Sequences and Their Properties

1. [latex]a_n = 0[/latex] if [latex]n[/latex] is odd and [latex]a_n = 2[/latex] if [latex]n[/latex] is even
2. [latex]{ a_n } = { 1, 3, 6, 10, 15, 21, \ldots }[/latex]
3. [latex]a_n = \dfrac{n (n+1)}{2}[/latex]
4. [latex]a_n = 4n - 7[/latex]
5. [latex]a_n = 3 \cdot 10^{1-n} = 30 \cdot 10^{-n}[/latex]
6. [latex]a_n = 2^n - 1[/latex]
7. [latex]a_n = \dfrac{(-1)^{n-1}}{2n-1}[/latex]
8. [latex]f(n) = 2^n[/latex]
9. [latex]f(n) = \dfrac{n!}{2^{n-2}}[/latex]
10. Terms oscillate above and below [latex]\dfrac{5}{3}[/latex] and appear to converge to [latex]\dfrac{5}{3}[/latex].
11. Terms oscillate above and below [latex]y \approx 1.57 \ldots[/latex] and appear to converge to a limit.
12. Graph oscillates and suggests no limit.
13. [latex]7[/latex]
14. [latex]0[/latex]
15. [latex]0[/latex]
16. [latex]1[/latex]
17. bounded, decreasing for [latex]n \ge 1[/latex]
18. bounded, not monotone
19. bounded, decreasing
20. not monotone, not bounded
21. [latex]a_n[/latex] is decreasing and bounded below by [latex]2[/latex]. The limit [latex]a[/latex] must satisfy [latex]a = \sqrt{2a}[/latex] so [latex]a = 2[/latex], independent of the initial value.
22. [latex]0[/latex]
23. [latex]0: |\sin{x}| \le |x|[/latex] and [latex]|\sin{x}| \le 1[/latex] so [latex]-\dfrac{1}{n} \le a_n \le \dfrac{1}{n}[/latex]
24. [latex]n^{\dfrac{1}{n}} \to 1[/latex] and [latex]2^{\dfrac{1}{n}} \to 1[/latex], so [latex]a_n \to 0[/latex]
25. Since [latex]\left(1 + \dfrac{1}{n}\right)^n \to e[/latex], one has [latex]\left(1 - \dfrac{2}{n}\right)^n \approx \left(1 + k\right)^{-2k} \to e^{-2}[/latex] as [latex]k \to \infty[/latex].
26. [latex]2^n + 3^n \le 2 \cdot 3^n[/latex] and [latex]\dfrac{3^n}{4^n} \to 0[/latex] as [latex]n \to \infty[/latex], so [latex]a_n \to 0[/latex] as [latex]n \to \infty[/latex].
27. [latex]0[/latex]
28. [latex]\dfrac{a_{n+1}}{a_n} = \dfrac{n!}{(n+1)(n+2) \ldots (2n)} = \dfrac{1 \cdot 2 \cdot 3 \ldots n}{(n+1)(n+2) \ldots (2n)} < \dfrac{1}{2^n}[/latex]. In particular, [latex]\dfrac{a_{n+1}}{a_n} \le \dfrac{1}{2}[/latex], so [latex]a_n \to 0[/latex] as [latex]n \to \infty[/latex].
29. a. Without losses, the population would obey [latex]P_n = 1.06 P_{n-1}[/latex]. The subtraction of [latex]150[/latex] accounts for fish losses.
    b. After [latex]12[/latex] months, we have [latex]P_{12} \approx 1494[/latex].
30. a. The student owes [latex]$9383[/latex] after [latex]12[/latex] months.
    b. The loan will be paid in full after [latex]139[/latex] months or eleven and a half years.

Introduction to Series

1. [latex]\displaystyle\sum _{n=1}^{\infty }\dfrac{1}{n}[/latex]
2. [latex]\displaystyle\sum _{n=1}^{\infty }\dfrac{{\left(-1\right)}^{n - 1}}{n}[/latex]
3. [latex]1,3,6,10[/latex]
4. [latex]1,1,0,0[/latex]
5. [latex]{a}{n}={S}{n}-{S}_{n - 1}=\dfrac{1}{n - 1}-\dfrac{1}{n}[/latex]. Series converges to [latex]S=1[/latex].
6. [latex]{a}{n}={S}{n}-{S}_{n - 1}=\sqrt{n}-\sqrt{n - 1}=\dfrac{1}{\sqrt{n - 1}+\sqrt{n}}[/latex]. Series diverges because partial sums are unbounded.
7. [latex]{S}{1}=\dfrac{1}{3}[/latex], [latex]{S}{2}=\dfrac{1}{3}+\dfrac{2}{4}>\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}[/latex], [latex]{S}{3}=\dfrac{1}{3}+\dfrac{2}{4}+\dfrac{3}{5}>3\cdot \left(\dfrac{1}{3}\right)=1[/latex]. In general [latex]{S}{k}>\dfrac{k}{3}[/latex]. Series diverges.
8. [latex]\begin{array}{l}{S}{1}=\dfrac{1}{\left(2.3\right)}=\dfrac{1}{6}=\dfrac{2}{3} - \dfrac{1}{2},\hfill \\ {S}{2}=\dfrac{1}{\left(2.3\right)}+\dfrac{1}{\left(3.4\right)}=\dfrac{2}{12}+\dfrac{1}{12}=\dfrac{1}{4}=\dfrac{3}{4} - \dfrac{1}{2},\hfill \\ {S}{3}=\dfrac{1}{\left(2.3\right)}+\dfrac{1}{\left(3.4\right)}+\dfrac{1}{\left(4.5\right)}=\dfrac{10}{60}+\dfrac{5}{60}+\dfrac{3}{60}=\dfrac{3}{10}=\dfrac{4}{5} - \dfrac{1}{2},\hfill \\ {S}{4}=\dfrac{1}{\left(2.3\right)}+\dfrac{1}{\left(3.4\right)}+\dfrac{1}{\left(4.5\right)}+\dfrac{1}{\left(5.6\right)}=\dfrac{10}{60}+\dfrac{5}{60}+\dfrac{3}{60}+\dfrac{2}{60}=\dfrac{1}{3}=\dfrac{5}{6} - \dfrac{1}{2}.\hfill \end{array}[/latex]The pattern is [latex]{S}_{k}=\dfrac{\left(k+1\right)}{\left(k+2\right)}-\dfrac{1}{2}[/latex] and the series converges to [latex]\dfrac{1}{2}[/latex].
9. [latex]0[/latex]
10. [latex]-3[/latex]
11. diverges, [latex]\displaystyle\sum _{n=1001}^{\infty }\dfrac{1}{n}[/latex]
12. convergent geometric series, [latex]r=\dfrac{1}{10}<1[/latex]
13. convergent geometric series, [latex]r=\dfrac{\pi}{{e}^{2}}<1[/latex]
14. [latex]\displaystyle\sum _{n=1}^{\infty }5\cdot {\left(-\dfrac{1}{5}\right)}^{n}[/latex], converges to [latex]-\dfrac{5}{6}[/latex]
15. [latex]\displaystyle\sum _{n=1}^{\infty }100\cdot {\left(\dfrac{1}{10}\right)}^{n}[/latex], converges to [latex]\dfrac{100}{9}[/latex]
16. [latex]x\displaystyle\sum _{n=0}^{\infty }{\left(\text{-}x\right)}^{n}=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}{x}^{n}[/latex]
17. [latex]\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{\sin}^{2n}\left(x\right)[/latex]
18. [latex]{S}_{k}=2-{2}^{\dfrac{1}{\left(k+1\right)}}\to 1[/latex] as [latex]k\to \infty[/latex].
19. [latex]{S}_{k}=1-\sqrt{k+1}[/latex] diverges
20. [latex]\displaystyle\sum {n=1}^{\infty }\text{ln}n-\text{ln}\left(n+1\right),{S}{k}=\text{-}\text{ln}\left(k+1\right)[/latex]
21. [latex]{a}{n}=\dfrac{1}{\text{ln}n}-\dfrac{1}{\text{ln}\left(n+1\right)}[/latex] and [latex]{S}{k}=\dfrac{1}{\text{ln}\left(2\right)}-\dfrac{1}{\text{ln}\left(k+1\right)}\to \dfrac{1}{\text{ln}\left(2\right)}[/latex]
22. [latex]\displaystyle\sum {n=1}^{\infty }{a}{n}=f\left(1\right)-f\left(2\right)[/latex]
23. [latex]{c}{0}+{c}{1}+{c}{2}+{c}{3}+{c}_{4}=0[/latex]
24. [latex]\dfrac{2}{{n}^{3}-1}=\dfrac{1}{n - 1}-\dfrac{2}{n}+\dfrac{1}{n+1}[/latex], [latex]{S}_{n}=\left(1 - 1+\dfrac{1}{3}\right)+\left(\dfrac{1}{2} - \dfrac{2}{3}+\dfrac{1}{4}\right)[/latex] [latex]+\left(\dfrac{1}{3} - \dfrac{2}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{4} - \dfrac{2}{5}+\dfrac{1}{6}\right)+\text{⋯ }=\dfrac{1}{2}[/latex]
25. [latex]{t}_{k}[/latex] converges to [latex]0.57721\text{⋯ }{t}_{k}[/latex] is a sum of rectangles of height [latex]\dfrac{1}{k}[/latex] over the interval [latex]\left[k,k+1\right][/latex] which lie above the graph of [latex]\dfrac{1}{x}[/latex].
26. [latex]N=22[/latex], [latex]{S}_{N}=6.1415[/latex]
27. [latex]N=3[/latex], [latex]{S}_{N}=1.559877597243667..[/latex].
28. a. The probability of any given ordered sequence of outcomes for [latex]n[/latex] coin flips is [latex]\dfrac{1}{{2}^{n}}[/latex]. b. The probability of coming up heads for the first time on the [latex]n[/latex] th flip is the probability of the sequence [latex]TT\text{$\ldots$ }TH[/latex] which is [latex]\dfrac{1}{{2}^{n}}[/latex]. The probability of coming up heads for the first time on an even flip is [latex]\displaystyle\sum _{n=1}^{\infty }\dfrac{1}{{2}^{2n}}[/latex] or [latex]\dfrac{1}{3}[/latex].
29. [latex]\dfrac{5}{9}[/latex]
30. The part of the first dose after [latex]n[/latex] hours is [latex]d{r}^{n}[/latex], the part of the second dose is [latex]d{r}^{n-N}[/latex], and, in general, the part remaining of the [latex]m\text{th}[/latex] dose is [latex]d{r}^{n-mN}[/latex], so [latex]A\left(n\right)=\displaystyle\sum _{l=0}^{m}d{r}^{n-lN}=\displaystyle\sum _{l=0}^{m}d{r}^{k+\left(m-l\right)N}=\displaystyle\sum _{q=0}^{m}d{r}^{k+qN}=d{r}^{k}\displaystyle\sum _{q=0}^{m}{r}^{Nq}=d{r}^{k}\dfrac{1-{r}^{\left(m+1\right)N}}{1-{r}^{N}},n=k+mN[/latex].

The Divergence and Integral Tests

1. [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=0[/latex]. Divergence test does not apply.
2. [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=2[/latex]. Series diverges.
3. [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=\infty[/latex] (does not exist). Series diverges.
4. [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=1[/latex]. Series diverges.
5. [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}[/latex] does not exist. Series diverges.
6. [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=\dfrac{1}{{e}^{2}}[/latex]. Series diverges.
7. [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=0[/latex]. Divergence test does not apply.
8. Series converges, [latex]p>1[/latex].
9. Series converges, [latex]p=\dfrac{4}{3}>1[/latex].
10. Series converges, [latex]p=2e-\pi >1[/latex].
11. Series diverges by comparison with [latex]{\displaystyle\int }_{1}^{\infty }\dfrac{dx}{{\left(x+5\right)}^{\frac{1}{3}}}[/latex].
12. Series diverges by comparison with [latex]{\displaystyle\int }_{1}^{\infty }\dfrac{x}{1+{x}^{2}}dx[/latex].
13. Series converges by comparison with [latex]{\displaystyle\int }_{1}^{\infty }\dfrac{2x}{1+{x}^{4}}dx[/latex].
14. [latex]{2}^{\text{-}\text{ln}n}=\dfrac{1}{{n}^{\text{ln}2}}[/latex]. Since [latex]\text{ln}2<1[/latex], diverges by [latex]p[/latex] -series.
15. [latex]{2}^{-2\text{ln}n}=\dfrac{1}{{n}^{2\text{ln}2}}[/latex]. Since [latex]2\text{ln}2 - 1<1[/latex], diverges by [latex]p[/latex] -series.
16. [latex]{R}{1000}\le {\displaystyle\int }{1000}^{\infty }\dfrac{dt}{{t}^{2}}=-\dfrac{1}{t}{|}_{1000}^{\infty }=0.001[/latex]
17. [latex]{R}{1000}\le {\displaystyle\int }{1000}^{\infty }\dfrac{dt}{1+{t}^{2}}={\tan}^{-1}\infty -{\tan}^{-1}\left(1000\right)=\dfrac{\pi}{2}-{\tan}^{-1}\left(1000\right)\approx 0.000999[/latex]
18. [latex]{R}{N}<{\displaystyle\int }{N}^{\infty }\dfrac{dx}{{x}^{2}}=\dfrac{1}{N},N>{10}^{4}[/latex]
19. [latex]{R}{N}<{\displaystyle\int }{N}^{\infty }\dfrac{dx}{{x}^{1.01}}=100{N}^{-0.01},N>{10}^{600}[/latex]
20. [latex]{R}{N}<{\displaystyle\int }{N}^{\infty }\dfrac{dx}{1+{x}^{2}}=\dfrac{\pi}{2}-{\tan}^{-1}\left(N\right),N>\tan\left(\dfrac{\pi}{2}-{10}^{-3}\right)\approx 1000[/latex]
21. [latex]{R}{N}<{\displaystyle\int }{N}^{\infty }\dfrac{dx}{{e}^{x}}={e}^{\text{-}N},N>5\text{ln}\left(10\right)[/latex], okay if [latex]N=12;\displaystyle\sum _{n=1}^{12}{e}^{\text{-}n}=0.581973...[/latex]. Estimate agrees with [latex]\dfrac{1}{\left(e - 1\right)}[/latex] to five decimal places.
22. [latex]{R}_{N}<{\displaystyle\int }_{N}^{\infty }\dfrac{dx}{{x}^{4}}=\dfrac{4}{{N}^{3}},N>{\left({4.10}^{4}\right)}^{\frac{1}{3}}[/latex], okay if [latex]N=35[/latex];[latex]\displaystyle\sum _{n=1}^{35}\dfrac{1}{{n}^{4}}=1.08231\text{...}[/latex]. Estimate agrees with the sum to four decimal places.
23. [latex]\text{ln}\left(2\right)[/latex]
24. [latex]T=0.5772..[/latex].
25. The expected number of random insertions to get [latex]B[/latex] to the top is [latex]n+\dfrac{n}{2}+\dfrac{n}{3}+\text{...}+\dfrac{n}{\left(n - 1\right)}[/latex]. Then one more insertion puts [latex]B[/latex] back in at random. Thus, the expected number of shuffles to randomize the deck is [latex]n\left(1+\dfrac{1}{2}+\text{...}+\dfrac{1}{n}\right)[/latex].
26. Set [latex]{b}{n}={a}{n+N}[/latex] and [latex]g\left(t\right)=f\left(t+N\right)[/latex] such that [latex]f[/latex] is decreasing on [latex]\left[t,\infty \right)[/latex].
27. The series converges for [latex]p>1[/latex] by integral test using change of variable.
28. [latex]N={e}^{{e}^{100}}\approx {e}^{{10}^{43}}[/latex] terms are needed.

Comparison Tests

1. Converges by comparison with [latex]\dfrac{1}{{n}^{2}}[/latex].
2. Diverges by comparison with harmonic series, since [latex]2n - 1\ge n[/latex].
3. [latex]{a}_{n}=\dfrac{1}{\left(n+1\right)\left(n+2\right)}<\dfrac{1}{{n}^{2}}[/latex]. Converges by comparison with p-series, [latex]p=2[/latex].
4. [latex]\sin\left(\dfrac{1}{n}\right)\le \dfrac{1}{n}[/latex], so converges by comparison with p-series, [latex]p=2[/latex].
5. [latex]\sin\left(\dfrac{1}{n}\right)\le 1[/latex], so converges by comparison with p-series, [latex]p=\dfrac{3}{2}[/latex].
6. Since [latex]\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\left(\sqrt{n+1}+\sqrt{n}\right)}\le \dfrac{2}{\sqrt{n}}[/latex], series converges by comparison with p-series for [latex]p=1.5[/latex].
7. Converges by limit comparison with p-series for [latex]p>1[/latex].
8. Converges by limit comparison with p-series, [latex]p=2[/latex].
9. Converges by limit comparison with [latex]{4}^{\text{-}n}[/latex].
10. Converges by limit comparison with [latex]\dfrac{1}{{e}^{1.1n}}[/latex].
11. Diverges by limit comparison with harmonic series.
12. Converges by limit comparison with p-series, [latex]p=3[/latex].
13. Converges by limit comparison with p-series, [latex]p=3[/latex].
14. Diverges by limit comparison with [latex]\dfrac{1}{n}[/latex].
15. Converges for [latex]p>1[/latex] by comparison with a [latex]p[/latex] series for slightly smaller [latex]p[/latex].
16. Converges for all [latex]p>0[/latex].
17. Converges for all [latex]r>1[/latex]. If [latex]r>1[/latex] then [latex]{r}^{n}>4[/latex], say, once [latex]n>\dfrac{\text{ln}\left(2\right)}{\text{ln}\left(r\right)}[/latex] and then the series converges by limit comparison with a geometric series with ratio [latex]\dfrac{1}{2}[/latex].
18. The numerator is equal to [latex]1[/latex] when [latex]n[/latex] is odd and [latex]0[/latex] when [latex]n[/latex] is even, so the series can be rewritten [latex]\displaystyle\sum _{n=1}^{\infty }\dfrac{1}{2n+1}[/latex], which diverges by limit comparison with the harmonic series.
19. [latex]{\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}[/latex] or [latex]{a}^{2}+{b}^{2}\ge 2ab[/latex], so convergence follows from comparison of [latex]2{a}{n}{b}{n}[/latex] with [latex]{a}^{2}{}{n}+{b}^{2}{}{n}[/latex]. Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.
20. [latex]{\left(\text{ln}n\right)}^{\text{-}\text{ln}n}={e}^{\text{-}\text{ln}\left(n\right)\text{ln}\text{ln}\left(n\right)}[/latex]. If [latex]n[/latex] is sufficiently large, then [latex]\text{ln}\text{ln}n>2[/latex], so [latex]{\left(\text{ln}n\right)}^{\text{-}\text{ln}n}<\dfrac{1}{{n}^{2}}[/latex], and the series converges by comparison to a [latex]p-\text{series}\text{.}[/latex]
21. [latex]{a}{n}\to 0[/latex], so [latex]{a}^{2}{}{n}\le |{a}_{n}|[/latex] for large [latex]n[/latex]. Convergence follows from limit comparison. [latex]\displaystyle\sum \dfrac{1}{{n}^{2}}[/latex] converges, but [latex]\displaystyle\sum \dfrac{1}{n}[/latex] does not, so the fact that [latex]\displaystyle\sum {n=1}^{\infty }{a}^{2}{}{n}[/latex] converges does not imply that [latex]\displaystyle\sum {n=1}^{\infty }{a}{n}[/latex] converges.
22. No. [latex]\displaystyle\sum {n=1}^{\infty }\dfrac{1}{n}[/latex] diverges. Let [latex]{b}{k}=0[/latex] unless [latex]k={n}^{2}[/latex] for some [latex]n[/latex]. Then [latex]\displaystyle\sum {k}\dfrac{{b}{k}}{k}=\displaystyle\sum \dfrac{1}{{k}^{2}}[/latex] converges.
23. [latex]|\sin{t}|\le |t|[/latex], so the result follows from the comparison test.
24. By the comparison test, [latex]x=\displaystyle\sum {n=1}^{\infty }\dfrac{{b}{n}}{{2}^{n}}\le \displaystyle\sum _{n=1}^{\infty }\dfrac{1}{{2}^{n}}=1[/latex].
25. If [latex]{b}_{1}=0[/latex], then, by comparison, [latex]x\le \displaystyle\sum _{n=2}^{\infty }\dfrac{1}{{2}^{n}}=\dfrac{1}{2}[/latex].
26. Yes. Keep adding [latex]1\text{-kg}[/latex] weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the [latex]1\text{-kg}[/latex] weights, and add [latex]0.1\text{-kg}[/latex] weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last [latex]0.1\text{-kg}[/latex] weight. Start adding [latex]0.01\text{-kg}[/latex] weights. If it balances, stop. If it tips to the side with the weights, remove the last [latex]0.01\text{-kg}[/latex] weight that was added. Continue in this way for the [latex]0.001\text{-kg}[/latex] weights, and so on. After a finite number of steps, one has a finite series of the form [latex]A+\displaystyle\sum {n=1}^{N}\dfrac{{s}{n}}{{10}^{n}}[/latex] where [latex]A[/latex] is the number of full kg weights and [latex]{d}_{n}[/latex] is the number of [latex]\dfrac{1}{{10}^{n}}\text{-kg}[/latex] weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the [latex]N\text{th}[/latex] partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most [latex]\dfrac{1}{{10}^{N}}[/latex].

Alternating Series

1. Does not converge by divergence test. Terms do not tend to zero.
2. Converges conditionally by alternating series test, since [latex]\dfrac{\sqrt{n+3}}{n}[/latex] is decreasing. Does not converge absolutely by comparison with p-series, [latex]p=\dfrac{1}{2}[/latex].
3. Converges absolutely by limit comparison to [latex]\dfrac{{3}^{n}}{{4}^{n}}[/latex], for example.
4. Diverges by divergence test since [latex]\underset{n\to \infty }{\text{lim}}|{a}_{n}|=e[/latex].
5. Does not converge. Terms do not tend to zero.
6. [latex]\underset{n\to \infty }{\text{lim}}{\cos}^{2}\left(\dfrac{1}{n}\right)=1[/latex]. Diverges by divergence test.
7. Converges by alternating series test.
8. Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, [latex]p=\pi -e[/latex]
9. Diverges; terms do not tend to zero.
10. Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.
11. Converges absolutely by limit comparison with p-series, [latex]p=\dfrac{3}{2}[/latex], after applying the hint.
12. Converges by alternating series test since [latex]n\left({\tan}^{-1}\left(n+1\right)\text{-}{\tan}^{-1}n\right)[/latex] is decreasing to zero for large [latex]n[/latex]. Does not converge absolutely by limit comparison with harmonic series after applying hint.
13. Converges absolutely, since [latex]{a}_{n}=\dfrac{1}{n}-\dfrac{1}{n+1}[/latex] are terms of a telescoping series.
14. Terms do not tend to zero. Series diverges by divergence test.
15. Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.
16. [latex]\text{ln}\left(N+1\right)>10[/latex], [latex]N+1>{e}^{10}[/latex], [latex]N\ge 22026[/latex]; [latex]{S}_{22026}=0.0257\text{...}[/latex]
17. [latex]{2}^{N+1}>{10}^{6}[/latex] or [latex]N+1>\dfrac{6\text{ln}\left(10\right)}{\text{ln}\left(2\right)}=19.93[/latex]. or [latex]N\ge 19[/latex]; [latex]{S}_{19}=0.333333969\text{...}[/latex]
18. [latex]{\left(N+1\right)}^{2}>{10}^{6}[/latex] or [latex]N>999[/latex]; [latex]{S}_{1000}\approx 0.822466[/latex].
19. True. [latex]{b}{n}[/latex] need not tend to zero since if [latex]{c}{n}={b}{n}-\text{lim}{b}{n}[/latex], then [latex]{c}{2n - 1}-{c}{2n}={b}{2n - 1}-{b}{2n}[/latex].
20. True. [latex]{b}{3n - 1}-{b}{3n}\ge 0[/latex], so convergence of [latex]\displaystyle\sum {b}_{3n - 2}[/latex] follows from the comparison test.
21. True. If one converges, then so must the other, implying absolute convergence.
22. Yes. Take [latex]{b}{n}=1[/latex] if [latex]{a}{n}\ge 0[/latex] and [latex]{b}{n}=0[/latex] if [latex]{a}{n}<0[/latex]. Then [latex]\displaystyle\sum {n=1}^{\infty }{a}{n}{b}{n}=\displaystyle\sum {n:{a}{n}\ge 0}{a}{n}[/latex] converges. Similarly, one can show [latex]\displaystyle\sum {n:{a}{n}<0}{a}_{n}[/latex] converges. Since both series converge, the series must converge absolutely.
23. Not decreasing. Does not converge absolutely.
24. Not alternating. Can be expressed as [latex]\displaystyle\sum _{n=1}^{\infty }\left(\dfrac{1}{3n - 2}+\dfrac{1}{3n - 1}-\dfrac{1}{3n}\right)[/latex], which diverges by comparison with [latex]\displaystyle\sum \dfrac{1}{3n - 2}[/latex].
25. Let [latex]{a}^{+}{}{n}={a}{n}[/latex] if [latex]{a}{n}\ge 0[/latex] and [latex]{a}^{+}{}{n}=0[/latex] if [latex]{a}{n}<0[/latex]. Then [latex]{a}^{+}{}{n}\le |{a}{n}|[/latex] for all [latex]n[/latex] so the sequence of partial sums of [latex]{a}^{+}{}{n}[/latex] is increasing and bounded above by the sequence of partial sums of [latex]|{a}_{n}|[/latex], which converges; hence, [latex]\displaystyle\sum {n=1}^{\infty }{a}^{+}{}{n}[/latex] converges.
26. For [latex]N=5[/latex] one has [latex]|{R}{N}|{b}{6}=\dfrac{{\theta }^{10}}{10\text{!}}[/latex]. When [latex]\theta =1[/latex], [latex]{R}{5}\le \dfrac{1}{10\text{!}}\approx 2.75\times {10}^{-7}[/latex]. When [latex]\theta =\dfrac{\pi}{6}[/latex], [latex]{R}{5}\le {\left(\dfrac{\pi}{6}\right)}^\dfrac{{10}}{10\text{!}}\approx 4.26\times {10}^{-10}[/latex]. When [latex]\theta =\pi[/latex], [latex]{R}_{5}\le \dfrac{{\pi }^{10}}{10\text{!}}=0.0258[/latex].
27. Let [latex]{b}_{n}=\dfrac{1}{\left(2n - 2\right)}\text{!}[/latex]. Then [latex]{R}_{N}\le \dfrac{1}{\left(2N\right)\text{!}}<0.00001[/latex] when [latex]\left(2N\right)\text{!}>{10}^{5}[/latex] or [latex]N=5[/latex] and [latex]1-\dfrac{1}{2\text{!}}+\dfrac{1}{4\text{!}}-\dfrac{1}{6\text{!}}+\dfrac{1}{8\text{!}}=0.540325\text{...}[/latex], whereas [latex]\cos1=0.5403023\text{...}[/latex]
28. Let [latex]T=\displaystyle\sum \dfrac{1}{{n}^{2}}[/latex]. Then [latex]T-S=\dfrac{1}{2}T[/latex], so [latex]S=\dfrac{T}{2}[/latex]. [latex]\sqrt{6\times \displaystyle\sum _{n=1}^{1000}\dfrac{1}{{n}^{2}}}=3.140638\text{...}[/latex]; [latex]\sqrt{12\times \displaystyle\sum _{n=1}^{1000}\dfrac{{\left(-1\right)}^{n - 1}}{{n}^{2}}}=3.141591\text{...}[/latex]; [latex]\pi =3.141592\text{...}[/latex]. The alternating series is more accurate for [latex]1000[/latex] terms.

Ratio and Root Tests

1. [latex]\dfrac{{a}{n+1}}{{a}{n}}\to 0[/latex]. Converges.
2. [latex]\dfrac{{a}{n+1}}{{a}{n}}=\dfrac{1}{2}{\left(\dfrac{n+1}{n}\right)}^{2}\to \dfrac{1}{2}<1[/latex]. Converges.
3. [latex]\dfrac{{a}{n+1}}{{a}{n}}\to \dfrac{1}{27}<1[/latex]. Converges.
4. [latex]\dfrac{{a}{n+1}}{{a}{n}}\to \dfrac{4}{{e}^{2}}<1[/latex]. Converges.
5. [latex]\dfrac{{a}{n+1}}{{a}{n}}\to 1[/latex]. Ratio test is inconclusive.
6. [latex]\dfrac{{a}{n}}{{a}{n+1}}\to \dfrac{1}{{e}^{2}}[/latex]. Converges.
7. [latex]{\left({a}_{k}\right)}^{\dfrac{1}{k}}\to 2>1[/latex]. Diverges.
8. [latex]{\left({a}_{n}\right)}^{\dfrac{1}{n}}\to \dfrac{1}{2}<1[/latex]. Converges.
9. [latex]{\left({a}_{k}\right)}^{\dfrac{1}{k}}\to \dfrac{1}{e}<1[/latex]. Converges.
10. [latex]{a}_{n}^{\dfrac{1}{n}}=\dfrac{1}{e}+\dfrac{1}{n}\to \dfrac{1}{e}<1[/latex]. Converges.
11. [latex]{a}_{n}^{\dfrac{1}{n}}=\dfrac{\left(\text{ln}\left(1+\text{ln}n\right)\right)}{\left(\text{ln}n\right)}\to 0[/latex] by L’Hôpital’s rule. Converges.
12. [latex]\dfrac{{a}{k+1}}{{a}{k}}=\dfrac{1}{2k+1}\to 0[/latex]. Converges by ratio test.
13. [latex]{\left({a}_{n}\right)}^{\dfrac{1}{n}}\to \dfrac{1}{e}[/latex]. Converges by root test.
14. [latex]{a}_{k}^{\dfrac{1}{k}}\to \text{ln}\left(3\right)>1[/latex]. Diverges by root test.
15. [latex]\dfrac{{a}{n+1}}{{a}{n}}=[/latex] [latex]\dfrac{{3}^{2n+1}}{{2}^{3{n}^{2}+3n+1}}\to 0[/latex]. Converge.
16. Converges by root test and limit comparison test since [latex]{x}_{n}\to \sqrt{2}[/latex].
17. Converges absolutely by limit comparison with [latex]p-\text{series,}[/latex] [latex]p=2[/latex].
18. [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=\dfrac{1}{{e}^{2}}\ne 0[/latex]. Series diverges.
19. Terms do not tend to zero: [latex]{a}_{k}\ge \dfrac{1}{2}[/latex], since [latex]{\sin}^{2}x\le 1[/latex].
20. [latex]{a}_{n}=\dfrac{2}{\left(n+1\right)\left(n+2\right)}[/latex], which converges by comparison with [latex]p-\text{series}[/latex] for [latex]p=2[/latex].
21. [latex]{a}_{k}=\dfrac{{2}^{k}1\cdot 2\text{...}k}{\left(2k+1\right)\left(2k+2\right)\text{...}3k}\le {\left(\dfrac{2}{3}\right)}^{k}[/latex] converges by comparison with geometric series.
22. [latex]{a}_{k}\approx {e}^{\text{-}\text{ln}{k}^{2}}=\dfrac{1}{{k}^{2}}[/latex]. Series converges by limit comparison with [latex]p-\text{series,}[/latex] [latex]p=2[/latex].
23. If [latex]{b}{k}=\dfrac{{c}^{1-k}}{\left(c - 1\right)}[/latex] and [latex]{a}{k}=k[/latex], then [latex]{b}{k+1}-{b}{k}=\text{-}{c}^{\text{-}k}[/latex] and [latex]\displaystyle\sum {n=1}^{\infty }\dfrac{k}{{c}^{k}}={a}{1}{b}_{1}+\dfrac{1}{c - 1}\displaystyle\sum _{k=1}^{\infty }{c}^{\text{-}k}=\dfrac{c}{{\left(c - 1\right)}^{2}}[/latex].
24. [latex]6+4+1=11[/latex]
25. [latex]|x|\le 1[/latex]
26. [latex]|x|<\infty[/latex]
27. All real numbers [latex]p[/latex] by the ratio test.
28. [latex]r<\dfrac{1}{p}[/latex]
29. [latex]0 < r < 1[/latex]. Note that the ratio and root tests are inconclusive. Using the hint, there are [latex]2k[/latex] terms [latex]{r}^{\sqrt{n}}[/latex] for [latex]{k}^{2}\le n<{\left(k+1\right)}^{2}[/latex], and for [latex]r<1[/latex] each term is at least [latex]{r}^{k}[/latex]. Thus, [latex]\displaystyle\sum _{n=1}^{\infty }{r}^{\sqrt{n}}=\displaystyle\sum _{k=1}^{\infty }\displaystyle\sum _{n={k}^{2}}^{{\left(k+1\right)}^{2}-1}{r}^{\sqrt{n}}[/latex] [latex]\ge \displaystyle\sum _{k=1}^{\infty }2k{r}^{k}[/latex], which converges by the ratio test for [latex]r<1[/latex]. For [latex]r\ge 1[/latex] the series diverges by the divergence test.
30. One has [latex]{a}_{1}=1[/latex], [latex]{a}_{2}={a}_{3}=\dfrac{1}{2}\text{...}{a}_{2n}={a}_{2n+1}=\dfrac{1}{{2}^{n}}[/latex]. The ratio test does not apply because [latex]\dfrac{{a}_{n+1}}{{a}_{n}}=1[/latex] if [latex]n[/latex] is even. However, [latex]\dfrac{{a}_{n+2}}{{a}_{n}}=\dfrac{1}{2}[/latex], so the series converges according to the previous exercise. Of course, the series is just a duplicated geometric series.