Chapter 9

Exercises 9.1

1. 1.

2. 3.

3. 5.

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

4. 7.

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

5. 9.

$a_{n}=3n+1$

6. 11.

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

7. 13.

$1/7$

8. 15.

$0$

9. 17.

diverges

10. 19.

converges to $0$

11. 21.

converges to $0$

12. 23.

diverges

13. 25.

converges to $e$

14. 27.

converges to 5

15. 29.

diverges

16. 31.

converges to 0

17. 33.

converges to 0

18. 35.

converges to $\ln 2$

19. 37.

converges to $0$

20. 39.

bounded

21. 41.

bounded below

22. 43.

monotonically increasing

23. 45.

never monotonic

24. 47.

never monotonic

25. 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 $\displaystyle\lim_{n\to\infty}a_{n}=0$.

26. 51.

27. 53.
(d) 2

Exercises 9.2

1. 1.

2. 3.

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

3. 5.

F

4. 7.
(a) $-1,-\frac{1}{2},-\frac{5}{6},-\frac{7}{12},-\frac{47}{60}$ (b) Plot omitted
5. 9.
(a) $-1,0,-1,0,-1$ (b) Plot omitted
6. 11.
(a) $1,\frac{3}{2},\frac{5}{3},\frac{41}{24},\frac{103}{60}$ (b) Plot omitted
7. 13.
(a) $-0.9,-0.09,-0.819,-0.1629,-0.75339$ (b) Plot omitted
8. 15.

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

9. 17.

Diverges by Theorem 9.2.4

10. 19.

Diverges

11. 21.

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

12. 23.

Diverges

13. 25.

Diverges

14. 27.

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

15. 29.

Converges

16. 31.

Converges

17. 33.
(a) $S_{n}=\frac{1-(1/4)^{n}}{3/4}$ (b) Converges to $4/3$.
18. 35.
(a) $S_{n}=\begin{cases}-\frac{n+1}{2}&n\text{ is odd }\\ \frac{n}{2}&n\text{ is even}\end{cases}$ (b) Diverges
19. 37.
(a) $S_{n}=\frac{1-(1/e)^{n+1}}{1-1/e}$. (b) Converges to $1/(1-1/e)=e/(e-1)$.
20. 39.
(a) With partial fractions, $a_{n}=\frac{1}{n}-\frac{1}{n+1}$. Thus $S_{n}=1-\frac{1}{n+1}$. (b) Converges to 1.
21. 41.
(a) 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 $S_{n}=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}$. (b) Converges to $1/2$.
22. 43.
(a) $S_{n}=1-\frac{1}{(n+1)^{2}}$ (b) Converges to 1.
23. 45.
(a) $a_{n}=1/2^{n}+1/3^{n}$ for $n\geq 0$. Thus $S_{n}=\frac{1-1/2^{2}}{1/2}+\frac{1-1/3^{n}}{2/3}$. (b) Converges to $2+3/2=7/2$.
24. 47.
(a) $S_{n}=\frac{1-(\sin 1)^{n+1}}{1-\sin 1}$ (b) Converges to $\frac{1}{1-\sin 1}$.
25. 49.

$(-3,3)$

26. 51.

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

27. 53.

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

28. 55.

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

29. 57.

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

30. 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.

31. 61.

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

Exercises 9.3

1. 1.

continuous, positive and decreasing

2. 3.

Converges

3. 5.

Diverges

4. 7.

Converges

5. 9.

Converges

6. 11.

$p>1$

7. 13.

$p>1$

8. 15.

Exercises 9.4

1. 1.

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

2. 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 $n>1$.

3. 5.

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

4. 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 $n\geq 2$.

5. 9.

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

6. 11.

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

7. 13.

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

8. 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 $n\geq 1$.
9. 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 $n\geq 1$.

10. 19.

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

11. 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 $n\geq 2$.
12. 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 $n\geq 3$.

13. 25.

Converges; Integral Test

14. 27.

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

15. 29.

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

16. 31.

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

17. 33.
(a) Converges; use Direct Comparison Test as $\frac{a_{n}}{n}. (b) Converges; since original series converges, we know $\lim_{n\to\infty}a_{n}=0$. Thus for large $n$, $a_{n}a_{n+1}. (c) Converges; similar logic to part (b) so $(a_{n})^{2}. (d) May converge; certainly $na_{n}>a_{n}$ but that does not mean it does not converge. (e) Does not converge, using logic from (b) and $n^{th}$ Term Test.

Exercises 9.5

1. 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. 3.

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

3. 5.
(a) converges (b) converges ($p$-Series) (c) absolute
4. 7.
(a) diverges (limit of terms is not 0) (b) diverges (c) n/a; diverges
5. 9.
(a) converges (b) diverges (Limit Comparison Test with $1/n$) (c) conditional
6. 11.
(a) diverges (limit of terms is not 0) (b) diverges (c) n/a; diverges
7. 13.
(a) diverges (terms oscillate between $\pm 1$) (b) diverges (c) n/a; diverges
8. 15.
(a) converges (b) converges (Geometric Series with $r=2/3$) (c) absolute
9. 17.
(a) converges (b) diverges ($p$-Series Test with $p=1/2$) (c) conditional
10. 19.

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

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

11. 21.

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

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

12. 23.

$n=5$

13. 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 $n\geq 249$.)

14. 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.

Exercises 9.6

1. 1.

algebraic, or polynomial.

2. 3.

Integral Test, Limit Comparison Test, and Root Test

3. 5.

Converges

4. 7.

Converges

5. 9.

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

6. 11.

Converges

7. 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.

8. 15.

Diverges

9. 17.

Converges

10. 19.

Converges

11. 21.

Diverges

12. 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 $n\to\infty$.)

13. 25.

Converges

14. 27.

Converges

Exercises 9.7

1. 1.

Diverges

2. 3.

Diverges

3. 5.

Diverges

4. 7.

Absolutely converges

5. 9.

Conditionally converges

6. 11.

Diverges

7. 13.

Absolutely converges

8. 15.

Absolutely converges

9. 17.

Absolutely converges

10. 19.

Conditionally converges

11. 21.

Absolutely converges

12. 23.

Absolutely converges

13. 25.

Diverges

14. 27.

Diverges

15. 29.

Absolutely converges

16. 31.

Diverges

17. 33.

Absolutely converges

18. 35.

Diverges

19. 37.

Absolutely converges

Exercises 9.8

1. 1.

1

2. 3.

5

3. 5.

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

4. 7.

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

5. 9.
(a) $R=\infty$ (b) $(-\infty,\infty)$
6. 11.
(a) $R=1$ (b) $(2,4]$
7. 13.
(a) $R=2$ (b) $(-2,2)$
8. 15.
(a) $R=1/5$ (b) $(4/5,6/5)$
9. 17.
(a) $R=1$ (b) $(-1,1)$
10. 19.
(a) $R=\infty$ (b) $(-\infty,\infty)$
11. 21.
(a) $R=1$ (b) $[-1,1]$
12. 23.
(a) $R=0$ (b) $x=0$
13. 25.
(a) $R=1$ (b) $[-\frac{1}{3},\frac{5}{3})$
14. 27.
(a) $R=\infty$ (b) $(-\infty,\infty)$
15. 29.

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

16. 31.

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

17. 33.
(a) $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}n^{2}x^{n-1}$;  $(-1,1)$ (b) $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{n}{n+1}x^% {n+1}$;  $(-1,1)$
18. 35.
(a) $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{n}{2^{n}}x^{n-1}$;  $(-2,2)$ (b) $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{1}{(n+1)2% ^{n}}x^{n+1}$;  $[-2,2)$
19. 37.
(a) $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n-1% )!}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}$;  $(-\infty,\infty)$ (b) $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x% ^{2n+1}}{(2n+1)!}$;  $(-\infty,\infty)$
20. 39.
(a) $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}n^{2}x^{n-1}$;  $(-1,1)$ (b) $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{n}{n+1}x^% {n+1}$;  $(-1,1)$
21. 41.
(a) $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{n}{2^{n}}x^{n-1}$;  $(-2,2)$ (b) $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{1}{(n+1)2% ^{n}}x^{n+1}$;  $[-2,2)$
22. 43.
(a) $\displaystyle f\,^{\prime}(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n-1% )!}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}$;  $(-\infty,\infty)$ (b) $\displaystyle\int f(x)\operatorname{d}\!x=C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x% ^{2n+1}}{(2n+1)!}$;  $(-\infty,\infty)$
23. 45.
(a) $\sum_{n=0}^{\infty}x^{n}(-1)^{n}(1+n)$; $R=1$ (b) $\sum_{n=1}^{\infty}x^{n-1}(-1)^{n-1}(1+n)n/2\lx@parboxnewline=\sum_{n=0}^{% \infty}x^{n}(-1)^{n}(2+n)(1+n)/2$ (c) $\sum_{n=1}^{\infty}x^{n+1}(-1)^{n-1}(1+n)n/2\lx@parboxnewline=\sum_{n=0}^{% \infty}x^{n+2}(-1)^{n}(2+n)(1+n)/2\lx@parboxnewline=\sum_{n=2}^{\infty}x^{n}(-% 1)^{n}n(-1+n)/2$
24. 47.

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

25. 49.

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

26. 51.

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

27. 53.

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

Exercises 9.9

1. 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. 3.

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

3. 5.

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

4. 7.

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

5. 9.

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

6. 11.

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

7. 13.

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

8. 15.

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

9. 17.

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

10. 19.

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

11. 21.

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

12. 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 $\frac{0.0015}{3!}\cdot 1^{3}=0.0003$.

13. 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 $0.0001$.

14. 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 $n=6$.

15. 29.

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

16. 31.

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

17. 33.

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

18. 35.

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

Exercises 9.10

1. 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. 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 $\displaystyle\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$
3. 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 $\displaystyle\sum_{n=0}^{\infty}x^{n}$
4. 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 $\displaystyle\sum_{n=0}^{\infty}(-1)^{n+1}\frac{(x-\pi/2)^{2n+1}}{(2n+1)!}$
5. 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 $\displaystyle\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n}}{n!}$.
6. 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 $\displaystyle\frac{1}{2}+\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^{n}}{2^{n+1}}$.
7. 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 and $f\,^{(n+1)}(z)=e^{z}. If $x<0$, then $x and $f\,^{(n+1)}(z)=e^{z}<1$. So given a fixed $x$ value, let $M=\max\{e^{x},1\}$; $f\,^{(n)}(z) 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

 $e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\quad\text{for all x}.$
8. 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, then $0 and $f\,^{(n+1)}(z)=\frac{n!}{(z+1)^{n+1}}. 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 $x<1$,

 $\lim_{n\to\infty}\frac{x^{n+1}}{n+1}=0.$
9. 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.
10. 19.
Given $\displaystyle\sin x=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$, $\displaystyle\frac{\operatorname{d}\!}{\operatorname{d}\!x}\big{(}\sin x\big{)% }=\frac{\operatorname{d}\!}{\operatorname{d}\!x}\left(\sum_{n=0}^{\infty}(-1)^% {n}\frac{x^{2n+1}}{(2n+1)!}\right)=\sum_{n=0}^{\infty}(-1)^{n}\frac{(2n+1)x^{2% n}}{(2n+1)!}=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}=\cos x$. (The summation still starts at $n=0$ as there was no constant term in the expansion of $\sin x$).
11. 21.

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

12. 23.

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

13. 25.

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

14. 27.

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

15. 29.

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

16. 31.

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