Chapter 9

Answers will vary.

Answers will vary.

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

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

$a_n=3n+1$

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

$1/7$

$0$

diverges

converges to $0$

converges to $0$

diverges

converges to $e$

converges to 5

diverges

converges to 0

converges to 0

converges to $\ln 2$

converges to $0$

bounded

bounded below

monotonically increasing

never monotonic

never monotonic

Let $\left\{a_n\right\}$ be given such that $\underset{n\to \mathrm{\infty }}{lim}\left|a_n\right|=0$. By the definition of the limit of a sequence, given any $ϵ>0$, there is a $m$ such that for all . Since $\left|\left|a_n\right|-0\right|=\left|a_n-0\right|$, this directly implies that for all $n>m$, , meaning that $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$.

Left to reader

(d) 2

Answers will vary.

One sequence is the sequence of terms $\left\{a_n\right\}$. The other is the sequence of $n^{\text{th}}$ partial sums, $\left\{S_n\right\}=\left\{{\sum }_{i=1}^n{a}_i\right\}$.

F

(a) $-1,-\frac{1}{2},-\frac{5}{6},-\frac{7}{12},-\frac{47}{60}$ (b) Plot omitted

(a) $-1,0,-1,0,-1$ (b) Plot omitted

(a) $1,\frac{3}{2},\frac{5}{3},\frac{41}{24},\frac{103}{60}$ (b) Plot omitted

(a) $-0.9,-0.09,-0.819,-0.1629,-0.75339$ (b) Plot omitted

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

Diverges by Theorem 9.2.4

Diverges

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=1$; by Theorem 9.2.4 the series diverges.

Diverges

Diverges

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=e$; by Theorem 9.2.4 the series diverges.

Converges

Converges

(a) $S_n=\frac{1-{\left(1/4\right)}^n}{3/4}$ (b) Converges to $4/3$.

(a) (b) Diverges

(a) $S_n=\frac{1-{\left(1/e\right)}^{n+1}}{1-1/e}$. (b) Converges to $1/\left(1-1/e\right)=e/\left(e-1\right)$.

(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.

(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 $1/2$.

(a) $S_n=1-\frac{1}{{\left(n+1\right)}^2}$ (b) Converges to 1.

(a) $a_n=1/2^n+1/3^n$ for $n\ge 0$. Thus $S_n=\frac{1-1/2^2}{1/2}+\frac{1-1/3^n}{2/3}$. (b) Converges to $2+3/2=7/2$.

(a) $S_n=\frac{1-{\left(\sin 1\right)}^{n+1}}{1-\sin 1}$ (b) Converges to $\frac{1}{1-\sin 1}$.

$(-3,3)$

$(-\mathrm{\infty },-4)\cup (4,\mathrm{\infty })$

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=3$; by Theorem 9.2.4 the series diverges.

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\mathrm{\infty }$; by Theorem 9.2.4 the series diverges.

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=1/2$; by Theorem 9.2.4 the series diverges.

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.

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

continuous, positive and decreasing

Converges

Diverges

Converges

Converges

$p>1$

$p>1$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{b}_n$ converges; we cannot conclude anything about ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{c}_n$

Converges; compare to ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{n^2}}$, as $1/\left(n^2+3n-5\right)\le 1/n^2$ for all $n>1$.

Diverges; compare to ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{n}}$, as $1/n\le \ln n/n$ for all $n\ge 3$.

Diverges; compare to ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{n}}$. Since $n=\sqrt{n^2}>\sqrt{n^2-1}$, $1/n\le 1/\sqrt{n^2-1}$ for all $n\ge 2$.

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

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

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

Diverges; compare to ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{n}}$: for all $n\ge 1$.

Converges; compare to ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\left({\displaystyle \frac{2}{5}}\right)}^n$, as for all $n\ge 1$.

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

Diverges; compare to ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{n}}$. 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 $n\ge 2$.

Converges; compare to ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{n^2}}$, as $1/\left(n^2\ln n\right)\le 1/n^2$ for all $n\ge 3$.

Converges; Integral Test

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

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

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

(a) Converges; use Direct Comparison Test as . (b) Converges; since original series converges, we know ${lim}_{n\to \mathrm{\infty }}{a}_n=0$. Thus for large $n$, . (c) Converges; similar logic to part (b) so . (d) May converge; certainly $n{a}_n>a_n$ but that does not mean it does not converge. (e) Does not converge, using logic from (b) and $n^{th}$ Term Test.

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.

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

(a) converges (b) converges ($p$-Series) (c) absolute

(a) diverges (limit of terms is not 0) (b) diverges (c) n/a; diverges

(a) converges (b) diverges (Limit Comparison Test with $1/n$) (c) conditional

(a) diverges (limit of terms is not 0) (b) diverges (c) n/a; diverges

(a) diverges (terms oscillate between $\pm 1$) (b) diverges (c) n/a; diverges

(a) converges (b) converges (Geometric Series with $r=2/3$) (c) absolute

(a) converges (b) diverges ($p$-Series Test with $p=1/2$) (c) conditional

$S_5=-1.1906$; $S_6=-0.6767$;

$-1.1906\le {\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{{\left(-1\right)}^n}{\ln \left(n+1\right)}}\le -0.6767$

$S_6=0.3681$; $S_7=0.3679$;

$0.3681\le {\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{\left(-1\right)}^n}{n!}}\le 0.3679$

$n=5$

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 $ϵ$ of $\pi /24$ when $n\ge 249$.)

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.

algebraic, or polynomial.

Integral Test, Limit Comparison Test, and Root Test

Converges

Converges

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

Converges

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

Diverges

Converges

Converges

Diverges

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 $n\to \mathrm{\infty }$.)

Converges

Converges

Diverges

Diverges

Diverges

Absolutely converges

Conditionally converges

Diverges

Absolutely converges

Absolutely converges

Absolutely converges

Conditionally converges

Absolutely converges

Absolutely converges

Diverges

Diverges

Absolutely converges

Diverges

Absolutely converges

Diverges

Absolutely converges

1

5

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

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

(a) $R=\mathrm{\infty }$ (b) $(-\mathrm{\infty },\mathrm{\infty })$

(a) $R=1$ (b) $(2,4]$

(a) $R=2$ (b) $(-2,2)$

(a) $R=1/5$ (b) $(4/5,6/5)$

(a) $R=1$ (b) $(-1,1)$

(a) $R=\mathrm{\infty }$ (b) $(-\mathrm{\infty },\mathrm{\infty })$

(a) $R=1$ (b) $[-1,1]$

(a) $R=0$ (b) $x=0$

(a) $R=1$ (b) $[-\frac{1}{3},\frac{5}{3})$

(a) $R=\mathrm{\infty }$ (b) $(-\mathrm{\infty },\mathrm{\infty })$

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

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

(a) $f^{\prime }\left(x\right)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{n}^2{x}^{n-1}$;  $(-1,1)$ (b) ${\displaystyle \int f\left(x\right)\mathrm{d}x}=C+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{n}{n+1}}{x}^{n+1}$;  $(-1,1)$

(a) $f^{\prime }\left(x\right)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{n}{2^n}}{x}^{n-1}$;  $(-2,2)$ (b) ${\displaystyle \int f\left(x\right)\mathrm{d}x}=C+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{1}{\left(n+1\right)2^n}}{x}^{n+1}$;  $[-2,2)$

(a) $f^{\prime }\left(x\right)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{{\left(-1\right)}^n{x}^{2n-1}}{\left(2n-1\right)!}}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{\left(-1\right)}^{n+1}{x}^{2n+1}}{\left(2n+1\right)!}}$;  $(-\mathrm{\infty },\mathrm{\infty })$ (b) ${\displaystyle \int f\left(x\right)\mathrm{d}x}=C+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{\left(-1\right)}^n{x}^{2n+1}}{\left(2n+1\right)!}}$;  $(-\mathrm{\infty },\mathrm{\infty })$

(a) $f^{\prime }\left(x\right)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{n}^2{x}^{n-1}$;  $(-1,1)$ (b) ${\displaystyle \int f\left(x\right)\mathrm{d}x}=C+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{n}{n+1}}{x}^{n+1}$;  $(-1,1)$

(a) $f^{\prime }\left(x\right)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{n}{2^n}}{x}^{n-1}$;  $(-2,2)$ (b) ${\displaystyle \int f\left(x\right)\mathrm{d}x}=C+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{1}{\left(n+1\right)2^n}}{x}^{n+1}$;  $[-2,2)$

(a) $f^{\prime }\left(x\right)={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{{\left(-1\right)}^n{x}^{2n-1}}{\left(2n-1\right)!}}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{\left(-1\right)}^{n+1}{x}^{2n+1}}{\left(2n+1\right)!}}$;  $(-\mathrm{\infty },\mathrm{\infty })$ (b) ${\displaystyle \int f\left(x\right)\mathrm{d}x}=C+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{\left(-1\right)}^n{x}^{2n+1}}{\left(2n+1\right)!}}$;  $(-\mathrm{\infty },\mathrm{\infty })$

(a) ${\sum }_{n=0}^{\mathrm{\infty }}{x}^n{\left(-1\right)}^n\left(1+n\right)$; $R=1$ (b) ${\sum }_{n=1}^{\mathrm{\infty }}{x}^{n-1}{\left(-1\right)}^{n-1}\left(1+n\right)n/2={\sum }_{n=0}^{\mathrm{\infty }}{x}^n{\left(-1\right)}^n\left(2+n\right)\left(1+n\right)/2$ (c) ${\sum }_{n=1}^{\mathrm{\infty }}{x}^{n+1}{\left(-1\right)}^{n-1}\left(1+n\right)n/2={\sum }_{n=0}^{\mathrm{\infty }}{x}^{n+2}{\left(-1\right)}^n\left(2+n\right)\left(1+n\right)/2={\sum }_{n=2}^{\mathrm{\infty }}{x}^n{\left(-1\right)}^{n}n\left(-1+n\right)/2$

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

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

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^n{\displaystyle \frac{x^{6n+5}}{2n+1}}$; $R=1$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{\left(n+2\right)\left(n+1\right)}{2^{n+4}}}{x}^{n+3}$; $R=2$

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.

$p_2\left(x\right)=6+3x-4x^2$.

$p_3\left(x\right)=1-x+\frac{1}{2}{x}^3-\frac{1}{6}{x}^3$

$p_8\left(x\right)=x+x^2+\frac{1}{2}{x}^3+\frac{1}{6}{x}^4+\frac{1}{24}{x}^5$

$p_4\left(x\right)=\frac{2x^4}{3}+\frac{4x^3}{3}+2x^2+2x+1$

$p_4\left(x\right)=x^4-x^3+x^2-x+1$

$p_4\left(x\right)=1+\frac{1}{2}\left(-1+x\right)-\frac{1}{8}{\left(-1+x\right)}^2+\frac{1}{16}{\left(-1+x\right)}^3-\frac{5}{128}{\left(-1+x\right)}^4$

$p_6\left(x\right)=\frac{1}{\sqrt{2}}-\frac{-\frac{\pi }{4}+x}{\sqrt{2}}-\frac{{\left(-\frac{\pi }{4}+x\right)}^2}{2\sqrt{2}}+\frac{{\left(-\frac{\pi }{4}+x\right)}^3}{6\sqrt{2}}+\frac{{\left(-\frac{\pi }{4}+x\right)}^4}{24\sqrt{2}}-\frac{{\left(-\frac{\pi }{4}+x\right)}^5}{120\sqrt{2}}-\frac{{\left(-\frac{\pi }{4}+x\right)}^6}{720\sqrt{2}}$

$p_5\left(x\right)=\frac{1}{2}-\frac{x-2}{4}+\frac{1}{8}{\left(x-2\right)}^2-\frac{1}{16}{\left(x-2\right)}^3+\frac{1}{32}{\left(x-2\right)}^4-\frac{1}{64}{\left(x-2\right)}^5$

$p_3\left(x\right)=\frac{1}{2}+\frac{1+x}{2}+\frac{1}{4}{\left(1+x\right)}^2$

$p_3\left(x\right)=x-\frac{x^3}{6}$; $p_3\left(0.1\right)=0.09983$. Error is bounded by $\frac{1}{4!}\cdot {0.1}^4\approx 0.000004167$.

$p_2\left(x\right)=3+\frac{1}{6}\left(-9+x\right)-\frac{1}{216}{\left(-9+x\right)}^2$; $p_2\left(10\right)=3.16204$. The third derivative of $f\left(x\right)=\sqrt{x}$ is bounded on $[9,10]$ by $0.0015$. Error is bounded by $\frac{0.0015}{3!}\cdot 1^3=0.0003$.

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

The $n^{\text{th}}$ derivative of $f\left(x\right)=\cos x$ is bounded by $1$ on intervals containing $0$ and $\pi /3$. Thus $\left|R_n\left(\pi /3\right)\right|\le \frac{1}{\left(n+1\right)!}{\left(\pi /3\right)}^{\left(n+1\right)}$. 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 $n=6$.

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

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

${\left(-1\right)}^n{x}^n$

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

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.

All derivatives of $e^x$ are $e^x$ which evaluate to 1 at $x=0$. The Taylor series starts $1+x+\frac{1}{2}{x}^2+\frac{1}{3!}{x}^3+\frac{1}{4!}{x}^4+\mathrm{\cdots }$; the Taylor series is ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{x^n}{n!}}$

The $n^{\text{th}}$ derivative of $1/\left(1-x\right)$ is $f^{\left(n\right)}\left(x\right)=\left(n\right)!/{\left(1-x\right)}^{n+1}$, which evaluates to $n!$ at $x=0$. The Taylor series starts $1+x+x^2+x^3+\mathrm{\cdots }$; the Taylor series is ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{x}^n$

The Taylor series starts $0-\left(x-\pi /2\right)+0x^2+\frac{1}{6}{\left(x-\pi /2\right)}^3+0x^4-\frac{1}{120}{\left(x-\pi /2\right)}^5$; the Taylor series is ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^{n+1}{\displaystyle \frac{{\left(x-\pi /2\right)}^{2n+1}}{\left(2n+1\right)!}}$

$f^{\left(n\right)}\left(x\right)={\left(-1\right)}^n{e}^{-x}$; at $x=0$, $f^{\left(n\right)}\left(0\right)=-1$ when $n$ is odd and $f^{\left(n\right)}\left(0\right)=1$ when $n$ is even. The Taylor series starts $1-x+\frac{1}{2}{x}^2-\frac{1}{3!}{x}^3+\mathrm{\cdots }$; the Taylor series is ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^n{\displaystyle \frac{x^n}{n!}}$.

$f^{\left(n\right)}\left(x\right)={\left(-1\right)}^{n+1}\frac{n!}{{\left(x+1\right)}^{n+1}}$; at $x=1$, $f^{\left(n\right)}\left(1\right)={\left(-1\right)}^{n+1}\frac{n!}{2^{n+1}}$ The Taylor series starts $\frac{1}{2}+\frac{1}{4}\left(x-1\right)-\frac{1}{8}{\left(x-1\right)}^2+\frac{1}{16}{\left(x-1\right)}^3\mathrm{\cdots }$; the Taylor series is ${\displaystyle \frac{1}{2}}+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\left(-1\right)}^{n+1}{\displaystyle \frac{{\left(x-1\right)}^n}{2^{n+1}}}$.

Given a value $x$, the magnitude of the error term $R_n\left(x\right)$ is bounded by $$\left|R_n\left(x\right)\right|\le \frac{\max \left|f^{\left(n+1\right)}\left(z\right)\right|}{\left(n+1\right)!}\left|x^{\left(n+1\right)}\right|,$$ where $z$ is between $0$ and $x$.

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

$$\left|R_n\left(x\right)\right|\le \frac{M}{\left(n+1\right)!}\left|x^{\left(n+1\right)}\right|.$$

For any $x$, $\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{M}{\left(n+1\right)!}}\left|x^{\left(n+1\right)}\right|=0$. Thus by the Squeeze Theorem, we conclude that $\underset{n\to \mathrm{\infty }}{lim}{R}_n\left(x\right)=0$ for all $x$, and hence

$$e^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\mathit{\hspace{1em}}\text{for all }x.$$

Given a value $x$, the magnitude of the error term $R_n\left(x\right)$ is bounded by $$\left|R_n\left(x\right)\right|\le \frac{\max \left|f^{\left(n+1\right)}\left(z\right)\right|}{\left(n+1\right)!}\left|x^{\left(n+1\right)}\right|,$$ where $z$ is between $0$ and $x$. Since $\left|f^{\left(n+1\right)}\left(z\right)\right|=\frac{n!}{{\left(z+1\right)}^{n+1}}$, $$\left|R_n\left(x\right)\right|\le \frac{1}{n+1}{\left(\frac{\left|x\right|}{\min z+1}\right)}^{n+1}.$$

If , then and . Thus

$$\left|R_n\left(x\right)\right|\le \frac{n!}{\left(n+1\right)!}\left|x^{\left(n+1\right)}\right|=\frac{x^{n+1}}{n+1}.$$

For a fixed ,

$$\underset{n\to \mathrm{\infty }}{lim}\frac{x^{n+1}}{n+1}=0.$$

Given $\cos x={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^n{\displaystyle \frac{x^{2n}}{\left(2n\right)!}}$, $\cos \left(-x\right)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^n{\displaystyle \frac{{\left(-x\right)}^{2n}}{\left(2n\right)!}}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^n{\displaystyle \frac{x^{2n}}{\left(2n\right)!}}=\cos x$, as all powers in the series are even.

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

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

$1+{\displaystyle \frac{x}{3}}-{\displaystyle \frac{x^2}{9}}+{\displaystyle \frac{5x^3}{81}}-{\displaystyle \frac{10x^4}{243}}$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^n{\displaystyle \frac{{\left(x^2\right)}^{2n}}{\left(2n\right)!}}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^n{\displaystyle \frac{x^{4n}}{\left(2n\right)!}}.$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left(-1\right)}^n{\displaystyle \frac{{\left(2x+3\right)}^{2n+1}}{\left(2n+1\right)!}}.$

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

${\displaystyle {\int }_0^{\sqrt{\pi }}}\sin \left(x^2\right)\mathrm{d}x\approx {\displaystyle {\int }_0^{\sqrt{\pi }}}\left(x^2-{\displaystyle \frac{x^6}{6}}+{\displaystyle \frac{x^{10}}{120}}-{\displaystyle \frac{x^{14}}{5040}}\right)\mathrm{d}x=0.8877$