqrt[3]{n^2}}{n^{k-1}} = \frac{n^{2/3}}{n^{k-1}} = n^{2/3 - (k-1)} = n^{2/3 - k + 1} = n^{5/3 - k} = \frac{1}{n^{k - 5/3}}$. This is a p-series with $p = k - \frac{5}{3}$. For a p-series to converge, the exponent p must be greater than 1. Therefore, we must have $k - \frac{5}{3} > 1$, which simplifies to $k > 1 + \frac{5}{3}$, or $k > \frac{8}{3}$.
14. 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?
    Answer: $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ For choice (B), $\sum \frac{1}{n^2+1}$, the Ratio Test gives $\lim_{n \to \infty} \frac{n^2+1}{(n+1)^2+1} = \lim_{n \to \infty} \frac{n^2+1}{n^2+2n+2} = 1$, so the test is inconclusive. However, we can use the Limit Comparison Test with the convergent p-series $\sum \frac{1}{n^2}$. 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.
15. Let $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ be the $n$-th harmonic number. If $H_{2n} - H_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}$, which inequality correctly relates this sum to $\frac{1}{2}$?
    Answer: $H_{2n} - H_n > \frac{1}{2}$ because there are $n$ terms each greater than $\frac{1}{2n}$ We have $H_{2n} - H_n = \sum_{k=n+1}^{2n} \frac{1}{k}$. This sum contains exactly $n$ terms: $\frac{1}{n+1}, \frac{1}{n+2}, \ldots, \frac{1}{2n}$. Each term satisfies $\frac{1}{k} > \frac{1}{2n}$ since $k \leq 2n$. Therefore: $H_{2n} - H_n = \sum_{k=n+1}^{2n} \frac{1}{k} > \sum_{k=n+1}^{2n} \frac{1}{2n} = n \cdot \frac{1}{2n} = \frac{1}{2}$. 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 $\frac{1}{2}$. Choice D correctly identifies that there are $n$ terms each greater than $\frac{1}{2n}$, giving the lower bound $\frac{1}{2}$.
16. Consider the modified p-series $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$ where $p > 0$. For which values of $p$ does this series have the same convergence behavior as the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$?
    Answer: $p \leq 1$ because the integral test gives the same result as for the harmonic series By the integral test, $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$ converges if and only if $\int_2^{\infty} \frac{1}{x(\ln x)^p} dx$ converges. Using substitution $u = \ln x$, $du = \frac{1}{x}dx$, this becomes $\int_{\ln 2}^{\infty} \frac{1}{u^p} du$. This integral converges when $p > 1$ and diverges when $p \leq 1$. Since the harmonic series diverges, the modified series has the same convergence behavior (divergence) when $p \leq 1$. Choice B is wrong because $p = 1$ should be included. Choice C is wrong because $p < 1$ also gives divergence. Choice D is wrong because $p > 1$ gives convergence, opposite to the harmonic series.
17. The series $\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}$ can be analyzed by comparing it to known p-series. Which statement correctly describes its convergence?
    Answer: Diverges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$ We can rewrite $\frac{1}{n^{1+\frac{1}{n}}} = \frac{1}{n \cdot n^{\frac{1}{n}}}$. Since $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$, we can use the limit comparison test with $\sum \frac{1}{n}$: $\lim_{n \to \infty} \frac{\frac{1}{n^{1+\frac{1}{n}}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{n^{1+\frac{1}{n}}} = \lim_{n \to \infty} \frac{1}{n^{\frac{1}{n}}} = 1$. Since this limit is positive and finite, and $\sum \frac{1}{n}$ diverges, our series also diverges. Choice A is wrong because having exponent > 1 doesn't guarantee convergence when the exponent varies. Choice C is wrong because no such comparison exists. Choice D gives the right answer but incorrect reasoning about 'approaching 1'.
18. The convergence of which of the following series can be established directly by applying the Integral Test with an integrand $f(x)$ that is positive, continuous, and decreasing for $x \ge 1$?
    Answer: $\sum_{n=1}^{\infty} \frac{1}{(n+3)^{1.2}}$ The Integral Test requires the function corresponding to the series terms to be positive, continuous, and decreasing on $[a, \infty)$. Choice (A) is an alternating series, so its terms are not always positive. For (C), the function $f(x) = (2+\sin(x))/x^2$ is not monotonically decreasing due to the oscillating $\sin(x)$ term. For (D), the function $f(x) = x^{1/x - 1}$ is not straightforward to analyze and is not decreasing on $[1, \infty)$. For (B), the function $f(x) = \frac{1}{(x+3)^{1.2}}$ is positive, continuous, and decreasing for all $x \ge 1$, so the Integral Test can be applied directly.
19. Let $S = \sum_{n=1}^{\infty} \frac{1}{n^p}$ and $T = \sum_{n=2}^{\infty} \frac{1}{(n-1)^p}$. If $S$ converges to a finite value, what can be concluded about $T$?
    Answer: $T$ converges and has the same sum as $S$. If $S = \sum_{n=1}^{\infty} \frac{1}{n^p}$ converges, it means $p > 1$ and the sum is a finite value. Let's re-index the series $T$. Let $k = n-1$. When $n=2$, $k=1$. As $n \to \infty$, $k \to \infty$. So, $T = \sum_{k=1}^{\infty} \frac{1}{k^p}$. This is exactly the same series as $S$. Therefore, $T$ converges and has the same sum as $S$.
20. The series $\sum_{n=1}^{\infty} \frac{1}{n^{1 + 1/n}}$ has terms that are similar to the harmonic series. Which statement correctly describes its convergence?
    Answer: The series diverges by limit comparison with the harmonic series $\sum \frac{1}{n}$. Although the exponent $1 + 1/n$ is always greater than 1, it approaches 1 as $n \to \infty$. This prevents a direct conclusion from the p-series test. The best approach is the Limit Comparison Test with the harmonic series, $b_n = 1/n$. Let $a_n = \frac{1}{n^{1 + 1/n}}$. Consider the limit $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1/n^{1 + 1/n}}{1/n} = \lim_{n \to \infty} \frac{n}{n \cdot n^{1/n}} = \lim_{n \to \infty} \frac{1}{n^{1/n}}$. Since $\lim_{n \to \infty} n^{1/n} = 1$, the limit of the ratio is 1. Because this limit is a finite positive number and the harmonic series diverges, the given series also diverges.

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