Convergent & Divergent Series - Calculus 2

1. It is known that the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges to $\frac{\pi^2}{6}$. What is the sum of the series $\sum_{n=3}^{\infty} \frac{1}{n^2}$?
   Answer: $\frac{\pi^2}{6} - \frac{5}{4}$ The sum of the series starting from $n=3$ is the total sum of the series minus the first two terms (for $n=1$ and $n=2$). $$\sum_{n=3}^{\infty} \frac{1}{n^2} = \left(\sum_{n=1}^{\infty} \frac{1}{n^2}\right) - \left(\frac{1}{1^2} + \frac{1}{2^2}\right) = \frac{\pi^2}{6} - \left(1 + \frac{1}{4}\right) = \frac{\pi^2}{6} - \frac{5}{4}.$$
2. Let $\sum_{n=1}^{\infty} a_n$ be an infinite series with its Nth partial sum given by $S_N = \frac{3N^2 - N}{2N^2 + 5}$. What is the sum of the series?
   Answer: The sum is $3/2$. By definition, the sum of a convergent infinite series is the limit of its sequence of partial sums. We need to compute $\lim_{N \to \infty} S_N$. $$\lim_{N \to \infty} S_N = \lim_{N \to \infty} \frac{3N^2 - N}{2N^2 + 5} = \lim_{N \to \infty} \frac{N^2(3 - 1/N)}{N^2(2 + 5/N^2)} = \frac{3 - 0}{2 + 0} = \frac{3}{2}.$$ Since the limit is a finite number, the series converges to this value.
3. Consider the series $\sum_{n=1}^{\infty} \left(1 - \frac{2}{n}\right)^n$. Which of the following statements is true?
   Answer: The series diverges because the limit of its terms is $e^{-2} \neq 0$. The n-th Term Test for Divergence states that if $\lim_{n \to \infty} a_n \neq 0$, the series $\sum a_n$ diverges. For this series, $a_n = \left(1 - \frac{2}{n}\right)^n$. The limit is a known form: $\lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k$. Here, $k = -2$, so $\lim_{n \to \infty} a_n = e^{-2}$. Since $e^{-2} \neq 0$, the series diverges by the n-th Term Test.
4. Suppose $\sum_{n=1}^{\infty} a_n$ is a convergent series and $\sum_{n=1}^{\infty} b_n$ is a divergent series. What can be concluded about the series $\sum_{n=1}^{\infty} (a_n - b_n)$?
   Answer: It must diverge. The difference of a convergent series and a divergent series must be divergent. This can be shown by contradiction. Assume $\sum (a_n - b_n)$ converges to a sum $L$. Since $\sum a_n$ converges to a sum $S$, we could write $\sum b_n = \sum (a_n - (a_n - b_n))$. By the properties of convergent series, this would be the difference of two convergent series, which must converge to $S - L$. This contradicts the given information that $\sum b_n$ diverges. Therefore, $\sum (a_n - b_n)$ must diverge.
5. Let $S_N = \sum_{n=1}^{N} a_n$. If the sequence of partial sums $\{S_N\}$ is given by $S_N = 5 - (N+1)e^{-N}$, what is $a_2$?
   Answer: $2e^{-1} - 3e^{-2}$ The n-th term of the series can be found using the relation $a_n = S_n - S_{n-1}$ for $n \ge 2$. We need to find $a_2$, which is $S_2 - S_1$. First, calculate $S_2$: $S_2 = 5 - (2+1)e^{-2} = 5 - 3e^{-2}$. Next, calculate $S_1$: $S_1 = 5 - (1+1)e^{-1} = 5 - 2e^{-1}$. Then, $a_2 = S_2 - S_1 = (5 - 3e^{-2}) - (5 - 2e^{-1}) = 2e^{-1} - 3e^{-2}$.
6. Find the sum of the infinite series $\sum_{n=1}^{\infty} \frac{2}{n(n+2)}$.
   Answer: 3/2 This is a telescoping series. First, use partial fraction decomposition: $\frac{2}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}$. Solving gives $A=1$ and $B=-1$, so the term is $\frac{1}{n} - \frac{1}{n+2}$. The Nth partial sum $S_N$ is: $$S_N = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{N} - \frac{1}{N+2}\right)$$ Most terms cancel, leaving $S_N = 1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}$. The sum of the series is $\lim_{N \to \infty} S_N = 1 + \frac{1}{2} - 0 - 0 = \frac{3}{2}$.
7. If the series $\sum_{n=1}^{\infty} (a_n - a_{n+1})$ converges, what can be concluded about the sequence $\{a_n\}$?
   Answer: The sequence $\{a_n\}$ must converge to a finite limit. The given series is a telescoping series. Its Nth partial sum is $S_N = (a_1 - a_2) + (a_2 - a_3) + \dots + (a_N - a_{N+1}) = a_1 - a_{N+1}$. For the series to converge, the limit of its partial sums must exist and be finite: $\lim_{N \to \infty} S_N = \lim_{N \to \infty} (a_1 - a_{N+1})$. This limit exists if and only if $\lim_{N \to \infty} a_{N+1}$ exists. Let $\lim_{N \to \infty} a_n = L$. Then the sum of the series is $a_1 - L$. Therefore, the sequence $\{a_n\}$ must converge to some finite limit $L$. It does not necessarily have to be 0.
8. Let $a_n$ be a sequence such that $a_n > 0$ for all $n$. If the series $\sum_{n=1}^{\infty} a_n$ diverges, what can be concluded about the series $\sum_{n=1}^{\infty} (1+a_n)$?
   Answer: It must diverge. We can use the n-th Term Test for Divergence on the series $\sum (1+a_n)$. Let the terms of this new series be $b_n = 1+a_n$. Since $a_n > 0$, we know that $b_n = 1+a_n > 1$. Therefore, the limit of the terms is $\lim_{n \to \infty} b_n = \lim_{n \to \infty} (1+a_n) \ge 1$. Since the limit of the terms is not 0, the series $\sum (1+a_n)$ must diverge. The fact that $\sum a_n$ diverges is consistent with this, as either $\lim a_n > 0$ or $\lim a_n = 0$, but in either case, $\lim(1+a_n) \neq 0$.
9. A series $\sum_{n=1}^{\infty} a_n$ converges to a sum $S$. A new series is formed, $\sum_{k=5}^{\infty} a_{k-2}$. What is the sum of this new series in terms of $S$, $a_1$, and $a_2$?
   Answer: $S - a_1 - a_2$ Let's re-index the new series. Let $n = k-2$. When the starting index is $k=5$, the new index is $n = 5-2 = 3$. As $k \to \infty$, $n \to \infty$. So the new series is $\sum_{n=3}^{\infty} a_n$. This is the original series with the first two terms removed. Therefore, its sum is the original sum $S$ minus the first two terms, $a_1$ and $a_2$. The sum is $S - (a_1 + a_2)$.
10. Given that $\sum_{n=1}^{\infty} a_n$ converges and $a_n > 0$ for all $n$. Which of the following statements must be true?
    Answer: The series $\sum_{n=1}^{\infty} \frac{1}{a_n}$ diverges. If $\sum a_n$ converges, then it is necessary that $\lim_{n \to \infty} a_n = 0$. Since $a_n > 0$, we can consider the terms of the series $\sum 1/a_n$. The limit of these terms is $\lim_{n \to \infty} \frac{1}{a_n}$, which approaches $+\infty$ as $a_n$ approaches 0. By the n-th Term Test for Divergence, since the limit of the terms is not zero, the series $\sum 1/a_n$ must diverge. For A, a counterexample is $a_n = 1/n^2$; $\sum a_n$ converges but $\sum \sqrt{a_n} = \sum 1/n$ diverges. C is false because with $a_n > 0$, the partial sums are increasing. D is not necessarily true.
11. Let $\sum a_n$ and $\sum b_n$ be series with positive terms. If $\sum a_n$ converges and $\sum b_n$ diverges, which of the following series must diverge?
    Answer: $\sum (a_n + b_n)$ If $\sum a_n$ converges to $L$ and $\sum b_n$ diverges, then $\sum (a_n + b_n)$ must diverge. If it were to converge to a sum $M$, then $\sum b_n = \sum ((a_n+b_n) - a_n) = M - L$, which would imply $\sum b_n$ converges, contradicting our assumption. For counterexamples to other choices: If $a_n = 1/n^2$ and $b_n = 1/n$, then $\sum a_n b_n = \sum 1/n^3$ converges. If $a_n = 1/n^2$ and $b_n = n$, then $\sum a_n/b_n = \sum 1/n^3$ converges. Choice D can also converge in some cases.
12. Suppose $\sum_{n=1}^{\infty} a_n$ is a series with positive terms and its sequence of partial sums is $S_N$. Which condition is sufficient to guarantee the convergence of the series?
    Answer: The sequence $\{S_N\}$ is bounded above. Since the series has positive terms ($a_n > 0$), the sequence of partial sums $S_N = S_{N-1} + a_N$ is strictly increasing. The Monotone Convergence Theorem states that a sequence that is both monotonic and bounded must converge. Since $\{S_N\}$ is already monotonic (increasing), the only additional condition needed to guarantee convergence is that it must be bounded. Because it's increasing, it is already bounded below by $S_1$, so being bounded above is sufficient. A is necessary but not sufficient. B is not sufficient (e.g., harmonic series). D is impossible since $a_n > 0$.
13. Let $f(x)$ be a function such that $f(1)=10$ and $\lim_{x\to\infty} f(x) = 2$. The series $\sum_{n=1}^\infty (f(n) - f(n+1))$ is a telescoping series. What is the sum of this series?
    Answer: 8 The Nth partial sum of this telescoping series is $S_N = (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(N) - f(N+1))$. The intermediate terms cancel out, leaving $S_N = f(1) - f(N+1)$. To find the sum of the infinite series, we take the limit of $S_N$ as $N \to \infty$. $$\text{Sum} = \lim_{N \to \infty} S_N = \lim_{N \to \infty} (f(1) - f(N+1)) = f(1) - \lim_{N \to \infty} f(N+1).$$ Given $f(1)=10$ and $\lim_{x\to\infty} f(x) = 2$, the sum is $10 - 2 = 8$.
14. The series $\sum_{n=1}^{\infty} c \cdot a_n$ converges if $\sum_{n=1}^{\infty} a_n$ converges. What must be true about the constant $c$?
    Answer: $c$ can be any real number. One of the fundamental properties of convergent series is that if $\sum a_n$ converges to a sum $S$, then for any real constant $c$, the series $\sum c \cdot a_n$ also converges, and its sum is $c \cdot S$. This property holds for any real number $c$, including $c=0$ (in which case the series converges to 0). The condition is not restricted to non-zero, positive, or small values of $c$.
15. If the infinite series $\sum_{n=1}^{\infty} a_n$ converges to a finite sum $L$, which statement regarding its sequence of partial sums, $\{S_N\}$, must be true?
    Answer: For any $\epsilon > 0$, there exists an integer $M$ such that for all $N > M$, $|S_N - L| < \epsilon$. The statement that the series $\sum a_n$ converges to $L$ is, by definition, equivalent to the statement that its sequence of partial sums $\{S_N\}$ converges to $L$. Choice C is the formal definition of the limit $\lim_{N \to \infty} S_N = L$. Choice A is only true if all terms $a_n$ have the same sign. Choice B confuses the limit of the partial sums with the limit of the terms $a_n$. Choice D is incorrect; $S_N - S_{N-1} = a_N$, and $\lim_{N \to \infty} a_N = 0$, not $L$.
16. Suppose $\sum_{n=1}^{\infty} |a_n|$ converges. Which of the following MUST be true about the series $\sum_{n=1}^{\infty} a_n$?
    Answer: The series $\sum a_n$ converges. This question relates to the concept of absolute convergence. A fundamental theorem of infinite series states that if a series converges absolutely (i.e., $\sum |a_n|$ converges), then the series itself ($\sum a_n$) must also converge. The sum is not necessarily 0. This is because $0 \le a_n + |a_n| \le 2|a_n|$. By comparison, $\sum (a_n + |a_n|)$ converges. Since $\sum a_n = \sum (a_n + |a_n|) - \sum |a_n|$, it is the difference of two convergent series and must therefore converge.
17. Which of the following statements about infinite series is FALSE?
    Answer: If $\lim_{n \to \infty} a_n = 0$, then $\sum a_n$ converges. Statement D is the converse of the n-th Term Test for Divergence, and it is false. While $\lim_{n \to \infty} a_n = 0$ is a necessary condition for convergence, it is not sufficient. The classic counterexample is the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$. For this series, $\lim_{n \to \infty} \frac{1}{n} = 0$, but the series diverges.
18. Which of the following series can be shown to be divergent using only the n-th Term Test for Divergence?
    Answer: $\sum_{n=1}^{\infty} n \sin(1/n)$ The n-th Term Test states that if $\lim_{n \to \infty} a_n \neq 0$, the series diverges. We test each option: A) $\lim_{n \to \infty} 1/n = 0$. Test is inconclusive. B) $\lim_{n \to \infty} (\ln n)/n = 0$ by L'Hopital's Rule. Test is inconclusive. C) $\lim_{n \to \infty} n \sin(1/n) = \lim_{x \to 0^+} \frac{\sin x}{x} = 1$ (by letting $x=1/n$). Since the limit is $1 \neq 0$, the series diverges by the n-th Term Test. D) $\lim_{n \to \infty} n/(n^2+1) = 0$. Test is inconclusive.
19. Consider the series $\sum_{n=0}^{\infty} \frac{3^n + 2^n}{3^{n+1} - 2^n}$. Which of the following is true?
    Answer: The series diverges. To determine convergence, we first apply the n-th Term Test for Divergence. We find the limit of the terms as $n \to \infty$. Divide the numerator and denominator by the dominant term, $3^n$: $$a_n = \frac{3^n + 2^n}{3 \cdot 3^n - 2^n} = \frac{1 + (2/3)^n}{3 - (2/3)^n}$$ Taking the limit: $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1 + (2/3)^n}{3 - (2/3)^n} = \frac{1 + 0}{3 - 0} = \frac{1}{3}.$$ Since the limit of the terms is not zero, the series diverges by the n-th Term Test.
20. The sum of a geometric series is given by $\frac{a}{1-r}$ when $|r|<1$. What is the value of the sum $\sum_{n=2}^{\infty} 3 \left(\frac{1}{2}\right)^n$?
    Answer: 3/2 This is a geometric series with common ratio $r = 1/2$. Since $|1/2| < 1$, the series converges. The standard formula $a/(1-r)$ applies when the series starts at $n=0$ or $n=1$. Here, the series starts at $n=2$. The first term is for $n=2$, which is $a = 3(1/2)^2 = 3/4$. We can use the formula for the sum as $S = \frac{\text{first term}}{1-r}$. $$S = \frac{3/4}{1 - 1/2} = \frac{3/4}{1/2} = \frac{3}{4} \cdot 2 = \frac{3}{2}.$$ Alternatively, find the sum from $n=0$, $\frac{3}{1-1/2} = 6$, and subtract the $n=0$ term (3) and the $n=1$ term (3/2), giving $6 - 3 - 3/2 = 3/2$.

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