Solutions To Selected Problems Chapter H Chapter J

Chapter I

Exercises I.1

1.

Answers will vary.
2.

natural
3.

Answers will vary.
4.

Answers will vary.
5.

$2,\frac{8}{3},\frac{8}{3},\frac{32}{15},\frac{64}{45}$
6.

$-\frac{3}{2},\frac{9}{4},-\frac{27}{8},\frac{81}{16},-\frac{243}{32}$
7.

$-\frac{1}{3},-2,-\frac{81}{5},-\frac{512}{3},-\frac{15625}{7}$
8.

$1,1,2,3,5$
9.

$a_{n}=3n+1$
10.

$a_{n}=(-1)^{n+1}\frac{3}{2^{n-1}}$
11.

$a_{n}=10\cdot 2^{n-1}$
12.

$a_{n}=1/(n-1)!$
13.

$1/7$
14.

$3e^{2}-1$
15.

$0$
16.

$e^{4}$
17.

diverges
18.

converges to $4/3$
19.

converges to $0$
20.

converges to 0
21.

converges to $0$
22.

converges to 0
23.

diverges
24.

converges to 3
25.

converges to $e$
26.

diverges
27.

converges to 5
28.

converges to 0
29.

diverges
30.

converges to 2
31.

converges to 0
32.

diverges
33.

converges to 0
34.

converges to $\ln 3$
35.

converges to $\ln 2$
36.

converges to $1$
37.

converges to $0$
38.

converges to $1$
39.

bounded
40.

bounded
41.

bounded below
42.

bounded above
43.

monotonically increasing
44.

monotonically increasing for $n\geq 3$
45.

never monotonic
46.

monotonically decreasing for $n\geq 3$
47.

never monotonic
48.

monotonically decreasing
49.

Let $\{a_{n}\}$ be given such that $\displaystyle\lim_{n\to\infty}\left\lvert a_{n}\right\rvert=0$ . By the definition of the limit of a sequence, given any $\epsilon>0$ , there is a $m$ such that for all $n>m,\left\lvert\ \left\lvert a_{n}\right\rvert-0\right\rvert<\epsilon$ . Since $\left\lvert\ \left\lvert a_{n}\right\rvert-0\right\rvert=\left\lvert a_{n}-0\right\rvert$ , this directly implies that for all $n>m$ , $\left\lvert a_{n}-0\right\rvert<\epsilon$ , meaning that $\displaystyle\lim_{n\to\infty}a_{n}=0$ .
50.
- Left to reader
  
  One possibility: $a_{n}=1/3^{n}$ and $b_{n}=1/2^{n}$
51.

Left to reader
52.
- $f(n)=Ar^{n}$
53.
- 2

Exercises I.2

1.

Answers will vary.
2.

Answers will vary.
3.

One sequence is the sequence of terms $\{a_{n}\}$ . The other is the sequence of $n^{\text{th}}$ partial sums, $\{S_{n}\}=\{\sum_{i=1}^{n}a_{i}\}$ .
4.

Answers will vary.
5.

F
6.

F
7.
- $-1,-\frac{1}{2},-\frac{5}{6},-\frac{7}{12},-\frac{47}{60}$
  
  Plot omitted
8.
- $1,\frac{5}{4},\frac{49}{36},\frac{205}{144},\frac{5269}{3600}$
  
  Plot omitted
9.
- $-1,0,-1,0,-1$
  
  Plot omitted
10.
- $1,3,6,10,15$
  
  Plot omitted
11.
- $1,\frac{3}{2},\frac{5}{3},\frac{41}{24},\frac{103}{60}$
  
  Plot omitted
12.
- $\frac{1}{3},\frac{4}{9},\frac{13}{27},\frac{40}{81},\frac{121}{243}$
  
  Plot omitted
13.
- $-0.9,-0.09,-0.819,-0.1629,-0.75339$
  
  Plot omitted
14.
- $0.1,0.11,0.111,0.1111,0.11111$
  
  Plot omitted
15.

Converges because it is a geometric series with $r=\frac{1}{5}$ .
16.

$\displaystyle\lim_{n\to\infty}a_{n}=3$ ; by Theorem 9.2.4 the series diverges.
17.

Diverges by Theorem 9.2.4
18.

$\displaystyle\lim_{n\to\infty}a_{n}=\infty$ ; by Theorem 9.2.4 the series diverges.
19.

Diverges
20.

$\displaystyle\lim_{n\to\infty}a_{n}=\ln\frac{4}{7}$ ; by Theorem 9.2.4 the series diverges.
21.

$\displaystyle\lim_{n\to\infty}a_{n}=1$ ; by Theorem 9.2.4 the series diverges.
22.

Diverges
23.

Diverges
24.

$\displaystyle\lim_{n\to\infty}a_{n}=1/2$ ; by Theorem 9.2.4 the series diverges.
25.

Diverges
26.

Diverges by the Test for Divergence
27.

$\displaystyle\lim_{n\to\infty}a_{n}=e$ ; by Theorem 9.2.4 the series diverges.
28.

Diverges by Theorem 9.2.4
29.

Converges
30.

Converges
31.

Converges
32.

Converges
33.
- $S_{n}=\frac{1-(1/4)^{n}}{3/4}$
  
  Converges to $4/3$ .
34.
- $S_{n}=\left(\frac{n(n+1)}{2}\right)^{2}$
  
  Diverges
35.
- $S_{n}=\begin{cases}-\frac{n+1}{2}&n\text{ is odd }\\ \frac{n}{2}&n\text{ is even}\end{cases}$
  
  Diverges
36.
- $S_{n}=5\frac{1-1/2^{n}}{1/2}$
  
  Converges to 10.
37.
- $S_{n}=\frac{1-(1/e)^{n+1}}{1-1/e}$ .
  
  Converges to $1/(1-1/e)=e/(e-1)$ .
38.
- $S_{n}=\frac{1-(-1/3)^{n}}{4/3}$
  
  Converges to $3/4$ .
39.
- With partial fractions, $a_{n}=\frac{1}{n}-\frac{1}{n+1}$ .
  Thus $S_{n}=1-\frac{1}{n+1}$ .
  
  Converges to 1.
40.
- With partial fractions, $a_{n}=\frac{3}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$ .
  Thus $S_{n}=\frac{3}{2}\left(\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)$ .
  
  Converges to 9/4
41.
- Use partial fraction decomposition to recognize the telescoping series: $a_{n}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$ , so that $S_{n}=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}$ .
  
  Converges to $1/2$ .
42.
- $S_{n}=\ln\big(1/(n+1)\big)$
  
  Diverges (to $-\infty$ ).
43.
- $S_{n}=1-\frac{1}{(n+1)^{2}}$
  
  Converges to 1.
44.
- $a_{n}=\frac{1}{n(n+3)}$ ; using partial fractions,
  the resulting telescoping sum reduces to
  $S_{n}=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-% \frac{1}{n+3}\right)$
  
  Converges to $11/18$ .
45.
- $a_{n}=1/2^{n}+1/3^{n}$ for $n\geq 0$ . Thus $S_{n}=\frac{1-1/2^{2}}{1/2}+\frac{1-1/3^{n}}{2/3}$ .
  
  Converges to $2+3/2=7/2$ .
46.
- With partial fractions, $a_{n}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$ . Thus $S_{n}=\frac{1}{2}\left(3/2-\frac{1}{n}-\frac{1}{n+1}\right)$ .
  
  Converges to 3/4.
47.
- $S_{n}=\frac{1-(\sin 1)^{n+1}}{1-\sin 1}$
  
  Converges to $\frac{1}{1-\sin 1}$ .
48.
- $S_{n}=1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}+\frac{\frac{5}{4}(1-(\frac{1}{% 4})^{n})}{1-\frac{1}{4}}$
  
  Converges to $\frac{19}{6}$ .
49.

$(-3,3)$
50.

$(-5,-1)$
51.

$(-\infty,-4)\cup(4,\infty)$
52.

$(-3,-1)$
53.

$\displaystyle\lim_{n\to\infty}a_{n}=3$ ; by Theorem 9.2.4 the series diverges.
54.

$\displaystyle\lim_{n\to\infty}a_{n}=\infty$ ; by Theorem 9.2.4 the series diverges.
55.

$\displaystyle\lim_{n\to\infty}a_{n}=\infty$ ; by Theorem 9.2.4 the series diverges.
56.

$\displaystyle\lim_{n\to\infty}a_{n}=1$ ; by Theorem 9.2.4 the series diverges.
57.

$\displaystyle\lim_{n\to\infty}a_{n}=1/2$ ; by Theorem 9.2.4 the series diverges.
58.

$\displaystyle\lim_{n\to\infty}a_{n}=e$ ; by Theorem 9.2.4 the series diverges.
59.

Using partial fractions, we can show that $a_{n}=\frac{1}{4}\left(\frac{1}{2n-1}+\frac{1}{2n+1}\right)$ . The series is effectively twice the sum of the odd terms of the Harmonic Series which was shown to diverge in Example 9.2.5. Thus this series diverges.
60.

$\sum_{n=1}^{\infty}12(100)^{-n}=\frac{12}{100}\cdot\frac{1}{1-\frac{1}{100}}=% \frac{12}{100-1}=\frac{12}{99}=\frac{4}{33}$
61.

$2\left(\frac{1+r}{1-r}\right)$

Exercises I.3

1.

continuous, positive and decreasing
2.

F
3.

Converges
4.

Converges
5.

Diverges
6.

Diverges
7.

Converges
8.

Converges
9.

Converges
10.

Converges
11.

$p>1$
12.

$p<-1$
13.

$p>1$
14.

$p>1$
15.

Exercises I.4

1.

$\displaystyle\sum_{n=0}^{\infty}b_{n}$ converges; we cannot conclude anything about $\displaystyle\sum_{n=0}^{\infty}c_{n}$
2.

$\displaystyle\sum_{n=0}^{\infty}c_{n}$ diverges; we cannot conclude anything about $\displaystyle\sum_{n=0}^{\infty}b_{n}$
3.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ , as $1/(n^{2}+3n-5)\leq 1/n^{2}$ for all $n>1$ .
4.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{4^{n}}$ , as $1/(4^{n}+n^{2}-n)\leq 1/4^{n}$ for all $n\geq 1$ .
5.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ , as $1/n\leq\ln n/n$ for all $n\geq 3$ .
6.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n!}$ , as $1/(n!+n)\leq 1/n!$ for all $n\geq 1$ .
7.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ . Since $n=\sqrt{n^{2}}>\sqrt{n^{2}-1}$ , $1/n\leq 1/\sqrt{n^{2}-1}$ for all $n\geq 2$ .
8.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ , as $1/\sqrt{n}\leq 1/(\sqrt{n}-2)$ for all $n\geq 5$ .
9.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ .
10.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{4^{n}}$ .
11.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n}$ .
12.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ .
13.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ .
14.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ . Just as $\displaystyle\lim_{n\to 0}\frac{\sin n}{n}=1$ , $\displaystyle\lim_{n\to\infty}\frac{\sin(1/n)}{1/n}=1$ .
15.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ :

$\frac{1}{n}=\frac{n^{2}}{n^{3}}<\frac{n^{2}+n+1}{n^{3}}<\frac{n^{2}+n+1}{n^{3}% -5},$

for all $n\geq 1$ .
16.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ .
17.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\left(\frac{2}{5}\right)^{n}$ , as $2^{n}/(5^{n}+10)<2^{n}/5^{n}$ for all $n\geq 1$ .
18.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ .
19.

Converges by Comparison Test with $\sum\frac{1}{n^{3}}$
20.

Converges by Comparison Test with $\sum\frac{1}{4^{n}}$
21.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ . Note that

$\frac{n}{n^{2}-1}=\frac{n^{2}}{n^{2}-1}\cdot\frac{1}{n}>\frac{1}{n},$

as $\frac{n^{2}}{n^{2}-1}>1$ , for all $n\geq 2$ .
22.

Diverges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{1/2}}$ .
23.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ , as $1/(n^{2}\ln n)\leq 1/n^{2}$ for all $n\geq 3$ .
24.

Converges; compare to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ .
25.

Converges; Integral Test
26.

Converges; Integral Test, $p$ -Series Test, Direct & Limit Comparison Tests can all be used.
27.

Diverges; the $n^{\text{th}}$ Term Test and Direct Comparison Test can be used.
28.

Converges; the Direct Comparison Test can be used with sequence $1/(n-1)!$ .
29.

Converges; the Direct Comparison Test can be used with sequence $1/3^{n}$ .
30.

Diverges; the $n^{\text{th}}$ Term Test can be used, along with the Limit Comparison Test (compare with 1/10).
31.

Diverges; the $n^{\text{th}}$ Term Test can be used, along with the Integral Test.
32.

Diverges; the Limit Comparison Test can be used with sequence $1/\sqrt{n}$ .
33.
- Converges; use Direct Comparison Test as $\frac{a_{n}}{n}<a_{n}$ .
  
  Converges; since original series converges, we know $\lim_{n\to\infty}a_{n}=0$ . Thus for large $n$ , $a_{n}a_{n+1}<a_{n}$ .
  
  Converges; similar logic to part (b) so $(a_{n})^{2}<a_{n}$ .
  
  May converge; certainly $na_{n}>a_{n}$ but that does not mean it does not converge.
  
  Does not converge, using logic from (b) and $n^{th}$ Term Test.
34.

Diverges

Exercises I.5

1.

The signs of the terms do not alternate; in the given series, some terms are negative and the others positive, but they do not necessarily alternate.
2.

positive, decreasing, 0
3.

Many examples exist; one common example is $a_{n}=(-1)^{n}/n$ .
4.

conditionally
5.
- converges
  
  converges ( $p$ -Series)
  
  absolute
6.
- converges
  
  converges (Geometric Series with $r=1/e$ )
  
  absolute
7.
- diverges (limit of terms is not 0)
  
  diverges
  
  n/a; diverges
8.
- diverges (limit of terms is not 0)
  
  diverges
  
  n/a; diverges
9.
- converges
  
  diverges (Limit Comparison Test with $1/n$ )
  
  conditional
10.
- converges
  
  diverges (Direct Comparison Test with $1/n$ )
  
  conditional
11.
- diverges (limit of terms is not 0)
  
  diverges
  
  n/a; diverges
12.
- converges
  
  converges (the sum in the denominator is $n^{2}$ )
  
  absolute
13.
- diverges (terms oscillate between $\pm 1$ )
  
  diverges
  
  n/a; diverges
14.
- converges
  
  diverges (Integral Test)
  
  conditional
15.
- converges
  
  converges (Geometric Series with $r=2/3$ )
  
  absolute
16.
- converges
  
  converges (Direct Comparison to $2^{-n}$ )
  
  absolute
17.
- converges
  
  diverges ( $p$ -Series Test with $p=1/2$ )
  
  conditional
18.
- converges
  
  converges (Integral Test)
  
  absolute
19.

$S_{5}=-1.1906$ ; $S_{6}=-0.6767$ ;

$\displaystyle-1.1906\leq\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\ln(n+1)}\leq-0.6767$
20.

$S_{4}=0.9459$ ; $S_{5}=0.9475$ ;

$\displaystyle 0.9459\leq\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{4}}\leq 0.9475$
21.

$S_{6}=0.3681$ ; $S_{7}=0.3679$ ;

$\displaystyle 0.3681\leq\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\leq 0.3679$
22.

$S_{9}=0.666016$ ; $S_{10}=0.666992$ ;

$\displaystyle 0.666016\leq\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}\leq 0% .666992$
23.

$n=5$
24.

$n=7$
25.

Using the theorem, we find $n=499$ guarantees the sum is within $0.001$ of $\pi/4$ . (Convergence is actually faster, as the sum is within $\epsilon$ of $\pi/24$ when $n\geq 249$ .)
26.

$n=5$ ( $(2n)!>10^{8}$ when $n\geq 6$ )
27.

Using 5 terms, the series in 23 gives $\pi\approx 3.142013$ . Using 499 terms, the series in 25 gives $\pi\approx 3.143597$ . The series in 23 gives the better approximation, and requires many fewer terms.
28.

Exercises I.6

1.

algebraic, or polynomial.
2.

factorial and/or exponential
3.

Integral Test, Limit Comparison Test, and Root Test
4.

raised to a power
5.

Converges
6.

Diverges
7.

Converges
8.

Converges
9.

The Ratio Test is inconclusive; the $p$ -Series Test states it diverges.
10.

The Ratio Test is inconclusive; the Direct Comparison Test with $1/n^{3}$ shows it converges.
11.

Converges
12.

Converges
13.

Converges; note the summation can be rewritten as $\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}n!}{3^{n}n!}$ , from which the Ratio Test can be applied.
14.

Converges; rewrite the summation as $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{5^{n}n!}$ then apply the Ratio Test.
15.

Diverges
16.

Converges
17.

Converges
18.

Converges
19.

Converges
20.

Converges
21.

Diverges
22.

Converges
23.

Diverges. The Root Test is inconclusive, but the $n^{\text{th}}$ -Term Test shows divergence. (The terms of the sequence approach $e^{-2}$ , not 0, as $n\to\infty$ .)
24.

Converges
25.

Converges
26.

Converges
27.

Converges

Exercises I.7

1.

Diverges
2.

Absolutely converges
3.

Diverges
4.

Absolutely converges
5.

Diverges
6.

Absolutely converges
7.

Absolutely converges
8.

Diverges
9.

Conditionally converges
10.

Absolutely converges
11.

Diverges
12.

Conditionally converges
13.

Absolutely converges
14.

Diverges
15.

Absolutely converges
16.

Absolutely converges
17.

Absolutely converges
18.

Diverges
19.

Conditionally converges
20.

Absolutely converges
21.

Absolutely converges
22.

Diverges
23.

Absolutely converges
24.

Absolutely converges
25.

Diverges
26.

Diverges
27.

Diverges
28.

Conditionally converges
29.

Absolutely converges
30.

Absolutely converges
31.

Diverges
32.

Absolutely converges
33.

Absolutely converges
34.

Conditionally converges
35.

Diverges
36.

Absolutely converges
37.

Absolutely converges
38.

Diverges

Exercises I.8

1.

1
2.

The radius of convergence is a value $R$ such that a power series, centered at $x=c$ , converges for all values of $x$ in $(c-R,c+R)$ . The interval of convergence is an interval on which the power series converges; it may differ from $(c-R,c+R)$ only at the endpoints.
3.

5
4.

5
5.

$1+2x+4x^{2}+8x^{3}+16x^{4}$
6.

$x+\frac{x^{2}}{4}+\frac{x^{3}}{9}+\frac{x^{4}}{16}+\frac{x^{5}}{25}$
7.

$1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}$
8.

$1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+\frac{x^{8}}{40320}$
9.
- $R=\infty$
  
  $(-\infty,\infty)$
10.
- $R=1$
  
  $(-1,1)$
11.
- $R=1$
  
  $(2,4]$
12.
- $R=\infty$
  
  $(-\infty,\infty)$
13.
- $R=2$
  
  $(-2,2)$
14.
- $R=10$
  
  $(-5,15)$
15.
- $R=1/5$
  
  $(4/5,6/5)$
16.
- $R=1/2$
  
  $(-1/2,1/2)$
17.
- $R=1$
  
  $(-1,1)$
18.
- $R=3$
  
  $(-3,3)$
19.
- $R=\infty$
  
  $(-\infty,\infty)$
20.
- $R=0$
  
  $x=10$
21.
- $R=1$
  
  $[-1,1]$
22.
- $R=1$
  
  $[-3,-1]$
23.
- $R=0$
  
  $x=0$
24.
- $R=4$
  
  $(-8,0)$
25.
- $R=1$
  
  $[-\frac{1}{3},\frac{5}{3})$
26.
- $R=5$
  
  $[-5,5]$
27.
- $R=\infty$
  
  $(-\infty,\infty)$
28.
- $R=\infty$
  
  $(-\infty,\infty)$
29.

$\displaystyle\sum_{n=0}^{\infty}8^{n}x^{n+1}$ , $R=1/8$
30.

$\displaystyle\sum_{n=0}^{\infty}6(-7)^{n}x^{4n}$ , $R=1/\sqrt[4]{7}$
31.

$\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+3}}{3^{n+1}}$ , $R=\sqrt{3}$
32.

$\displaystyle\sum_{n=0}^{\infty}\frac{3\cdot 2^{n}x^{n/3+2}}{5^{n+1}}$ , $R=125/8$
33.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}n^{2}x^{n-1}$ ; $(-1,1)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{n}{n+1}x^% {n+1}$ ; $(-1,1)$
34.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}x^{n-1}$ ; $(-1,1)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=1}^{\infty}\frac{1}{n(n+1)% }x^{n+1}$ ; $[-1,1]$
35.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{n}{2^{n}}x^{n-1}$ ; $(-2,2)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{1}{(n+1)2% ^{n}}x^{n+1}$ ; $[-2,2)$
36.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}n(-3)^{n}x^{n-1}$ ; $(-1/3,1/3)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{(-3)^{n}}% {n+1}x^{n+1}$ ; $(-1/3,1/3]$
37.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n-1% )!}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}$ ; $(-\infty,\infty)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x% ^{2n+1}}{(2n+1)!}$ ; $(-\infty,\infty)$
38.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n-1}}{(n-1)!% }=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{n}}{n!}$ ; $(-\infty,\infty)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x% ^{n+1}}{(n+1)!}$ ; $(-\infty,\infty)$
39.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}n^{2}x^{n-1}$ ; $(-1,1)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{n}{n+1}x^% {n+1}$ ; $(-1,1)$
40.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}x^{n-1}$ ; $(-1,1)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=1}^{\infty}\frac{1}{n(n+1)% }x^{n+1}$ ; $[-1,1]$
41.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{n}{2^{n}}x^{n-1}$ ; $(-2,2)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{1}{(n+1)2% ^{n}}x^{n+1}$ ; $[-2,2)$
42.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}n(-3)^{n}x^{n-1}$ ; $(-1/3,1/3)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{(-3)^{n}}% {n+1}x^{n+1}$ ; $(-1/3,1/3]$
43.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n-1% )!}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}$ ; $(-\infty,\infty)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x% ^{2n+1}}{(2n+1)!}$ ; $(-\infty,\infty)$
44.
- $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n-1}}{(n-1)!% }=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{n}}{n!}$ ; $(-\infty,\infty)$
  
  $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x% ^{n+1}}{(n+1)!}$ ; $(-\infty,\infty)$
45.
- $\sum_{n=0}^{\infty}x^{n}(-1)^{n}(1+n)$ ; $R=1$
  
  $\sum_{n=1}^{\infty}x^{n-1}(-1)^{n-1}(1+n)n/2\\ =\sum_{n=0}^{\infty}x^{n}(-1)^{n}(2+n)(1+n)/2$
  
  $\sum_{n=1}^{\infty}x^{n+1}(-1)^{n-1}(1+n)n/2\\ =\sum_{n=0}^{\infty}x^{n+2}(-1)^{n}(2+n)(1+n)/2\\ =\sum_{n=2}^{\infty}x^{n}(-1)^{n}n(-1+n)/2$
46.
- Converges
  
  Diverges
  
  Converges
  
  Diverges
47.

$\displaystyle\ln 3-\sum_{n=1}^{\infty}3^{-n}x^{n}/n$ , $R=3$
48.

$\displaystyle\sum_{n=1}^{\infty}n(-9)^{n-1}x^{n}$ , $R=1/9$
49.

$\displaystyle\sum_{n=0}^{\infty}\frac{2}{2n+1}x^{2n+1}$ , $R=1$
50.

$\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}$ ; $R=1$
51.

$\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{6n+5}}{2n+1}$ ; $R=1$
52.

$\displaystyle\sum_{n=0}^{\infty}(2n+1)x^{n}$ ; $R=1$
53.

$\displaystyle\sum_{n=0}^{\infty}\frac{(n+2)(n+1)}{2^{n+4}}x^{n+3}$ ; $R=2$

Exercises I.9

1.

The Maclaurin polynomial is a special case of Taylor polynomials. Taylor polynomials are centered at a specific $x$ -value; when that $x$ -value is 0, it is a Maclaurin polynomial.
2.

T
3.

$p_{2}(x)=6+3x-4x^{2}$ .
4.

$f\,^{\prime\prime\prime}(0)=30$
5.

$p_{3}(x)=1-x+\frac{1}{2}x^{3}-\frac{1}{6}x^{3}$
6.

$p_{8}(x)=x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}-\frac{1}{5040}x^{7}$
7.

$p_{8}(x)=x+x^{2}+\frac{1}{2}x^{3}+\frac{1}{6}x^{4}+\frac{1}{24}x^{5}$
8.

$p_{6}(x)=\frac{2x^{5}}{15}+\frac{x^{3}}{3}+x$
9.

$p_{4}(x)=\frac{2x^{4}}{3}+\frac{4x^{3}}{3}+2x^{2}+2x+1$
10.

$p_{4}(x)=x^{4}+x^{3}+x^{2}+x+1$
11.

$p_{4}(x)=x^{4}-x^{3}+x^{2}-x+1$
12.

$p_{7}(x)=-\frac{x^{7}}{7}+\frac{x^{5}}{5}-\frac{x^{3}}{3}+x$
13.

$p_{4}(x)=1+\frac{1}{2}(-1+x)-\frac{1}{8}(-1+x)^{2}+\frac{1}{16}(-1+x)^{3}-% \frac{5}{128}(-1+x)^{4}$
14.

$p_{4}(x)=\ln(2)+\frac{1}{2}(-1+x)-\frac{1}{8}(-1+x)^{2}+\frac{1}{24}(-1+x)^{3}% -\frac{1}{64}(-1+x)^{4}$
15.

$p_{6}(x)=\frac{1}{\sqrt{2}}-\frac{-\frac{\pi}{4}+x}{\sqrt{2}}-\frac{\left(-% \frac{\pi}{4}+x\right)^{2}}{2\sqrt{2}}+\frac{\left(-\frac{\pi}{4}+x\right)^{3}% }{6\sqrt{2}}+\frac{\left(-\frac{\pi}{4}+x\right)^{4}}{24\sqrt{2}}-\frac{\left(% -\frac{\pi}{4}+x\right)^{5}}{120\sqrt{2}}-\frac{\left(-\frac{\pi}{4}+x\right)^% {6}}{720\sqrt{2}}$
16.

$p_{5}(x)=\frac{1}{2}+\frac{1}{2}\sqrt{3}\left(-\frac{\pi}{6}+x\right)-\frac{1}% {4}\left(-\frac{\pi}{6}+x\right)^{2}-\frac{\left(-\frac{\pi}{6}+x\right)^{3}}{% 4\sqrt{3}}+\frac{1}{48}\left(-\frac{\pi}{6}+x\right)^{4}+\frac{\left(-\frac{% \pi}{6}+x\right)^{5}}{80\sqrt{3}}$
17.

$p_{5}(x)=\frac{1}{2}-\frac{x-2}{4}+\frac{1}{8}(x-2)^{2}-\frac{1}{16}(x-2)^{3}+% \frac{1}{32}(x-2)^{4}-\frac{1}{64}(x-2)^{5}$
18.

$p_{8}(x)=1-2(-1+x)+3(-1+x)^{2}-4(-1+x)^{3}+5(-1+x)^{4}-6(-1+x)^{5}+7(-1+x)^{6}% -8(-1+x)^{7}+9(-1+x)^{8}$
19.

$p_{3}(x)=\frac{1}{2}+\frac{1+x}{2}+\frac{1}{4}(1+x)^{2}$
20.

$p_{2}(x)=-\pi^{2}-2\pi(x-\pi)+\frac{1}{2}\left(\pi^{2}-2\right)(x-\pi)^{2}$
21.

$p_{3}(x)=x-\frac{x^{3}}{6}$ ; $p_{3}(0.1)=0.09983$ . Error is bounded by $\frac{1}{4!}\cdot 0.1^{4}\approx 0.000004167$ .
22.

$p_{4}(x)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}$ ; $p_{4}(1)=13/24\approx 0.54167$ . Error is bounded by $\frac{1}{5!}\cdot 1^{5}\approx 0.00833$ .
23.

$p_{2}(x)=3+\frac{1}{6}(-9+x)-\frac{1}{216}(-9+x)^{2}$ ; $p_{2}(10)=3.16204$ . The third derivative of $f(x)=\sqrt{x}$ is bounded on $[9,10]$ by $0.0015$ . Error is bounded by $\frac{0.0015}{3!}\cdot 1^{3}=0.0003$ .
24.

$p_{3}(x)=-1+x-\frac{1}{2}(-1+x)^{2}+\frac{1}{3}(-1+x)^{3}$ ; $p_{3}(1.5)=0.41667$ . The absolute value of the fourth derivative of $f(x)=\ln x$ is bounded on $[1,1.5]$ by $6$ . Error is bounded by $\frac{6}{4!}\cdot.5^{4}=0.016$ .
25.

The $n^{\text{th}}$ derivative of $f(x)=e^{x}$ is bounded by $e$ on $[0,1]$ . Thus $\left\lvert R_{n}(1)\right\rvert\leq\frac{e}{(n+1)!}1^{(n+1)}$ . When $n=7$ , this is less than $0.0001$ .
26.

The $n^{\text{th}}$ derivative of $f(x)=\sqrt{x}$ has a maximum on $[3,4]$ of $(2n-3)!!(-1)^{n+1}3^{1/2}6^{-n}$ . Thus $\left\lvert R_{n}(3)\right\rvert\leq\frac{3^{1/2}2^{-2}}{3^{n}n(n+1)}$ . When $n=5$ , this is less than $0.0001$ .
27.

The $n^{\text{th}}$ derivative of $f(x)=\cos x$ is bounded by $1$ on intervals containing $0$ and $\pi/3$ . Thus $\left\lvert R_{n}(\pi/3)\right\rvert\leq\frac{1}{(n+1)!}(\pi/3)^{(n+1)}$ . When $n=7$ , this is less than $0.0001$ . Since the Maclaurin polynomial of $\cos x$ only uses even powers, we can actually just use $n=6$ .
28.

The $n^{\text{th}}$ derivative of $f(x)=\sin x$ is bounded by $1$ on intervals containing $0$ and $\pi$ . Thus $\left\lvert R_{n}(\pi)\right\rvert\leq\frac{1}{(n+1)!}(\pi)^{(n+1)}$ . When $n=12$ , this is less than $0.0001$ . Since the Maclaurin polynomial of $\sin x$ only uses odd powers, we can actually just use $n=11$ .
29.

$\frac{1}{n!}x^{n}$
30.

When $n$ is even, $\frac{(-1)^{n/2}}{n!}x^{n}$ ; when $n$ is odd, $0$ .
31.

When $n$ even, 0; when $n$ is odd, $\frac{(-1)^{(n-1)/2}}{n!}x^{n}$ .
32.

$x^{n}$
33.

$(-1)^{n}x^{n}$
34.

$(-1)^{n+1}\frac{(x-1)^{n}}{n}$
35.

$\displaystyle 1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{24}x^{4}$
36.

$\displaystyle 1+2x-2x^{2}+4x^{3}-10x^{4}$

Exercises I.10

1.

A Taylor polynomial is a polynomial, containing a finite number of terms. A Taylor series is a series, the summation of an infinite number of terms.
2.

Theorem 9.10.1, entitled “Function and Taylor Series Equality”
3.

All derivatives of $e^{x}$ are $e^{x}$ which evaluate to 1 at $x=0$ .

The Taylor series starts $1+x+\frac{1}{2}x^{2}+\frac{1}{3!}x^{3}+\frac{1}{4!}x^{4}+\dotsb$ ;

the Taylor series is $\displaystyle\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$
4.

All derivatives of $\sin x$ are either $\pm\cos x$ or $\pm\sin x$ , which evaluate to $\pm 1$ or $0$ at $x=0$ . The Taylor series starts $0+x+0x^{2}-\frac{1}{6}x^{3}+0x^{4}+\frac{1}{120}x^{5}$ ;

the Taylor series is $\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$
5.

The $n^{\text{th}}$ derivative of $1/(1-x)$ is $f\,^{(n)}(x)=(n)!/(1-x)^{n+1}$ , which evaluates to $n!$ at $x=0$ .

The Taylor series starts $1+x+x^{2}+x^{3}+\dotsb$ ;

the Taylor series is $\displaystyle\sum_{n=0}^{\infty}x^{n}$
6.

The derivative of $\tan^{-1}x$ is $1/(1+x^{2})$ . Taking successive derivatives using the Quotient Rule, the derivatives of $\tan^{-1}x$ fall into two categories in terms of their evaluation at $x=0$ .

When $n$ is even, $\displaystyle f\,^{(n)}(x)=(-1)^{(n-1)/2}\frac{p(x)}{(1+x^{2})^{n}}$ , where $p(x)$ is a polynomial such that $p(0)=0$ . Hence $f\,^{(n)}(0)=0$ when $n$ is even.

When $n$ is odd, $\displaystyle f\,^{(n)}(x)=(-1)^{(n-1)/2}\frac{p(x)}{(1+x^{2})^{n}}$ , where $p(x)$ is a polynomial such that $p(0)=(n-1)!$ . Hence $f\,^{(n)}(0)=(-1)^{(n-1)/2}(n-1)!$ when $n$ is odd. (The unusual power of $(-1)$ is such that every other odd term is negative.)

The Taylor series starts $x-\frac{1}{3}x^{3}+\frac{1}{5}x^{5}+\dotsb$ ; by reindexing to only obtain odd powers of $x$ , we get

the Taylor series is $\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}$ .
7.

The Taylor series starts $0-(x-\pi/2)+0x^{2}+\frac{1}{6}(x-\pi/2)^{3}+0x^{4}-\frac{1}{120}(x-\pi/2)^{5}$ ;

the Taylor series is $\displaystyle\sum_{n=0}^{\infty}(-1)^{n+1}\frac{(x-\pi/2)^{2n+1}}{(2n+1)!}$
8.

The Taylor series starts $1-(x-1)+(x-1)^{2}-(x-1)^{3}+(x-1)^{4}-(x-1)^{5}$ ;

the Taylor series is $\displaystyle\sum_{n=0}^{\infty}(-1)^{n}(x-1)^{n}$
9.

$f\,^{(n)}(x)=(-1)^{n}e^{-x}$ ; at $x=0$ , $f\,^{(n)}(0)=-1$ when $n$ is odd and $f\,^{(n)}(0)=1$ when $n$ is even.

The Taylor series starts $1-x+\frac{1}{2}x^{2}-\frac{1}{3!}x^{3}+\dotsb$ ;

the Taylor series is $\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n}}{n!}$ .
10.

$f\,^{(n)}(x)=(-1)^{n+1}\frac{(n-1)!}{(1+x)^{n}}$ ; at $x=0$ , $f\,^{(n)}(0)=(-1)^{n+1}(n-1)!$

The Taylor series starts $x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\dotsb$ ;

the Taylor series is $\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n}}{n}$ .
11.

$f\,^{(n)}(x)=(-1)^{n+1}\frac{n!}{(x+1)^{n+1}}$ ; at $x=1$ , $f\,^{(n)}(1)=(-1)^{n+1}\frac{n!}{2^{n+1}}$

The Taylor series starts $\frac{1}{2}+\frac{1}{4}(x-1)-\frac{1}{8}(x-1)^{2}+\frac{1}{16}(x-1)^{3}\dotsb$ ;

the Taylor series is $\displaystyle\frac{1}{2}+\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^{n}}{2^{n+1}}$ .
12.

The derivatives of $\sin x$ are $\pm\cos x$ and $\pm\sin x$ ; at $x=\pi/4$ , these derivatives evaluate to $\pm\sqrt{2}/2$ .

The Taylor series starts $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}(x-\pi/4)-\frac{\sqrt{2}}{2}\frac{(x-\pi/% 4)^{2}}{2}-\frac{\sqrt{2}}{2}\frac{(x-\pi/4)^{3}}{3!}+\frac{\sqrt{2}}{2}\frac{% (x-\pi/4)^{4}}{4!}+\frac{\sqrt{2}}{2}\frac{(x-\pi/4)^{5}}{5!}\dotsb$ . Note how the signs are “ $++--++\dots$ .” We saw signs like these in Example 9.1.1; one way of producing such signs is to raise $(-1)$ to a special quadratic power. While many possibilities exist, one such quadratic is $(n+3)(n+4)/2$ .

Thus the Taylor series is $\displaystyle\sum_{n=0}^{\infty}(-1)^{\frac{(n+3)(n+4)}{2}}\frac{\sqrt{2}}{2}% \frac{(x-\pi/4)^{n}}{n!}$ .
13.

Given a value $x$ , the magnitude of the error term $R_{n}(x)$ is bounded by

$\left\lvert R_{n}(x)\right\rvert\leq\frac{\max\left\lvert f\,^{(n+1)}(z)\right% \rvert}{(n+1)!}\left\lvert x^{(n+1)}\right\rvert,$

where $z$ is between $0$ and $x$ .

If $x>0$ , then $z<x$ and $f\,^{(n+1)}(z)=e^{z}<e^{x}$ . If $x<0$ , then $x<z<0$ and $f\,^{(n+1)}(z)=e^{z}<1$ . So given a fixed $x$ value, let $M=\max\{e^{x},1\}$ ; $f\,^{(n)}(z)<M.$ This allows us to state

$\left\lvert R_{n}(x)\right\rvert\leq\frac{M}{(n+1)!}\left\lvert x^{(n+1)}% \right\rvert.$

For any $x$ , $\displaystyle\lim_{n\to\infty}\frac{M}{(n+1)!}\left\lvert x^{(n+1)}\right% \rvert=0$ . Thus by the Squeeze Theorem, we conclude that $\displaystyle\lim_{n\to\infty}R_{n}(x)=0$ for all $x$ , and hence

$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\quad\text{for all $x$}.$
14.

The following argument is essentially the same as that given for $f(x)=\cos x$ in Example 9.10.3.

Given a value $x$ , the magnitude of the error term $R_{n}(x)$ is bounded by

$\left\lvert R_{n}(x)\right\rvert\leq\frac{\max\left\lvert f\,^{(n+1)}(z)\right% \rvert}{(n+1)!}\left\lvert x^{(n+1)}\right\rvert.$

Since all derivatives of $\sin x$ are $\pm\cos x$ or $\pm\sin x$ , whose magnitudes are bounded by $1$ , we can state

$\left\lvert R_{n}(x)\right\rvert\leq\frac{1}{(n+1)!}\left\lvert x^{(n+1)}% \right\rvert.$

For any $x$ , $\displaystyle\lim_{n\to\infty}\frac{x^{n+1}}{(n+1)!}=0$ . Thus by the Squeeze Theorem, we conclude that $\displaystyle\lim_{n\to\infty}R_{n}(x)=0$ for all $x$ , and hence

$\sin x=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\quad\text{for all $% x$}.$
15.

Given a value $x$ , the magnitude of the error term $R_{n}(x)$ is bounded by

$\left\lvert R_{n}(x)\right\rvert\leq\frac{\max\left\lvert f\,^{(n+1)}(z)\right% \rvert}{(n+1)!}\left\lvert x^{(n+1)}\right\rvert,$

where $z$ is between $0$ and $x$ . Since $\left\lvert f\,^{(n+1)}(z)\right\rvert=\frac{n!}{(z+1)^{n+1}}$ ,

$\left\lvert R_{n}(x)\right\rvert\leq\frac{1}{n+1}\left(\frac{\left\lvert x% \right\rvert}{\min z+1}\right)^{n+1}.$

If $0<x<1$ , then $0<z<x$ and $f\,^{(n+1)}(z)=\frac{n!}{(z+1)^{n+1}}<n!$ . Thus

$\left\lvert R_{n}(x)\right\rvert\leq\frac{n!}{(n+1)!}\left\lvert x^{(n+1)}% \right\rvert=\frac{x^{n+1}}{n+1}.$

For a fixed $x<1$ ,

$\lim_{n\to\infty}\frac{x^{n+1}}{n+1}=0.$
16.

Given a value $x$ , the magnitude of the error term $R_{n}(x)$ is bounded by

$\left\lvert R_{n}(x)\right\rvert\leq\frac{\max\left\lvert f\,^{(n+1)}(z)\right% \rvert}{(n+1)!}\left\lvert x^{(n+1)}\right\rvert,$

where $z$ is between $0$ and $x$ .

Note that $\left\lvert f\,^{(n+1)}(x)\right\rvert=\frac{(n+1)!}{(1-x)^{n+2}}$ .

If $-1<x<0$ , then $x<z<0$ and $f\,^{(n+1)}(z)=\frac{(n+1)!}{(1-z)^{n+2}}<(n+1)!$ . Thus

$\left\lvert R_{n}(x)\right\rvert\leq\frac{(n+1)!}{(n+1)!}\left\lvert x^{n+1}% \right\rvert=\left\lvert x^{n+1}\right\rvert.$

For a fixed $x$ ,

$\lim_{n\to\infty}\left\lvert x^{n+1}\right\rvert=0\text{ as }\left\lvert x% \right\rvert<1.$
17.

Given $\displaystyle\cos x=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}$ ,

$\displaystyle\cos(-x)=\sum_{n=0}^{\infty}(-1)^{n}\frac{(-x)^{2n}}{(2n)!}=\sum_% {n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}=\cos x$ , as all powers in the series are even.
18.

Given $\displaystyle\sin x=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$ ,

$\displaystyle\sin(-x)=\sum_{n=0}^{\infty}(-1)^{n}\frac{(-x)^{2n+1}}{(2n+1)!}=% \sum_{n=0}^{\infty}(-1)^{n}\frac{-x^{2n+1}}{(2n+1)!}=-\sin x$ , as all powers in the series are odd.
19.

Given $\displaystyle\sin x=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$ ,

$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\big(\sin x\big)=% \frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\sum_{n=0}^{\infty}(-1)^{n% }\frac{x^{2n+1}}{(2n+1)!}\right)=\sum_{n=0}^{\infty}(-1)^{n}\frac{(2n+1)x^{2n}% }{(2n+1)!}=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}=\cos x$ . (The summation still starts at $n=0$ as there was no constant term in the expansion of $\sin x$ ).
20.

Given $\displaystyle\cos x=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}$ ,

$\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\big(\cos x\big)=% \frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\sum_{n=0}^{\infty}(-1)^{n% }\frac{x^{2n}}{(2n)!}\right)=\sum_{n=1}^{\infty}(-1)^{n}\frac{(2n)x^{2n-1}}{(2% n)!}=\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n-1}}{(2n-1)!}$ . We can re-index this summation to start at $n=0$ by replacing $n$ with $n+1$ in the summation:

$\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n-1}}{(2n-1)!}=\sum_{n=0}^{\infty}(-1)^{n% +1}\frac{x^{2n+1}}{(2n+1)!}.$

Note that this series has the opposite sign of the Taylor series for $\sin x$ ; thus $\frac{\operatorname{d}\!}{\operatorname{d}\!x}(\cos x)=-\sin x$ .
21.

$\displaystyle 1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\frac{5x^{4}}{128}$
22.

$\displaystyle 1-\frac{x}{2}+\frac{3x^{2}}{8}-\frac{5x^{3}}{16}+\frac{35x^{4}}{% 128}$
23.

$\displaystyle 1+\frac{x}{3}-\frac{x^{2}}{9}+\frac{5x^{3}}{81}-\frac{10x^{4}}{2% 43}$
24.

$\displaystyle 1+4x+6x^{2}+4x^{3}+x^{4}$ (note the series is finite, and the formula still applies)
25.

$\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{(x^{2})^{2n}}{(2n)!}=\sum_{n=0}^% {\infty}(-1)^{n}\frac{x^{4n}}{(2n)!}.$
26.

$\displaystyle\sum_{n=0}^{\infty}\frac{(-x)^{n}}{n!}.$
27.

$\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{(2x+3)^{2n+1}}{(2n+1)!}.$
28.

$\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{(x/2)^{2n+1}}{(2n+1)}.$
29.

$\displaystyle x+x^{2}+\frac{x^{3}}{3}-\frac{x^{5}}{30}$
30.

$\displaystyle 1+\frac{x}{2}-\frac{5x^{2}}{8}-\frac{3x^{3}}{16}$
31.

$\displaystyle\int_{0}^{\sqrt{\pi}}\sin\big(x^{2}\big)\operatorname{d}\!x% \approx\int_{0}^{\sqrt{\pi}}\left(x^{2}-\frac{x^{6}}{6}+\frac{x^{10}}{120}-% \frac{x^{14}}{5040}\right)\operatorname{d}\!x=0.8877$
32.

$\displaystyle\int_{0}^{\pi^{2}/4}\cos\big(\sqrt{x}\big)\operatorname{d}\!x% \approx\int_{0}^{\pi^{2}/4}\left(1-\frac{x}{2}+\frac{x^{2}}{24}-\frac{x^{3}}{7% 20}\right)\operatorname{d}\!x=1.1412$ . (Actual answer: $\pi-2$ )