The rational numbers are dense in the reals$\,;$ that is, if $a$ and $b$ are real numbers with $a<b,$ there is a rational number $p/q$ such that $a<p/q<b$.


The set of irrational numbers is dense in the reals$\,;$ that is, if $a$ and $b$ are real numbers with $a<b,$  there is an irrational number $t$ such that $a<t<b.$


Let $n_0$ be any integer $($positive$,$  negative$,$ or zero$)$$.$ Let $P_{n_0},$ $P_{n_0+1},$ \dots$,$ $P_n,$ \dots\ be propositions$,$  one for each integer $n\ge n_0,$ such that \begin{alist} \item % (a)  $P_{n_0}$ is true$\,;$ \item % (b)  for each integer $n\ge n_0,$ $P_n$  implies $P_{n+1}.$ \end{alist} Then $P_n$ is true for every integer  $n\ge n_0.$


Let $n_0$ be any integer $($positive$,$  negative$,$ or zero$)$$.$ Let $P_{n_0},$ $P_{n_0+1}, $\dots$,$ $P_n,$ \dots\ be propositions$,$  one for each integer $n\ge n_0,$ such that \begin{alist} \item % (a)  $P_{n_0}$ is true$\,;$ \item % (b) for $n\ge n_0,$ $P_{n+1}$ is true if $P_{n_0},$ $P_{n_0+1}, $\dots$,$ $P_n$ are all true. \end{alist} Then $P_n$ is true for $n\ge n_0.$


If ${\mathcal H}$ is an open covering of a closed and bounded subset $S$ of the real line$,$ then $S$ has an open covering $\widetilde{\mathcal H}$ consisting of finitely many open sets belonging to ${\mathcal H}.$


Every bounded infinite set of real numbers has at least one limit point$.$


then it is unique$\,;$ that is$,$ if \begin{equation} \label{eq:2.1.7} \lim_{x\to x_0} f(x)=L_1\mbox{\quad and \quad}\lim_{x\to x_0} f(x)= L_2, \end{equation} then $L_1=L_2.$


\begin{equation}\label{eq:2.1.9} \lim_{x\to x_0} f(x)=L_1\mbox{\quad and \quad}\lim_{x\to x_0} g(x)= L_2, \end{equation} then \begin{eqnarray} \lim_{x\to x_0} (f+g)(x)\ar= L_1+L_2,\label{eq:2.1.10}\\ \lim_{x\to x_0} (f-g)(x)\ar= L_1-L_2,\label{eq:2.1.11}\\ \lim_{x\to x_0} (fg)(x)\ar= L_1L_2,\label{eq:2.1.12}\\ \arraytext{and, if $L_2\ne0$,}\\ \lim_{x\to x_0}\left(\frac{f}{g}\right)(x)\ar= \frac{L_1}{ L_2}.\label{eq:2.1.13} \end{eqnarray}


If $f$ is bounded on $[a,x_0),$ then $\beta=\limsup_{x\to x_0-}f(x)$ exists and is the unique real number with the following properties$\,:$ \begin{alist} \item % (a) If $\epsilon>0$, there is an $a_1$ in $[a,x_0)$ such that \begin{equation} \label{eq:2.1.22} f(x)<\beta+\epsilon\mbox{\quad if \quad} a_1\le x<x_0. \end{equation} \item % (b) If $\epsilon>0$ and $a_1$ is in $[a,x_0),$ then $$ f(\overline x)>\beta-\epsilon\mbox{\quad for some }\overline x\in[a_1,x_0). $$ \end{alist}


If $f$ is continuous on a finite closed interval $[a,b],$ then $f$ is bounded on $[a,b].$


If $f$ is monotonic and nonconstant on $[a,b],$ then $f$ is continuous on $[a,b]$ if and only if its range $R_f=\set{f(x)}{x\in[a,b]}$ is the closed interval with endpoints $f(a)$ and $f(b).$


Suppose that $f$ is increasing and continuous on $[a,b],$ and let $f(a)=c$ and $f(b)=d.$ Then there is a unique function $g$ defined on $[c,d]$ such that \begin{equation}\label{eq:2.2.17} g(f(x))=x,\quad a\le x\le b, \end{equation} and \begin{equation}\label{eq:2.2.18} f(g(y))=y,\quad c\le y\le d. \end{equation} Moreover$,$ $g$ is continuous and increasing on $[c,d].$


If $f$ and $g$ are differentiable at $x_0,$ then so are $f+g,$ $f-g,$ and $fg,$ with \begin{alist} \item % (a)  $(f+g)'(x_0)=f'(x_0)+g'(x_0);$ \item % (b)  $(f-g)'(x_0)=f'(x_0)-g(x_0);$ \item % (c)  $(fg)'(x_0)=f'(x_0)g(x_0)+f(x_0)g'(x_0).$ \end{alist}  The quotient $f/g$ is differentiable at $x_0$ if $g(x_0)\ne0,$ with \begin{alist} \setcounter{lcal}{3} \item % (d) $\dst{\left(\frac{f}{g}\right)' (x_0)= \frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{\left[g(x_0)\right]^2}}.$ \end{alist}


The Chain Rule Suppose that  $g$ is differentiable at $x_0$ and $f$ is differentiable at $g(x_0).$ Then the composite function $h=f\circ g,$ defined by $$ h(x)=f(g(x)), $$ is differentiable at $x_0,$ with $$ h'(x_0)=f'(g(x_0))g'(x_0). $$


If $f$ is differentiable at a local extreme point $x_0\in D_{f}^{0},$ then $f'(x_0)=~0.$


Suppose that  $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b),$ and $f(a)=f(b).$ Then $f'(c)=0$ for some $c$ in the open interval $(a,b).$


Intermediate Value Theorem for Derivatives  Suppose that  $f$ is differentiable on $[a,b],$  $f'(a)\ne f'(b),$ and $\mu$ is between $f'(a)$ and $f'(b).$ Then $f'(c)=\mu$ for some $c$ in $(a,b).$


Mean Value Theorem If $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b),$ then $$ f'(c)=\frac{f(b)-f(a)}{ b-a} $$ for some $c$ in $(a,b).$


Suppose that $f$ and $g$ are differentiable and $g'$ has no zeros on $(a,b).$ Let \begin{equation}\label{eq:2.4.1} \lim_{x\to b-}f(x)=\lim_{x\to b-}g(x)=0 \end{equation} \newpage \noindent or \begin{equation}\label{eq:2.4.2} \lim_{x\to b-}f(x)=\pm\infty\mbox{\quad and \quad} \lim_{x\to b-}g(x)=\pm\infty, \end{equation} and suppose that \begin{equation}\label{eq:2.4.3} \lim_{x\to b-}\frac{f'(x)}{ g'(x)}=L\quad\mbox{$($finite or $\pm \infty)$}. \end{equation} Then \begin{equation}\label{eq:2.4.4} \lim_{x\to b-}\frac{f(x)}{ g(x)}=L. \end{equation}


Extended Mean Value Theorem Suppose that $f$ is continuous on a finite closed interval $I$ with endpoints $a$ and $b$ $($that is, either $I=(a,b)$ or $I=(b,a)),$ $f^{(n+1)}$ exists on the open interval $I^0,$ and$,$ if $n>0,$ that $f'$, \dots, $f^{(n)}$ exist and are continuous at $a.$ Then \begin{equation}\label{eq:2.5.17} f(b)-\sum_{r=0}^n\frac{f^{(r)}(a)}{ r!}(b-a)^r=\frac{f^{(n+1)}(c)}{(n+1)!} (b-a)^{n+1} \end{equation} for some $c$ in $I^0.$


If $f$ is unbounded on $[a,b],$ then $f$ is not integrable on $[a,b].$


Let $f$ be bounded on $[a,b]$,  and let $P$ be a partition of $[a,b].$  Then \begin{alist} \item % (a)  The upper sum $S(P)$ of $f$ over $P$ is the supremum  of the set of all Riemann sums of $f$ over $P.$ \item % (b)  The lower sum $s(P)$ of $f$ over $P$ is the infimum  of the set of all Riemann sums of $f$ over $P.$ \end{alist}


If $f$ is integrable on $[a,b],$ then $$ \underline{\int_a^b}f(x)\,dx=\overline{\int_a^b}f(x)\,dx=\int_a^b f(x)\,dx. $$


If $f$ is bounded on $[a,b]$ and \begin{equation} \label{eq:3.2.16} \underline{\int_a^b} f(x)\,dx=\overline{\int_a^b}f(x)\,dx=L, \end{equation} then $f$ is integrable on $[a,b]$ and \begin{equation} \label{eq:3.2.17} \int_a^b f(x)\,dx=L. \end{equation}


If $f$ is bounded on $[a,b],$ then $f$ is integrable on $[a,b]$  if and only if for each $\epsilon>0$ there is a partition $P$ of $[a,b]$ for which \begin{equation} \label{eq:3.2.19} S(P)-s(P)<\epsilon. \end{equation}


If $f$ is continuous on $[a,b],$ then $f$ is integrable on $[a,b]$.


If $f$ is monotonic on $[a,b],$ then $f$ is integrable on $[a,b]$.


If $f$ and $g$ are integrable on $[a,b],$ then so is $f+g,$ and \vskip4pt $$ \int_a^b (f+g)(x)\,dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx. $$


If $f$ and $g$ are integrable on $[a,b]$ and $f(x)\le g(x)$ for $a\le x\le b,$ then \begin{equation}\label{eq:3.3.1} \int_a^b f(x)\,dx\le\int_a^b g(x)\,dx. \end{equation}


If $f$ is integrable on $[a,b],$ then so is $|f|$, and \begin{equation} \label{eq:3.3.3} \left|\int_a^b f(x)\,dx\right|\le\int_a^b |f(x)|\,dx. \end{equation}


If $f$ and $g$ are integrable on $[a,b],$ then so is the product $fg.$


First Mean Value Theorem for Integrals Suppose that $u$ is continuous and $v$ is integrable and nonnegative on $[a,b].$ Then \begin{equation} \label{eq:3.3.8} \int_a^b u(x)v(x)\,dx=u(c)\int_a^b v(x)\,dx \end{equation} for some $c$ in $[a,b]$.


If $f$ is integrable on $[a,b]$ and $a\le a_1<b_1\le b,$ then $f$ is integrable on $[a_1,b_1].$


If $f$ is integrable on $[a,b]$ and $a\le c\le b,$ then the function $F$ defined by $$  F(x)=\int_c^x f(t)\,dt $$  satisfies a Lipschitz condition on $[a,b],$ and is therefore continuous on $[a,b].$


If $f$ is integrable on $[a,b]$ and $a\le c\le b,$ then $F(x)=\int_c^x f(t)\,dt$ is differentiable at any point $x_0$ in $(a,b)$ where $f$ is continuous$,$ with $F'(x_0)=f(x_0).$ If $f$ is continuous from the right at $a,$ then $F_+'(a)=f(a)$. If $f$ is continuous from the left at $b,$ then $F_-'(b)=f(b).$


Suppose that $F$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b),$ and $f$ is integrable on $[a,b].$ Suppose  also that $$ F'(x)=f(x),\quad a<x<b. $$ Then \begin{equation}\label{eq:3.3.14} \int_a^b f(x)\,dx=F(b)-F(a). \end{equation}


Fundamental Theorem of  Calculus If $f$ is continuous on $[a,b],$ then $f$ has an antiderivative on $[a,b].$ Moreover$,$ if $F$ is any antiderivative of $f$ on $[a,b],$ then $$ \int_a^b f(x)\,dx=F(b)-F(a). $$


Integration by Parts If $u'$ and $v'$ are integrable on $[a,b],$ then \begin{equation}\label{eq:3.3.16} \int_a^b u(x)v'(x)\,dx=u(x)v(x)\bigg|^b_a-\int_a^b v(x)u'(x)\,dx. \end{equation}


Second Mean Value Theorem for Integrals Suppose that $f'$ is nonnegative and integrable and $g$ is continuous on $[a,b].$  Then \begin{equation}\label{eq:3.3.17} \int_a^b f(x)g(x)\,dx=f(a)\int_a^c g(x)\,dx+f(b)\int_c^b g(x)\,dx \end{equation} for some $c$ in $[a,b].$


Suppose that the transformation $x=\phi(t)$ maps the interval $c\le t\le d$ into the interval $a\le x\le b,$ with $\phi(c)=\alpha$ and $\phi(d)=\beta,$ and let $f$ be continuous on $[a,b].$ Let $\phi'$ be integrable on $[c,d].$ Then \begin{equation}\label{eq:3.3.21} \int_\alpha^\beta f(x)\,dx=\int_c^d f(\phi(t))\phi'(t)\,dt. \end{equation}


Suppose that $\phi'$ is integrable and $\phi$ is monotonic on $[c,d],$ and the transformation $x=\phi(t)$ maps $[c,d]$ onto $[a,b].$ Let $f$ be bounded on $[a,b].$ Then $$ g(t)=f(\phi(t))\phi'(t) $$ is integrable on $[c,d]$ if and only if $f$ is integrable over $[a,b],$ and in this case $$ \int_a^b f(x)\,dx=\int_c^d f(\phi(t))|\phi'(t)|\,dt. $$


Suppose that   $f_1,$ $f_2,$ \dots$,$ $f_n$  are locally integrable on $[a,b)$ and that $\int_a^bf_1(x)\,dx,$ $\int_a^bf_2(x)\,dx,$ \dots$,$ $\int_a^bf_n(x)\,dx$ converge$.$ Let $c_1,$ $c_2,$ \dots$,$ $c_n$ be constants$.$ Then  $\int_a^b(c_1f+c_2f_1+\cdots+c_nf_n)(x)\,dx$ converges and \begin{eqnarray*} \int_a^b (c_1f_1+c_2f_2+\cdots+c_nf_n)(x)\,dx\ar=c_1\int_a^b f_1(x)\,dx +c_2\int_a^b f_2(x)\,dx\\ \ar{}+\cdots+c_n\int_a^b f_n(x)\,dx. \end{eqnarray*}


If $f$ is nonnegative and locally integrable on $[a,b),$ then $\int_a^b f(x)\,dx$ converges if the function $$ F(x)=\int_a^x f(t)\,dt $$ is bounded on $[a,b)$, and $\int_a^b f(x)\,dx=\infty$ if it is not. These are the only possibilities, and $$ \int_a^b f(t)\,dt=\sup_{a\le x<b}F(x) $$ in either case$.$


Comparison Test If $f$ and $g$ are locally integrable on $[a,b)$ and \begin{equation}\label{eq:3.4.2} 0\le f(x)\le g(x),\quad a\le x<b, \end{equation} then \vskip3pt \noindent \part{a}\phantom{xxxxxxxxxxxxxxxxxxxx} $\dst\int_a^b f(x)\,dx<\infty \mbox{\quad if\quad}\dst\int_a^b g(x)\,dx<\infty$ \\ \vskip3pt \noindent and\\ \vskip3pt \noindent \part{b}\phantom{xxxxxxxxxxxxxxxxxxxx} $\dst\int_a^b g(x)\,dx= \infty\mbox{\quad if\quad}\dst\int_a^b f(x)\,dx=\infty$.


Suppose that  $f$ and $g$ are locally integrable on $[a,b),$ $g(x)>0$ and $f(x)\ge0$ on some subinterval $[a_1,b)$ of $[a,b),$ and \begin{equation}\label{eq:3.4.3} \lim_{x\to b-}\frac{f(x)}{ g(x)}=M. \end{equation} \begin{alist} \item % (a) If $0<M<\infty,$ then $\int_a^b f(x)\,dx$ and $\int_a^b g(x)\,dx$ converge or diverge together. \item % (b)  If $M=\infty$ and $\int_a^b g(x)\,dx=\infty,$ then $\int_a^b f(x)\,dx=\infty$. \item % (c)  If $M=0$ and $\int_a^b g(x)\,dx<\infty,$ then $\int_a^b f(x)\,dx<\infty$. \end{alist}


If $f$ is locally integrable on $[a,b)$ and $\int_a^b |f(x)|\,dx<\infty,$ then $\int_a^b f(x)\,dx$ converges$;$ that is$,$ an absolutely convergent integral is convergent$.$


Suppose that  $f$ is continuous  and its antiderivative $F(x)=\int_a^x f(t)\,dt$ is bounded on $[a,b).$ Let $g'$ be absolutely integrable on $[a,b),$ and   suppose that \begin{equation}\label{eq:3.4.9} \lim_{x\to b-} g(x)=0. \end{equation} Then $\int_a^b f(x)g(x)\,dx$ converges$.$


Suppose that  $u$ is continuous on $[a,b)$ and $\int_a^bu(x)\,dx$   diverges$.$ Let $v$ be positive and differentiable on $[a,b),$ and  suppose that  $\lim_{x\to b-}v(x)=\infty$ and $v'/v^2$ is absolutely integrable on $[a,b).$ Then $\int_a^b u(x)v(x)\,dx$  diverges$.$


A bounded function $f$ is integrable on a finite interval $[a,b]$ if and only if the set $S$ of discontinuities of $f$ in $[a,b]$ is of Lebesgue measure zero$.$


The limit of a convergent sequence is unique$.$


\begin{alist} \item % (a)  If $\{s_n\}$ is nondecreasing$,$ then $\lim_{n\to\infty}s_n=\sup\{s_n\}.$ \item % (b If $\{s_n\}$ is nonincreasing$,$ then $\lim_{n\to\infty}s_n= \inf\{s_n\}.$ \end{alist}


Let \begin{equation}\label{eq:4.1.4} \lim_{n\to\infty} s_n=s\mbox{\quad and\quad}\lim_{n\to\infty} t_n=t, \end{equation} where $s$ and $t$ are finite$.$  Then \begin{equation}\label{eq:4.1.5} \lim_{n\to\infty} (cs_n)=cs \end{equation} if $c$ is a constant$;$ \begin{eqnarray} \lim_{n\to\infty}(s_n+t_n)\ar=s+t,\label{eq:4.1.6}\\ \lim_{n\to\infty}(s_n-t_n)\ar=s-t, \label{eq:4.1.7}\\ \lim_{n\to\infty}(s_nt_n)\ar=st,\label{eq:4.1.8}\\ \arraytext{and}\nonumber\\ \lim_{n\to\infty}\frac{s_n}{ t_n}\ar=\frac{s}{ t}\label{eq:4.1.9} \end{eqnarray} if $t_n$ is nonzero for all $n$ and $t\ne0$.


\begin{alist} \item % (a) If $\{s_n\}$ is bounded above and does not diverge to $-\infty,$ then there is a unique real number $\overline{s}$ such that$,$ if $\epsilon>0,$ \begin{equation}\label{eq:4.1.16} s_n<\overline{s}+\epsilon\mbox{\quad for large $n$} \end{equation} and \begin{equation}\label{eq:4.1.17} s_n>\overline{s}-\epsilon\mbox{\quad for infinitely many  $n$}. \end{equation} \item % (b) If $\{s_n\}$ is bounded below and does not diverge to $\infty,$ then there is a unique real number $\underline{s}$ such that$,$ if $\epsilon >0,$ \begin{equation}\label{eq:4.1.18} s_n>\underline{s}-\epsilon\mbox{\quad for large $n$} \end{equation} and \begin{equation}\label{eq:4.1.19} s_n<\underline{s}+\epsilon\mbox{\quad for infinitely many $n$}. \end{equation} \end{alist}


Every sequence $\{s_n\}$ of real numbers has a unique limit superior$,$ $\overline{s},$ and a unique limit inferior$,$ $\underline{s}$, in the extended reals$,$ and \begin{equation}\label{eq:4.1.21} \underline{s}\le \overline{s}. \end{equation}


If $\{s_n\}$ is a sequence of real numbers, then \begin{equation}\label{eq:4.1.22} \lim_{n\to\infty} s_n=s \end{equation} if and only if \begin{equation}\label{eq:4.1.23} \limsup_{n\to\infty}s_n=\liminf_{n\to\infty} s_n=s. \end{equation}


A sequence $\{s_n\}$ of real numbers converges if and only if$,$ for every  $\epsilon>0,$ there is an integer $N$ such that \begin{equation}\label{eq:4.1.24} |s_n-s_m|<\epsilon\mbox{\quad if\quad} m,n\ge N. \end{equation}


If $\{s_n\}$ is monotonic and has a subsequence $\{s_{n_k}\}$ such that $$ \lim_{k\to\infty} s_{n_k}=s\quad (-\infty\le s\le\infty), $$ then $$ \lim_{n\to\infty} s_n=s. $$


\vspace*{3pt} \begin{alist} \item % (a)   If $\{x_n\}$ is bounded$,$ then $\{x_n\}$ has a convergent subsequence$.$ \vspace*{3pt} \item % (b)   If $\{x_n\}$ is unbounded above$,$  then $\{x_n\}$ has a subsequence $\{x_{n_k}\}$ such that $$ \lim_{k\to\infty} x_{n_k}=\infty. $$ \vspace*{3pt} \item % (c)   If $\{x_n\}$ is unbounded below$,$ then $\{x_n\}$ has a subsequence $\{x_{n_k}\}$ such that $$ \lim_{k\to\infty} x_{n_k}=-\infty. $$ \end{alist}


If $f$ is continuous on a closed interval $[a,b],$ then $f$ is bounded on $[a,b].$


Cauchy's Convergence Criterion for Series A series $\sum a_n$ converges if and only if for every $\epsilon>0$ there is an integer $N$ such that \begin{equation}\label{eq:4.3.3} |a_n+a_{n+1}+\cdots+a_m|<\epsilon\mbox{\quad if\quad} m\ge n\ge N. \end{equation}


If $a_n\ge0$ for $n\ge k,$ then $\sum a_n$ converges if its partial sums are bounded$,$ or diverges to $\infty$ if they are not$.$ These are the only possibilities and$,$ in either case$,$ $$ \sum_{n=k}^\infty a_n =\,\sup\set{A_n}{n\ge k}\negthickspace, $$ where $$ A_n=a_k+a_{k+1}+\cdots+a_n,\quad n\ge k. $$


The Comparison Test Suppose that \begin{equation}\label{eq:4.3.5} 0\le a_n\le b_n,\quad n\ge k. \end{equation} Then \begin{alist} \item % (a)  $\sum a_n<\infty$ if  $\sum b_n<\infty$$.$ \item % (b)  $\sum b_n=\infty$  if  $\sum a_n=\infty.$ \end{alist}


The Integral Test Let \begin{equation}\label{eq:4.3.7} c_n=f(n),\quad n\ge k, \end{equation} where $f$ is  positive$,$ nonincreasing$,$ and locally integrable on $[k,\infty).$ Then \begin{equation}\label{eq:4.3.8} \sum c_n<\infty \end{equation} if and only if \begin{equation}\label{eq:4.3.9} \int^\infty_k f(x)\,dx<\infty. \end{equation}


Suppose that  $a_n\ge0$ and $b_n>0$ for $n\ge k.$ Then \begin{alist} \item % (a) $\dst{\sum a_n<\infty\mbox{\quad if\quad}\sum b_n< \infty\mbox{\quad and\quad}\limsup_{n\to\infty} a_n/b_n<\infty}.$ \item % (b)  $\dst{\sum a_n=\infty\mbox{\quad if\quad}\sum b_n= \infty\mbox{\quad and\quad}\liminf_{n\to\infty} a_n/b_n>0}.$ \end{alist}


Suppose that  $a_n>0,$ $b_n>0,$ and \begin{equation}\label{eq:4.3.12} \frac{a_{n+1}}{ a_n}\le \frac{b_{n+1}}{ b_n}. \end{equation} Then \begin{alist} \item % (a)  $\sum a_n<\infty$ if  $\sum b_n<\infty.$ \item % (b)  $\sum b_n=\infty$  if  $\sum a_n=\infty.$ \end{alist}


The Ratio Test Suppose that  $a_n>0$ for $n\ge k.$  Then \vspace*{5pt} \begin{alist} \vspace*{5pt} \item % (a) $\sum a_n<\infty$ if\, $\limsup_{n\to\infty} a_{n+1}/a_n<1.$ \vspace*{5pt} \item % (b)  $\sum a_n=\infty$  if\, $\liminf_{n\to\infty} a_{n+1}/a_n>1.$ \end{alist} \vspace*{5pt} \noindent If \begin{equation}\label{eq:4.3.13} \liminf_{n\to\infty}\frac{a_{n+1}}{ a_n}\le1\le \limsup_{n\to\infty}\frac{a_{n+1}}{ a_n}, \end{equation} then the test is inconclusive$;$ that is$,$ $\sum a_n$ may converge or diverge$.$


Suppose that  $a_n>0$ for large $n.$  Let $$ M=\limsup_{n\to\infty} n\left(\frac{a_{n+1}}{ a_n}- 1\right)\mbox{\quad and\quad} m=\liminf_{n\to\infty} n \left(\frac{a_{n+1}}{ a_n}-1\right). $$ Then \begin{alist} \item % (a)  $\sum a_n<\infty$ if $M<-1.$ \item % (b)  $\sum a_n=\infty$  if $m>-1.$ \end{alist} The test is inconclusive if $m\le-1\le M.$


Cauchy's Root Test If $a_n\ge 0$ for $n\ge k,$ then \begin{alist} \item % (a)  $\sum a_n<\infty$ if $\limsup_{n\to\infty} a^{1/n}_n<1.$ \item % (b)  $\sum a_n=\infty$ if $\limsup_{n\to\infty} a^{1/n}_n>1.$ \end{alist} The test is inconclusive if $\limsup_{n\to\infty} a^{1/n}_n= 1.$


Dirichlet's Test for Series The series $\sum ^\infty_{n=k} a_nb_n$ converges if $\lim_{n\to\infty} a_n= 0,$ \begin{equation}\label{eq:4.3.18} \sum |a_{n+1}-a_n|<\infty, \end{equation} and \begin{equation}\label{eq:4.3.19} |b_k+b_{k+1}+\cdots+b_n|\le M,\quad n\ge k, \end{equation} for some constant $M.$


Suppose that  $\sum_{n=k}^\infty a_n=A,$ where $-\infty \le A\le\infty.$ Let $\{n_j\}_1^\infty$ be an increasing sequence of integers, with $n_1\ge k$. Define \begin{eqnarray*} b_1\ar=a_k+\cdots+a_{n_1},\\ b_2\ar=a_{{n_1}+1}+\cdots+a_{n_2},\\ &\vdots\\ b_r\ar=a_{n_{r-1}+1}+\cdots+a_{n_r}. \end{eqnarray*} Then $$ \sum_{j=1}^\infty b_{n_j}=A. $$


If $\sum_{n=1}^\infty b_n$ is a rearrangement of an absolutely convergent series $\sum_{n=1}^\infty a_n,$ then $\sum_{n=1}^\infty b_n$ also converges absolutely$,$ and to the same sum$.$


Suppose that $\sum_{n=1}^\infty a_n$ is conditionally convergent and  $\mu$ and $\nu$ are arbitrarily given in  the extended reals$,$ with $\mu\le\nu.$ Then the terms of $\sum_{n=1}^\infty a_n$ can be rearranged  to form a series $\sum_{n=1}^\infty b_n$ with partial sums $$ B_n=b_1+b_2+\cdots+b_n,\quad n\ge1, $$ such that \begin{equation}\label{eq:4.3.25} \limsup_{n\to\infty}B_n=\nu\mbox{\quad and\quad} \liminf_{n\to\infty}B_n=\mu. \end{equation}


Let $$ \sum_{n=0}^\infty a_n=A\mbox{\quad and\quad}\sum_{n=0}^\infty b_n=B, $$ where $A$ and $B$ are finite, and at least one term of each series is nonzero. Then $\sum_{n=0}^\infty p_n=AB$ for every sequence $\{p_n\}$ obtained by ordering the products in $\eqref{eq:4.3.33}$ if and only if $\sum a_n$ and $\sum b_n$ converge absolutely$.$ Moreover$,$ in this case, $\sum p_n$ converges absolutely$.$


If $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ converge absolutely to sums $A$ and $B,$ then the Cauchy product of $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ converges absolutely to $AB.$


Cauchy's Uniform Convergence Criterion A sequence of functions $\{F_n\}$ converges uniformly on a set $S$ if and only if for each $\epsilon>0$ there is an integer $N$ such that \begin{equation} \label{eq:4.4.2} \|F_n-F_m\|_S<\epsilon\mbox{\quad if\quad} n, m\ge N. \end{equation}


Suppose that  $F'_n$ is continuous on $[a,b]$ for all $n\ge1$ and $\{F'_n\}$ converges uniformly on $[a,b].$ Suppose also that  $\{F_n(x_0)\}$ converges for some $x_0$ in $[a,b].$ Then $\{F_n\}$ converges uniformly on $[a,b]$ to a differentiable limit function $F,$ and \begin{equation} \label{eq:4.4.11} F'(x)=\lim_{n\to\infty}F'_n(x),\quad a<x<b, \end{equation} while \begin{equation} \label{eq:4.4.12} F'_+(a)=\lim_{n\to\infty}F'_n(a+)\mbox{\quad and\quad} F'_-(b)= \lim_{n\to\infty}F'_n(b-). \end{equation}


Cauchy's  Uniform Convergence Criterion A series $\sum f_n$ converges uniformly on a set $S$ if and only if for each $\epsilon>0$ there is an integer $N$ such that \vskip0pt \begin{equation} \label{eq:4.4.16} \|f_n+f_{n+1}+\cdots+f_m\|_S<\epsilon\mbox{\quad if\quad} m\ge n\ge N. \end{equation}


Weierstrass's Test The series $\sum f_n$ converges uniformly on $S$ if \begin{equation} \label{eq:4.4.17} \|f_n\|_S\le M_n,\quad n\ge k, \end{equation} where $\sum M_n<\infty.$


Dirichlet's Test for Uniform Convergence The series $$ \sum_{n=k}^\infty  f_ng_n $$  converges uniformly on $S$ if  $\{f_n\}$ converges uniformly to zero on $S,$  $\sum (f_{n+1}-f_n)$ converges absolutely uniformly on $S,$ and \begin{equation} \label{eq:4.4.19} \|g_k+g_{k+1}+\cdots+g_n\|_S\le M,\quad n\ge k, \end{equation} for some constant $M.$


In connection with the power series $\eqref{eq:4.5.1},$ define $R$ in the extended reals by \begin{equation}\label{eq:4.5.2} \frac{1}{ R}=\limsup_{n\to\infty} |a_n|^{1/n}. \end{equation} In particular$,$ $R=0$ if $\limsup_{n\to\infty} |a_n|^{1/n}= \infty$, and $R=\infty$ if $\limsup_{n\to\infty} |a_n|^{1/n}=0.$ Then the power series converges \begin{alist} \item % (a)  only for $x=x_0$ if $R=0;$ \item % (b)  for all $x$ if $R=\infty,$ and absolutely uniformly in every bounded set$;$ \item % (c) for $x$ in $(x_0-R, x_0+R)$ if $0<R<\infty,$ and absolutely uniformly in every closed subset of this interval. \end{alist} The series diverges if $|x-x_0|>R.$ No general statement can be made concerning convergence at the endpoints $x=x_0+R$ and $x=x_0-R:$ the series may converge absolutely or conditionally at both$,$ converge conditionally at one and diverge at the other$,$ or diverge at both$.$


The radius of convergence of $\sum a_n(x-x_0)^n$ is given by $$ \frac{1}{ R}=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| $$ if the limit exists in the extended reals$.$


A power series $$ f(x)=\sum^\infty_{n=0} a_n(x-x_0)^n $$ \newpage \noindent with positive radius of convergence $R$ is continuous and differentiable in its interval of convergence$,$ and its derivative can be obtained by differentiating term by term$;$ that is$,$ \begin{equation} \label{eq:4.5.9} f'(x)=\sum^\infty_{n=1} na_n(x-x_0)^{n-1}, \end{equation} which can also be written as \begin{equation} \label{eq:4.5.10} f'(x)=\sum^\infty_{n=0}(n+1)a_{n+1} (x-x_0)^n. \end{equation} This series also has radius of convergence $R.$


A power series $$ f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n $$ with positive radius of convergence $R$  has derivatives of all orders in its interval of convergence$,$ which can be obtained by repeated term by term differentiation$;$ thus$,$ \begin{equation}\label{eq:4.5.11} f^{(k)}(x)=\sum^\infty_{n=k} n(n-1)\cdots (n-k+1)a_n(x-x_0)^{n-k}. \end{equation} The radius of convergence of each of these series is $R.$


If $f$ and $g$ are given by $\eqref{eq:4.5.22}$ and $\eqref{eq:4.5.23},$ then \begin{eqnarray} f(x)g(x)\ar=\sum^\infty_{n=0} c_n(x-x_0)^n,\quad|x-x_0|<R,\label{eq:4.5.24}\\ \arraytext{where}\nonumber c_n\ar=\sum^n_{r=0} a_rb_{n-r}=\sum^n_{r=0} a_{n-r}b_r\nonumber \end{eqnarray} and $R\ge\min\{R_1,R_2\}.$


Triangle Inequality If $\mathbf{X}$ and $\mathbf{Y}$ are in $\R^n,$ then \begin{equation}\label{eq:5.1.6} |\mathbf{X}+\mathbf{Y}|\le |\mathbf{X}|+|\mathbf{Y}|, \end{equation} with equality if and only if one of the vectors is a nonnegative multiple of the other$.$


Principle of Nested Sets If $S_1,$ $S_2,$ \dots\ are closed nonempty subsets of $\R^n$ such that \begin{equation}\label{eq:5.1.14} S_1\supset S_2\supset\cdots\supset S_r\supset\cdots \end{equation} and \begin{equation}\label{eq:5.1.15} \lim_{r\to\infty} d(S_r)=0, \end{equation} then the intersection $$ I=\bigcap^\infty_{r=1}S_r $$ contains exactly one point$.$


Heine--Borel Theorem If ${\mathcal H}$ is an open covering of a compact subset $S,$  then $S$ can be covered by finitely many sets from ${\mathcal H}.$


An open set $S$ in $\R^n$ is connected if and only if it is polygonally connected$.$


Intermediate Value Theorem Let $f$ be continuous on a region $S$ in $\R^n.$ Suppose that $\mathbf{A}$ and $\mathbf{B}$ are in $S$ and $$ f(\mathbf{A})<u<f(\mathbf{B}). $$ Then $f(\mathbf{C})=u$ for some $\mathbf{C}$ in $S.$


Suppose that  $f$ is defined in a neighborhood of $\mathbf{X}_0$ in $\R^n$  and $f_{x_1}(\mathbf{X}_0),$ $f_{x_2}(\mathbf{X}_{0}),$  \dots$,$ $f_{x_n}(\mathbf{X}_{0})$  exist$.$ Let $\mathbf{X}_0$ be a local extreme point of $f.$ Then \begin{equation}\label{eq:5.3.42} f_{x_i}(\mathbf{X}_0)=0,\quad 1\le i\le n. \end{equation}


The Chain Rule Suppose that  the real-valued function $f$ is differentiable at $\mathbf{X}_0$ in $\R^n,$ the vector-valued function $\mathbf{G} =(g_1,g_2, \dots,g_n)$ is differentiable at $\mathbf{U}_0$ in $\R^m,$ and $\mathbf{X}_{0}  = \mathbf{G}(\mathbf{U}_0).$ Then the real-valued composite function $h=f\circ\mathbf{G}$ defined by \begin{equation} \label{eq:5.4.3} h(\mathbf{U})=f(\mathbf{G}(\mathbf{U})) \end{equation} is differentiable at $\mathbf{U}_0,$ and \begin{equation} \label{eq:5.4.4} d_{\mathbf{U}_0}h=f_{x_1}(\mathbf{X}_0) d_{\mathbf{U}_0}g_1+f_{x_2} (\mathbf{X}_0) d_{\mathbf{U}_0}g_2+\cdots +f_{x_n} (\mathbf{X}_0) d_{\mathbf{U}_0}g_n. \end{equation}


Mean Value Theorem for Functions of $\mathbf n$ Variables Let $f$ be continuous at $\mathbf{X}_1=(x_{11},x_{21}, \dots, x_{n1})$ and $\mathbf{X}_2=(x_{12},x_{22}, \dots,x_{n2})$ and differentiable on the line segment $L$ from $\mathbf{X}_1$ to $\mathbf{X}_2.$ Then \begin{equation} \label{eq:5.4.21} f(\mathbf{X}_2)-f(\mathbf{X}_1)=\sum_{i=1}^n f_{x_i} (\mathbf{X}_0)(x_{i2}-x_{i1})=(d_{\mathbf{X}_0}f)(\mathbf{X}_2 -\mathbf{X}_1) \end{equation} for some $\mathbf{X}_0$ on $L$ distinct from $\mathbf{X}_1$ and $\mathbf{X}_2$.


Taylor's Theorem for Functions of $\mathbf n$ Variables Suppose \\that  $f$ and its partial derivatives of order $\le k$ are differentiable at $\mathbf{X}_0$ and $\mathbf{X}$ in $\R^n$ and on the line segment $L$ connecting them$.$ Then \begin{equation} \label{eq:5.4.25} f(\mathbf{X})=\sum_{r=0}^k\frac{1}{ r!} (d^{(r)}_{\mathbf{X}_0}f)  (\mathbf{X}-\mathbf{X})+\frac{1}{(k+1)!} (d^{(k+1)}_{\widetilde{\mathbf{X}}} f)(\mathbf{X}-{\mathbf{X}_0}) \end{equation} for some $\widetilde{\mathbf{X}}$  on $L$ distinct from $\mathbf{X}_0$ and $\mathbf{X}$.


Suppose that  $f$ and its partial derivatives of order $\le k-1$ are differentiable in a neighborhood $N$ of a point $\mathbf{X}_0$ in $\R^n$ and all $k$th-order partial derivatives of $f$ are continuous at $\mathbf{X}_0.$ Then \begin{equation} \label{eq:5.4.31} \lim_{\mathbf{X}\to\mathbf{X}_0} \frac{f(\mathbf{X})-T_k(\mathbf{X})}{ |\mathbf{X}-\mathbf{X}_0|^k}=0. \end{equation}


Suppose that  $f$ satisfies the hypotheses of Theorem~$\ref{thmtype:5.4.9}$ with $k\ge2,$ and  \begin{equation} \label{eq:5.4.38} d^{(r)}_{\mathbf{X}_0} f\equiv0\quad (1\le r\le k-1),\quad d^{(k)}_\mathbf{X_0} f\not\equiv0. \end{equation} Then \begin{alist} \item % (a) $\mathbf{X}_0$ is not a local extreme point of $f$ unless $d^{(k)}_{\mathbf{X}_0}f$ is semidefinite as a polynomial in $\mathbf{X}-\mathbf{X}_0.$ In particular$,$  $\mathbf{X}_0$ is not a local extreme point of $f$ if $k$ is odd$.$ \item % (b)  $\mathbf{X}_0$ is a local minimum point of $f$ if $d^{(k)}_{\mathbf{X}_0} f$ is positive definite$,$ or a local maximum point if $d^{(k)}_{\mathbf{X}_0}f$ is negative definite$.$ \item % (c)  If $d^{(k)}_{\mathbf{X}_0}f$ is semidefinite$,$ then $\mathbf{X}_0$ may be a local extreme point of $f,$ but it need not be$.$ \end{alist}


Let $\mathbf{A}$ be an $n\times n$ matrix$.$ If $\det(\mathbf{A})=0,$ then $\mathbf{A}$ is singular$.$ If $\det(\mathbf{A})\ne0,$ then $\mathbf{A}$ is nonsingular$,$ and $\mathbf{A}$ has the unique inverse \begin{equation} \label{eq:6.1.10} \mathbf{A}^{-1}=\frac{1}{\det(\mathbf{A})}\adj(\mathbf{A}). \end{equation}


The system $\eqref{eq:6.1.11}$ has a solution $\mathbf{X}$ for any given $\mathbf{Y}$ if and only if $\mathbf{A}$ is nonsingular$.$ In this case$,$ the solution is unique and is given by $\mathbf{X}=\mathbf{A}^{-1}\mathbf{Y}$.


If $\mathbf{A}=[a_{ij}]$ is nonsingular$,$ then the solution of  the system \begin{eqnarray*} a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n\ar=y_1\\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n\ar=y_2\\ &\vdots& \\ a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n\ar=y_n \end{eqnarray*} $($or$,$ in matrix form$,$ $\mathbf{AX}=\mathbf{Y}$$)$ is given by $$ x_i=\frac{D_i}{\det(\mathbf{A})},\quad 1\le i\le n, $$ where $D_i$ is the determinant of the matrix obtained by replacing the $i$th column of $\mathbf{A}$ with $\mathbf{Y};$ thus$,$ $$ D_1=\left|\begin{array}{cccc} y_1&a_{12}&\cdots&a_{1n}\\ y_2&a_{22}&\dots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ y_n&a_{n2}&\cdots&a_{nn}\end{array}\right|,\quad D_2=\left|\begin{array}{ccccc} a_{11}&y_1&a_{13}&\cdots&a_{1n}\\ a_{21}&y_2&a_{23}&\cdots&a_{2n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a_{n1}&y_n&a_{n3}&\cdots&a_{nn}\end{array}\right|,\quad\cdots, $$ $$ D_n=\left|\begin{array}{cccc}  a_{11}&\cdots&a_{1,n-1}&y_1\\ a_{21}&\cdots&a_{2,n-1}&y_2\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{n,n-1}&y_n\end{array}\right|. $$


A transformation $\mathbf{F}=(f_1,f_2, \dots,f_m)$ defined in a neighborhood of $\mathbf{X}_0\in\R^n$  is differentiable at $\mathbf{X}_0$ if and only if there is a constant $m\times n$ matrix $\mathbf{A}$ such that \begin{equation}\label{eq:6.2.2} \lim_{\mathbf{X}\to\mathbf{X}_0} \frac{ \mathbf{F}(\mathbf{X})-\mathbf{F}(\mathbf{X}_0)-\mathbf{A} (\mathbf{X}-\mathbf{X}_0)} {|\mathbf{X}-\mathbf{X}_0|}=\mathbf{0}.  \end{equation} If $\eqref{eq:6.2.2}$ holds$,$ then $\mathbf{A}$ is given uniquely by \begin{equation}\label{eq:6.2.3} \mathbf{A}=\left[\frac{\partial f_i(\mathbf{X}_0)}{\partial x_j}\right]= \left[\begin{array}{cccc}\dst{\frac{\partial f_1(\mathbf{X}_0)}{\partial x_1}}& \dst{\frac{\partial f_1(\mathbf{X}_0)}{\partial x_2}}&\cdots& \dst{\frac{\partial f_1(\mathbf{X}_0)}{\partial x_n}}\\ [3\jot] \dst{\frac{\partial f_2(\mathbf{X}_0)}{\partial x_1}}& \dst{\frac{\partial f_2(\mathbf{X}_0)}{\partial x_2}}& \cdots&\dst{\frac{\partial f_2(\mathbf{X}_0)}{\partial x_n}}\\ \vdots&\vdots&\ddots&\vdots\\ \dst{\frac{\partial f_m(\mathbf{X}_0)}{\partial x_1}}& \dst{\frac{\partial f_m(\mathbf{X}_0)}{\partial x _2}}& \cdots&\dst{\frac{\partial f_m(\mathbf{X}_0)}{\partial x_n}} \end{array}\right]. \end{equation}


Let  $\mathbf{F}=(f_1,f_2, \dots,f_m):\R^n\to\R^m,$ and suppose that the partial derivatives \begin{equation}\label{eq:6.2.7} \frac{\partial f_i}{\partial x_j},\quad 1\le i\le m,\quad 1\le j\le n, \end{equation} exist on a neighborhood of $\mathbf{X}_0$ and are continuous at $\mathbf{X}_0.$ Then $\mathbf{F}$ is differentiable at $\mathbf{X}_0.$


Suppose that $\mathbf{F}: \R^n\to\R^m$ is differentiable at $\mathbf{X}_0,$ $\mathbf{G}:\R^k\to\R^n$ is differentiable at $\mathbf{U}_0,$ and $\mathbf{X}_0=\mathbf{G}(\mathbf{U}_0).$ Then the composite function $\mathbf{H}=\mathbf{F}\circ\mathbf{G}:\R^k\to\R^m,$ defined by $$ \mathbf{H}(\mathbf{U})=\mathbf{F}(\mathbf{G}(\mathbf{U})), $$ is differentiable at $\mathbf{U}_0.$ Moreover$,$ \begin{equation}\label{eq:6.2.22} \mathbf{H}'(\mathbf{U}_0)=\mathbf{F}'(\mathbf{G}(\mathbf{U}_0)) \mathbf{G}'(\mathbf{U}_0) \end{equation} and \begin{equation}\label{eq:6.2.23} d_{\mathbf{U}_0}\mathbf{H}=d_{\mathbf{X}_0}\mathbf{F}\circ d_{\mathbf{U}_0}\mathbf{G}, \end{equation} where $\circ$ denotes composition$.$


Suppose that $\mathbf{F}: \R^n\to \R^n$ is regular on an open set $S,$ and let $\mathbf{G}=\mathbf{F}^{-1}_S.$ Then $\mathbf{F}(S)$ is open$,$ $\mathbf{G}$ is continuously differentiable on $\mathbf{F}(S),$ and $$ \mathbf{G}'(\mathbf{U})=(\mathbf{F}'(\mathbf{X}))^{-1}, \mbox{\quad where\quad}\mathbf{U}=\mathbf{F}(\mathbf{X}). $$ Moreover$,$ since $\mathbf{G}$ is one-to-one on $\mathbf{F}(S),$  $\mathbf{G}$ is regular on $\mathbf{F}(S).$


The Inverse Function Theorem Let $\mathbf{F}: \R^n\to \R^n$ be continuously differentiable on an open set $S,$ and suppose that $J\mathbf{F}(\mathbf{X})\ne0$ on $S.$ Then$,$ if $\mathbf{X}_0\in S,$ there is an open neighborhood $N$ of $\mathbf{X}_0$ on which $\mathbf{F}$ is regular$.$ Moreover$,$ $\mathbf{F}(N)$ is open and $\mathbf{G}= \mathbf{F}^{-1}_N$ is continuously differentiable on $\mathbf{F}(N),$ with $$ \mathbf{G}'(\mathbf{U})=\left[\mathbf{F}'(\mathbf{X})\right]^{-1}\quad \mbox{ $($where $\mathbf{U}=\mathbf{F}(\mathbf{X})$$)$},\quad \mathbf{U}\in\mathbf{F}(N). $$


The Implicit Function Theorem Suppose that  $\mathbf{F}:\R^{n+m}\to \R^m$ is continuously differentiable on an open set $S$ of $\R^{n+m}$ containing $(\mathbf{X}_0,\mathbf{U}_0).$ Let $\mathbf{F}(\mathbf{X}_0,\mathbf{U}_0)=\mathbf{0},$ and suppose that  $\mathbf{F}_\mathbf{U}(\mathbf{X}_0,\mathbf{U}_0)$ is nonsingular$.$ Then there is a neighborhood $M$ of  $(\mathbf{X}_0,\mathbf{U}_{0}),$  contained in $S,$ on which  $\mathbf{F}_\mathbf{U}(\mathbf{X},\mathbf{U})$  is nonsingular  and a neighborhood $N$ of $\mathbf{X}_0$ in $\R^n$ on which a unique  continuously differentiable  transformation $\mathbf{G}: \R^n\to \R^m$ is defined$,$ such that $\mathbf{G}(\mathbf{X}_0)=\mathbf{U}_0$ and \begin{equation} \label{eq:6.4.6} (\mathbf{ X},\mathbf{G}(\mathbf{X}))\in M\mbox{\quad and \quad} \mathbf{F}(\mathbf{X},\mathbf{G}(\mathbf{X}))=0\mbox{\quad  if}\quad\mathbf{X}\in N. \end{equation} Moreover$,$ \begin{equation} \label{eq:6.4.7} \mathbf{G}'(\mathbf{X})=-[\mathbf{F}_\mathbf{U}(\mathbf{X},\mathbf{G}(\mathbf{X}))]^{-1} \mathbf{F}_\mathbf{X}(\mathbf{X},\mathbf{G}(\mathbf{X})),\quad \mathbf{X}\in N. \end{equation}


If $f$ is unbounded on the nondegenerate rectangle $R$ in $\R^n,$ then $f$ is not integrable on $R.$


If $f$ is continuous on a rectangle $R$ in $\R^n,$ then $f$ is integrable on~$R.$


Suppose that $f$ is bounded on a rectangle \begin{equation}\label{eq:7.1.30} R=[a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_n,b_n] \end{equation} and continuous except on a subset $E$ of $R$ with zero content$.$ Then $f$ is integrable on $R.$


Suppose that $f$ is bounded on a bounded set $S$ and continuous except on a subset $E$ of $S$ with zero content. Suppose also that $\partial S$ has zero content$.$ Then $f$ is integrable on $S.$


A differentiable surface in $\R^n$ has zero content$.$


If $f$ is integrable on disjoint sets $S_1$ and $S_2,$ then $f$ is integrable on $S_1\cup S_2,$ and \begin{equation}\label{eq:7.1.39} \int_{S_1\cup S_2} f(\mathbf{X})\,d\mathbf{X}= \int_{S_1} f(\mathbf{X})\,d\mathbf{X}+ \int_{S_2} f(\mathbf{X})\,d\mathbf{X}. \end{equation}


$R= [a,b]\times [c,d]$ and $$  F(y)=\int_a^b f(x,y)\,dx $$ exists for each $y$ in $[c,d].$ Then $F$ is integrable on $[c,d],$ and \begin{equation}\label{eq:7.2.1} \int_c^d F(y)\,dy=\int_R f(x,y)\,d(x,y); \end{equation} that is$,$ \begin{equation}\label{eq:7.2.2} \int_c^d dy\int_a^b f(x,y)\,dx=\int_R f(x,y)\,d(x,y). \end{equation}


Let $I_1,$ $I_2,$ \dots$,$ $I_n$ be closed intervals and suppose that  $f$ is integrable on $R=I_1\times I_2\times\cdots\times I_n.$ Suppose that  there is an integer $p$ in $\{1,2, \dots,n-1\}$ such that $$ F_p(x_{p+1},x_{p+2}, \dots,x_n)=\int_{I_1\times I_2\times\cdots\times I_p} f(x_1,x_2, \dots,x_n)\,d(x_1,x_2, \dots,x_p) $$ exists for each $(x_{p+1},x_{p+2}, \dots,x_n)$ in $I_{p+1}\times I_{p+2}\times\cdots\times I_n.$ Then $$ \int_{I_{p+1}\times I_{p+2}\times\cdots\times I_n} F_p(x_{p+1}, x_{p+2}, \dots,x_n)\,d(x_{p+1},x_{p+2}, \dots,x_n) $$ exists and equals $\int_R f(\mathbf{X})\,d\mathbf{X}$.


Let $I_j=[a_j,b_j],$ $1\le j\le n$, and suppose that $f$ is integrable on $R=I_1\times I_2 \times\cdots\times I_n.$ Suppose also that  the integrals $$ F_p(x_{p+1}, \dots,x_n)=\int_{I_1\times I_2\cdots\times I_p} f(\mathbf{X}) \,d(x_1,x_2, \dots,x_p),\quad1\le p\le n-1, $$ exist for all $$ (x_{p+1}, \dots,x_n)\mbox{\quad in\quad} I_{p+1}\times\cdots\times I_n. $$ Then the iterated integral $$ \int^{b_n}_{a_n} dx_n\int^{b_{n-1}}_{a_{n-1}} dx_{n-1}\cdots \int^{b_2}_{a_2} dx_2\int^{b_1}_{a_1} f(\mathbf{X})\,dx_1 $$ exists and equals $\int_R f(\mathbf{X})\,d\mathbf{X}.$


A bounded set $S$ is Jordan measurable if and only if the boundary of $S$ has zero content$.$


Suppose that  $\mathbf{G}:\R^n\to \R^n$ is regular on a compact Jordan measurable set $S.$ Then $\mathbf{G}(S)$ is compact and Jordan measurable$.$


\E^n\to \R^n$ is regular on a compact Jordan measurable set $S$ and $f$ is continuous on $\mathbf{G}(S).$ Then \begin{equation}\label{eq:7.3.28} \int_{\mathbf{G}(S)} f(\mathbf{X})\,d\mathbf{X}= \int_S f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}. \end{equation}


Suppose that  $\mathbf{G}: \E^n\to \R^n$ is continuously differentiable on a bounded open set $N$ containing the compact Jordan measurable set $S,$ and regular on $S^0.$ Suppose   also that $\mathbf{G}(S)$ is Jordan measurable$,$ $f$ is continuous on $\mathbf{G}(S),$ and $G(C)$ is Jordan measurable for every cube $C\subset N$. Then \begin{equation}\label{eq:7.3.50} \int_{\mathbf{G}(S)} f(\mathbf{X})\,d\mathbf{X}= \int_S f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}. \end{equation}


If $x$ and $y$ are vectors in  a normed vector space $(A,N),$ then \begin{equation} \label{eq:8.1.2} |N(x)-N(y)|\le N(x-y). \end{equation}


Any two norms  $N_1$ and $N_2$ on $\R^n$ induce equivalent metrics on~$\R^n.$


If  $T$ is  compact$,$  then every Cauchy sequence $\{t_n\}_{n=1}^\infty$ in $T$ converges to a limit in $T.$


If $T$ is compact$,$ then $T$ is closed and bounded.


If $T$  is compact$,$ then $T$ is totally bounded.


If $(A,\rho)$ is complete and $T$  is closed and totally bounded$,$ then $T$ is compact.


A nonempty subset $T$ of $C[a,b]$ is compact if and only if it is closed$,$ uniformly bounded$,$ and equicontinuous.


If $f$ is continuous on a compact set $T,$ then $f(T)$ is compact.


If $f$ is continuous on a compact set $T,$ then $f$ is uniformly continuous on $T$.


Contraction Mapping Theorem If $f$ is a contraction of a complete metric space $(A,\rho),$ then the equation \begin{equation} \label{eq:8.3.8} f(u)=u \end{equation} has a unique solution$.$


If $f$ is differentiable at $x_0,$ then \begin{equation}\label{eq:2.3.3} f(x)=f(x_0)+[f'(x_0)+E(x)](x-x_0), \end{equation} where $E$ is defined on a neighborhood of $x_0$ and $$ \lim_{x\to x_0} E(x)=E(x_0)=0. $$


If $f$ is bounded on $[a,b]$ and  $\epsilon>0,$   there is a $\delta>0$ such that \begin{equation} \label{eq:3.2.12} \overline{\int_a^b}f(x)\,dx\le S(P)<\overline{\int_a^b}f(x)\,dx+\epsilon \end{equation} and $$ \underline{\int_a^b} f(x)\,dx\ge s(P)>\underline{\int_a^b} f(x)\,dx-\epsilon $$ if $\|P\|<\delta$.


If $w_f(x)<\epsilon$ for $a\le x \le b,$ then there is a $\delta>0$ such that $W_f[a_1,b_1]\le\epsilon,$ provided that $[a_1,b_1]\subset [a,b]$ and $b_1-a_1<\delta.$


Let $f$ be bounded on $[a,b]$ and define $$ E_\rho=\set{x\in [a,b]}{w_f(x)\ge\rho}. $$ Then $E_\rho$ is closed$,$ and  $f$ is integrable on $[a,b]$ if and only if for every pair of positive numbers $\rho$ and $\delta,$ $E_\rho$ can be covered by finitely many open intervals $I_1,$ $I_2, $\dots$,$ $I_p$ such that \begin{equation} \label{eq:3.5.3} \sum_{j=1}^p L(I_j)<\delta. \end{equation}


Suppose that for $n$ sufficiently large  $($that is$,$ for $n \ge\mbox{some integer }N$$)$  the terms of $\sum_{n=k}^\infty a_n$ satisfy  some condition that implies convergence of an infinite series$.$ Then $\sum_{n=k}^\infty a_n$ converges$.$ Similarly, suppose that for $n$  sufficiently large  the terms $\sum_{n=k}^\infty a_n$ satisfy  some condition that implies divergence of an infinite series$.$ Then $\sum_{n=k}^\infty a_n$ diverges$.$


Suppose that  $\mathbf{G}=(g_1,g_2, \dots,g_n)$ is differentiable at $$  \mathbf{U}_0=(u_{10}, u_{20}, \dots,u_{m0}), $$  and  define $$ M=\left(\sum_{i=1}^n\sum_{j=1}^m\left(\frac{\partial g_i(\mathbf{U}_0} {\partial u_j}\right)^2\right)^{1/2}. $$ Then$,$ if $\epsilon>0,$ there is a $\delta>0$ such that $$ \frac{|\mathbf{G}(\mathbf{U})-\mathbf{G}(\mathbf{U}_0)|} {|\mathbf{U}-\mathbf{U}_{0}|} <M+\epsilon\mbox{\quad if \quad}0<|\mathbf{U}-\mathbf{U}_0|<\delta. $$


Suppose that $\mathbf{F}:\R^n\to\R^m$ is continuously differentiable on a neighborhood $N$ of $\mathbf{X}_0.$ Then$,$ for every $\epsilon>0,$ there is a $\delta>0$ such that \begin{equation}\label{eq:6.2.8} |\mathbf{F}(\mathbf{X})-\mathbf{F}(\mathbf{Y})|< (\|\mathbf{F}'(\mathbf{X}_{0})\| +\epsilon) |\mathbf{X}-\mathbf{Y}| \mbox{\quad if\quad}\mathbf{A},\mathbf{Y}\in B_\delta (\mathbf{X}_0). \end{equation}


If $\mathbf{F}:\R^n\to\R^m$ is continuously differentiable on an open set containing a compact set $D,$ then there is a constant $M$ such that \begin{equation}\label{eq:6.2.18} |\mathbf{F}(\mathbf{Y})-\mathbf{F}(\mathbf{X})|\le M|\mathbf{Y}-\mathbf{X}| \mbox{\quad if\quad}\mathbf{X},\mathbf{Y}\in D. \end{equation}


Suppose that $|f(\mathbf{X})|\le M$ if $\mathbf{X}$ is in the rectangle $$ R=[a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_n,b_n]. $$ Let ${\bf P}=P_1\times P_2\times\cdots\times P_n$ and ${\bf P}'= P_1'\times P_2'\times\cdots\times P_n'$ be partitions of $R,$ where $P_j'$ is obtained by adding $r_j$ partition points to $P_j,$ $1\le j\le n.$ Then \begin{equation}\label{eq:7.1.16} S({\bf P})\ge S({\bf P}')\ge S({\bf P})-2MV(R)\left(\sum_{j=1}^n \frac{r_j}{ b_j-a_j}\right)\|{\bf P}\| \end{equation} and \begin{equation}\label{eq:7.1.17} s({\bf P})\le s({\bf P}')\le s({\bf P})+2MV(R)\left(\sum_{j=1}^n \frac{r_j }{ b_j-a_j}\right)\|{\bf P}\|. \end{equation}


The union of finitely many sets with zero content has zero content$.$


Suppose that $S$ is contained in a bounded set $T$ and $f$ is integrable on $S.$ Then  $f_S$ $($see $\eqref{eq:7.1.36})$ is integrable on $T,$ and $$ \int_T f_S(\mathbf{X})\,d\mathbf{X}=\int_S f(\mathbf{X})\,d\mathbf{X}. $$


Suppose that  $\mathbf{G}: \R^n\to \R^n$ is continuously differentiable on a bounded  open set $S,$ and let $K$ be a closed subset of $S$ with zero content$.$ Then $\mathbf{G}(K)$ has zero content.


A nonsingular $n\times n$ matrix $\mathbf{A}$ can be written as \begin{equation}\label{eq:7.3.10} \mathbf{A}=\mathbf{E}_k\mathbf{E}_{k-1}\cdots\mathbf{E}_1, \end{equation} where each $\mathbf{E}_i$ is a matrix that can be obtained from the $n\times n$ identity matrix $\mathbf{I}$ by one of the following operations$:$ \begin{alist} \item % (a) interchanging two rows of $\mathbf{I};$ \item % (b) multiplying a row of $\mathbf{I}$ by a nonzero constant$;$ \item % (c) adding a multiple of one row of $\mathbf{I}$ to another$.$ \end{alist}


Suppose that  $\mathbf{G}:\E^n\to \R^n$ is regular  on a cube $C$ in $\E^n,$ and let $\mathbf{A}$ be a nonsingular $n\times n$ matrix$.$ Then \begin{equation}\label{eq:7.3.29} V(\mathbf{G}(C))\le |\det(\mathbf{A})|\left[\max \set{\|\mathbf{A}^{-1}\mathbf{G}'(\mathbf{Y})\|_\infty}{\mathbf{Y}\in C} \right]^n V(C). \end{equation}


If $\mathbf{G}:\E^n\rightarrow \R^n$  is regular on a cube $C$ in $\R^n,$ then \begin{equation}\label{eq:7.3.32} V(\mathbf{G}(C))\le\int_C |JG(\mathbf{Y})|\,d\mathbf{Y}. \end{equation}


Suppose that  $\mathbf{G}: \E^n\to \R^n$ is regular on a compact Jordan measurable set $S$ and $f$ is continuous and nonnegative on $\mathbf{G}(S).$ Let \begin{equation}\label{eq:7.3.37} Q(S)=\int_{\mathbf{G}(S)} f(\mathbf{X})\,d\mathbf{X}-\int_S  f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}. \end{equation} Then $Q(S)\le0.$


Under the assumptions of Lemma~$\ref{thmtype:7.3.13},$ $Q(S)\ge0.$


If $f$ is continuous on a set $T,$ then  $f$ is  uniformly continuous on any finite closed interval contained in $T.$


If $f'$ is integrable on $[a,b],$ then $$ \int_a^b f'(x)\,dx=f(b)-f(a). $$


If $\sum a_n$ converges$,$ then for each $\epsilon>0$ there is an integer $K$ such that $$ \left|\sum_{n=k}^\infty a_n\right|<\epsilon\mbox{\quad if\quad} k\ge K; $$ that is$,$ $$ \lim_{k\to\infty}\sum_{n=k}^\infty a_n=0. $$


Suppose that  $a_n\ge0$ and $b_n>0$ for $n\ge k,$ and $$ \lim_{n\to\infty}\frac{a_n}{ b_n}=L, $$ where $0<L<\infty.$ Then $\sum a_n$ and $\sum b_n$ converge or diverge together$.$


Suppose that  $a_n>0\ (n\ge k)$ and $$ \lim_{n\to\infty}\frac{a_{n+1}}{ a_n}=L. $$ \vskip-1em Then \begin{alist} \item % (a)  $\sum a_n<\infty$ if  $L<1.$ \item % (b)  $\sum a_n=\infty$  if  $L>1.$ \end{alist} The test is inconclusive if $L=1.$


If $\sum f_n$ converges uniformly on $S,$ then $\lim_{n\to\infty}\|f_n\|_S=0.$


The series $\sum_{n=k}^\infty  f_ng_n$ converges uniformly on $S$ if $$ f_{n+1}(x)\le f_n(x),\quad x\in S,\quad n\ge k, $$ $\{f_n\}$ converges uniformly to zero on $S,$ and $$ \|g_k+g_{k+1}+\cdots+g_n\|_S\le M,\quad n\ge k, $$ for some constant $M.$


Uniqueness of Power Series If \begin{equation}\label{eq:4.5.13} \sum^\infty_{n=0} a_n(x-x_0)^n=\sum^\infty_{n=0} b_n(x-x_0)^n \end{equation} for all $x$ in some interval $(x_0-r,x_0+r),$ then \begin{equation}\label{eq:4.5.14} a_n=b_n,\quad n\ge0. \end{equation}


If $\mathbf{X},$ $\mathbf{Y},$ and $\mathbf{Z}$ are  in $\R^n,$ then $$ |\mathbf{X}-\mathbf{Z}|\le |\mathbf{X}-\mathbf{Y}|+|\mathbf{Y}-\mathbf{Z}|. $$


If $\mathbf{X}$ and $\mathbf{Y}$ are  in $\R^n,$ then $$ |\mathbf{X}-\mathbf{Y}|\ge\left| |\mathbf{X}|-|\mathbf{Y}|\right|. $$


Under the assumptions of Theorem~$\ref{thmtype:5.4.3},$ \begin{equation} \label{eq:5.4.8} \frac{\partial h(\mathbf{U}_0)}{\partial u_i}=\sum_{j=1}^n \frac{\partial f(\mathbf{X}_0) }{\partial x_j} \frac{\partial g_j(\mathbf{U}_0)}{\partial u_i},\quad 1\le i \le m. \end{equation}


If $f_{x_1},$ $f_{x_2},$ \dots$,$ $f_{x_n}$ are identically zero in an open region $S$ of $\R^n,$ then $f$ is constant in $S.$


Suppose that  $f,$ $f_x,$ and $f_y$ are differentiable in a neigborhood of a critical point $\mathbf{X}_0=(x_0,y_0)$ of $f$ and $f_{xx},$ $f_{yy},$ and $f_{xy}$ are continuous at $(x_0,y_0).$ Let $$ D=f_{xx}(x_0,y_0)f_{xy}(x_0,y_0)-f^2_{xy}(x_0,y_0). $$ Then \begin{alist} \item % (a)  $(x_0,y_0)$ is a local extreme point of $f$ if $D>0;$ $(x_0,y_0)$ is a local minimum point if $f_{xx}(x_0,y_0)>0$, or a local maximum point if $f_{xx}(x_0,y_0)<0.$ \item % (b) $(x_0,y_0)$ is not a local extreme point of $f$ if $D<0.$ \end{alist}


If $\mathbf{F}$ is continuously differentiable on a neighborhood  of $\mathbf{X}_0$ and $J\mathbf{F}(\mathbf{X}_0)\ne 0,$ then there is an open neighborhood $N$ of $\mathbf{X}_0$ on which the conclusions of Theorem~$\ref{thmtype:6.3.4}$ hold$.$


Suppose that  $f:\R^{n+1}\to \R$ is continuously differentiable on an open set containing $(\mathbf{X}_0,u_0),$ with $f(\mathbf{X}_0,u_0)=0$ and $f_u(\mathbf{X}_0,u_0)\ne0$. Then there is a neighborhood $M$ of $(\mathbf{X}_0,u_0),$ contained in $S,$ and a neighborhood $N$ of $\mathbf{X}_0$ in $\R^n$ on which is defined a unique continuously differentiable function $u=u(\mathbf{X}):\R^n\to \R$ such that $$ (\mathbf{X},u(\mathbf{X}))\in M\mbox{\quad and \quad}  f_u(\mathbf{X},u(\mathbf{X}))\ne0,\quad\mathbf{X}\in N, $$ $$ u(\mathbf{X}_0)=u_0, \mbox{\quad and \quad} f(\mathbf{X},u(\mathbf{X}))=0,\quad\mathbf{X}\in N. $$ The partial derivatives of $u$ are given by $$ u_{x_i}(\mathbf{X})=-\frac{f_{x_i}(\mathbf{X},u(\mathbf{X}))}{ f_u(\mathbf{X},u(\mathbf{X}))},\quad 1\le i\le n. $$


Suppose that  $f$ is integrable on sets $S_1$ and $S_2$ such that $S_1\cap S_2$ has zero content$.$ Then $f$ is integrable on $S_1\cup S_2,$ and $$ \int_{S_1\cup S_2} f(\mathbf{X})\,d\mathbf{X}= \int_{S_1} f(\mathbf{X})\,d\mathbf{X}+ \int_{S_2} f(\mathbf{X})\,d\mathbf{X}. $$


If $f$ is integrable on $[a,b] \times [c,d],$ then $$ \int_a^b dx\int_c^d f(x,y)\,dy=\int_c^d dy\int_a^b f(x,y)\,dx, $$ provided that $\int_c^d f(x,y)\,dy$ exists for $a\le x\le b$ and $\int_a^b f(x,y)\,dx$ exists for $c\le y\le d.$ In particular$,$ these hypotheses hold if $f$ is continuous on $[a,b]\times [c,d].$