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All poles lie outside some circle $|z|=R$ with $R>1$. The rational function is holomorphic in that disc, so its Taylor series at zero converges uniformly on $|z|\le 1$. Its polynomial partial sums therefore approximate uniformly on $K$. If there are no finite poles, the function is already a polynomial.

Let $\mathcal{A}$ be the uniform closure on $K$ of polynomials. It is a closed algebra: uniform convergence on a compact set is preserved under sums and products. For $\lambda \in \Lambda$, put $q(z)=(z-\lambda {)}^{-1}\in \mathcal{A}$ and $d=\mathrm{dist}(\lambda ,K)>0$. If $|\mu -\lambda |<d/2$, then $\mu \notin K$ andThe series converges uniformly, its ratio having modulus at most one half. Each partial sum is in $\mathcal{A}$, so its limit is too. Hence $\mu \in \Lambda$ in this neighbourhood. For empty $K$, approximation is vacuous and the assertion is immediate.

A subset is nowhere dense when its closure has empty interior. It is of first category, or meagre, when it is a countable union of nowhere-dense subsets; otherwise it is of second category.

The Baire category theorem says that in a complete metric space every countable intersection of open dense subsets is dense. To prove it, start inside an arbitrary nonempty open set and choose a closed ball of positive radius there and in the first dense open set. Inductively choose a closed ball inside the interior of the preceding ball and the next dense open set, with radius less than $2^{-n}$. The centres are Cauchy. Completeness gives a limit lying in all the nested closed balls, hence in all the dense open sets and the original open set. Thus a nonempty complete metric space is not meagre in itself.

For $p<r\le \infty$, put $E_m=\{x\in {\ell }^p:\Vert x{\Vert }_p\le m\}$, viewed inside ${\ell }^r$. Necessarily $p<\infty$. Each $E_m$ is closed: convergence in the ${\ell }^r$ norm gives coordinatewise convergence, and every finite partial sum of $\sum |x_i{|}^p$ is bounded by $m^p$ in the limit. Taking the supremum of partial sums proves the assertion.

It has empty interior. Given a point in it and any small ${\ell }^r$ radius, perturb a block of $N$ coordinates by entries of modulus $\delta N^{-1/r}$, choosing signs to avoid cancelling the existing coordinates. The ${\ell }^r$ distance is $\delta$, while the resulting ${\ell }^p$ norm is at least $\delta N^{1/p-1/r}$, exceeding $m$ for large $N$. For $r=\infty$, use entries of modulus $\delta$ on that block. Hence each $E_m$ is nowhere dense, andFor $r=p$, including $p=\infty$, the space is Banach and so second category in itself by Baire. There is no case $r>p$ when $p=\infty$.

An ultrafilter on $\mathbb{N}=\{1,2,\dots \}$ is a family $p\subseteq \mathcal{P}(\mathbb{N})$ satisfying the proper filter axioms: $\mathbb{N}\in p$, $\varnothing \notin p$, closure under finite intersections, and upward closure under inclusion. In addition, for every $A\subseteq \mathbb{N}$, exactly one of $A$ and $A^c$ belongs to $p$. Equivalently, an ultrafilter is a maximal proper filter. The principal ultrafilter at $n$ consists of all subsets containing $n$; an ultrafilter not of this form is a nonprincipal ultrafilter.

To prove existence, start with the cofinite filter and order the proper filters containing it by inclusion. The union of a chain is still a proper filter: finitely many of its members lie in one filter of the chain, and no constituent contains the empty set. Thus Zorn lemma supplies a maximal such filter $p$. If $A\notin p$, adjoining $A$ cannot generate a proper filter; otherwise maximality fails. Consequently some $B\in p$ has $B\cap A=\varnothing$, so $A^c\in p$ by upward closure. This proves the ultrafilter alternative. No finite $A$ can belong to $p$, because $A^c$ is cofinite and already belongs to $p$. In particular $p$ contains no singleton and is a nonprincipal ultrafilter. The same argument, starting from any proper filter, proves the ultrafilter lemma used below.

Define $\beta \mathbb{N}$ to be the set of all ultrafilters on $\mathbb{N}$, with basic open setsThey form a topological basis because $\hat{\mathbb{N}}=\beta \mathbb{N}$ and $\hat{A}\cap \hat{B}=\widehat{A\cap B}$. Moreover $(\hat{A}{)}^c=\widehat{A^c}$, so they are clopen sets. This is the compact Hausdorff topology on ultrafilters. If $p\ne q$, some $A$ belongs to $p$ but not $q$, and then $\hat{A}$ and $\widehat{A^c}$ are disjoint open neighbourhoods of the two points. Thus $\beta \mathbb{N}$ is Hausdorff.

For compactness, take an open cover and refine it by basic open sets $\widehat{A_{\lambda }}$. If no finite subfamily covers, then for every finite collection of indices the intersection $\bigcap A_{\lambda }^c$ is nonempty. Indeed, if that intersection were empty, the corresponding $A_{\lambda }$ would cover $\mathbb{N}$, and every ultrafilter would contain at least one of them by the finite ultrafilter alternative. The complements therefore have the finite intersection property and generate a proper filter. Extend it to an ultrafilter $q$ by the proved ultrafilter lemma. Then $A_{\lambda }^c\in q$ for every $\lambda$, so $q$ belongs to none of the covering $\widehat{A_{\lambda }}$, a contradiction. A finite basic subcover lies inside a finite subcover of the original cover. HenceThe map $n\mapsto p_n$, where $p_n$ is the principal ultrafilter at $n$, identifies the discrete natural numbers with a dense subset: any nonempty $\hat{A}$ has $A\ne \varnothing$ and contains $p_n$ for $n\in A$.

Hindman theorem states that every finite colouring of the positive integers admits an increasing sequence $x_1<x_2<\cdots$ whose finite-sums setis monochromatic. We now give the Idempotent-ultrafilter proof of Hindman's theorem. For $A\subseteq \mathbb{N}$, write $A-x=\{y\in \mathbb{N}:x+y\in A\}$. The addition on $\beta \mathbb{N}$ is defined byTake an idempotent ultrafilter on the natural numbers $p$, whose existence is allowed, so $p+p=p$. It is a nonprincipal ultrafilter: addition of principal points is $p_m+p_n=p_{m+n}$, whereas no positive integer satisfies $2m=m$.

Exactly one colour class $A$ belongs to $p$. DefineBy idempotence, $A^{\ast }\in p$. The idempotent-ultrafilter star-set lemma needed for the recursion is that $A^{\ast }-x\in p$ whenever $x\in A^{\ast }$. To prove it, put $D=A-x\in p$. Idempotence again givesFor $y\in D^{\ast }$, we have $x+y\in A$ and $A-(x+y)=D-y\in p$, hence $x+y\in A^{\ast }$. Thus $D^{\ast }\subseteq A^{\ast }-x$, proving the lemma by upward closure.

Choose $x_1\in A^{\ast }$. Suppose $F_n=FS(x_1,\dots ,x_n)\subseteq A^{\ast }$ has been constructed. Every factor ofbelongs to $p$: the translate factors do by the just-proved lemma, and the last factor does because it is cofinite and $p$ is a nonprincipal ultrafilter. The finite intersection is therefore nonempty. Choose $x_{n+1}$ in it. Then $x_{n+1}$ and every $s+x_{n+1}$ with $s\in F_n$ lie in $A^{\ast }$, so $F_{n+1}\subseteq A^{\ast }$. This constructs a strictly increasing sequence, indeed $x_{n+1}>{\sum }_{i\le n}{x}_i$, and proves Hindman theorem.

Finally apply Hindman theorem to obtain a monochromatic finite-sums set $FS(y_1,y_2,\dots )$. We construct a divisibility chain inside a finite-sums set using disjoint successive blocks of indices. Start with $x_1=y_1$. If $x_i=m$, take $2m$ unused terms after all indices used so far, and split them into two groups of $m$ consecutive terms. In each group its $m+1$ partial sums, including $0$, contain two in the same congruence class modulo $m$ by the pigeonhole principle. Their difference is a nonempty consecutive sum divisible by $m$, and positivity makes it at least $m$. Let $x_{i+1}$ be the sum of these two selected sums. Then $m\mid x_{i+1}$ and $x_{i+1}\ge 2m>m$.

The index set defining $x_{i+1}$ lies strictly after all previous index sets. Discard the other terms in these two groups before continuing. Because the chosen index sets are disjoint, every finite sum of distinct $x_i$ is a finite sum of distinct $y_j$, retaining the same colour. Therefore

The ratio test states that, for a series $\sum a_n$ with nonzero terms eventually, if $|a_{n+1}/a_n|\to L<1$, the series converges absolutely; if $L>1$, including $L=\infty$, it diverges because its terms do not tend to zero. At $L=1$ the test is inconclusive: the harmonic series diverges while $\sum n^{-2}$ converges.

For $|x|<1$, the ratio test applied to $x^n/n$ gives absolute convergence when $x\ne 0$, and at $x=0$ every such term vanishes. ThusAt $x=1$ all terms are zero, so the series converges to zero. At $x=-1$, the even terms vanish and the odd terms are $-2/n$, whose partial sums tend to $-\infty$. Finally, if $|x|>1$, then $|x^n-1|/n\ge (|x{|}^n-1)/n\to \infty$, so the term test for divergence applies. Consequently

Suppose all infinite binary sequences could be listed as ${ϵ}^{(1)},{ϵ}^{(2)},\dots$, where ${ϵ}^{(j)}=({ϵ}_1^{(j)},{ϵ}_2^{(j)},\dots )$. Define another binary sequence by $d_j=1-{ϵ}_j^{(j)}$. It differs from the $j$th listed sequence at its $j$th entry, so it is absent from the list. This contradiction is Cantor's diagonal argument, proving thatTo transfer this result to $[0,1]$, map a binary sequence toThis lies in $[0,1]$ because ${\sum }_{j\ge 1}2/3^j=1$. If two sequences first differ at index $j$, with ${ϵ}_j=1$ and ${\eta }_j=0$, thenThus $f$ is an injection, avoiding the ambiguous expansions that would arise from unrestricted binary digits in base two. An injection from an uncountable set into $[0,1]$ proves that $\boxed{[0,1]\text{ is uncountable}}$.

For $\Sigma (\mathbb{Z})$, a nondecreasing sequence bounded above can take only finitely many integer values, all lying between its first term and an integer upper bound. Every strict increase raises its value by at least one, so only finitely many strict increases occur. It is therefore eventually constant. Every such sequence has the formfor some finite list of integers. The set of finite integer lists is a countable set: at stage $j$, list the tuples of length at most $j$ whose entries have absolute value at most $j$, omitting earlier duplicates. Each stage is finite and every finite tuple occurs at some stage. Constant sequences give an injection of $\mathbb{Z}$ into the collection, soThis is the discrete-order phenomenon that bounded nondecreasing integer sequences are eventually constant.

For $\Sigma (\mathbb{Q})$, use a different encoding, into rational partial sums:Each $s_n$ is a rational number, the sequence is nondecreasing, and $s_n\le 1/2$. If two binary inputs first differ at $j$, their $j$th partial sums differ by $3^{-j}$, so their output sequences differ. This defines an injection of the uncountable binary-sequence set into $\Sigma (\mathbb{Q})$. Thus bounded nondecreasing rational sequences are uncountable. Finally, every such rational sequence is also an admissible real sequence, so inclusion into $\Sigma (\mathbb{R})$ is injective. ConsequentlyBoundedness and monotonicity force eventual constancy for integer values, but not for rational or real values; countability of the set of possible individual terms does not imply countability of all infinite sequences of those terms.

The assumed differentiability makes $e$ continuous. The identity $e^{\prime }=e$ then implies inductively that $e$ has continuous derivatives of every order andFix $x\ne 0$. By the extreme value theorem, the continuous function $|e|$ is bounded on the closed interval between $0$ and $x$; let $M_x$ be its maximum. Applying the Taylor theorem with Lagrange remainder about $0$ gives some ${\xi }_n$ between $0$ and $x$ such thatFor the last factor, successive ratios are $|x|/(n+2)$, which are at most $1/2$ once $n$ is sufficiently large. A geometric sequence bound therefore proves that $|x{|}^{n+1}/(n+1)!\to 0$. The partial sums converge to $e(x)$, soAt $x=0$ the equality follows directly from $e(0)=1$. The proof obtains this Taylor series solely from the differential identity, continuity on a fixed finite interval, and the Taylor remainder estimate.

For a bounded real function on $[a,\infty )$, convergence of its improper integral means that its restriction to every $[a,b]$, $b>a$, is Riemann integrable and that the finite limitexists. Boundedness by itself does not guarantee either local Riemann integrability or existence of this limit.

A monotone function on $[a,b]$ is bounded between its endpoint values. Divide the interval into $m$ equal pieces with endpoints $x_j=a+j(b-a)/m$. If $f$ is nondecreasing, its upper and lower Darboux sums differ byFor nonincreasing $f$, the same calculation with reversed endpoint values gives $(b-a)(f(a)-f(b))/m$. Thus in either case $U-L=(b-a)|f(b)-f(a)|/m\to 0$. For every $\epsilon >0$ one can choose $m$ to make this gap smaller than $\epsilon$; the Riemann integrability criterion proves that $f$ is Riemann integrable.

Now let $f$ be decreasing on $[1,\infty )$ with limit zero. For each $x$, $f(x)\ge {\lim }_{y\to \infty }f(y)=0$. On $[n,n+1]$ its monotonicity givesSumming these bounds, with $S_N={\sum }_{n=1}^{N}f(n)$, yieldsBoth the partial sums $S_N$ and the truncated integrals are nondecreasing because $f\ge 0$. The inequalities show that one family is bounded above if and only if the other is. The bounded monotone sequence theorem then gives convergence of the sums exactly when the integer-endpoint integrals converge. For real $b$ between $N$ and $N+1$, the integral to $b$ lies between the integrals to these integer endpoints; the squeeze theorem supplies the same limit. If either family is unbounded, both tend to $+\infty$. This proves the integral test for convergence in the required form.

For $\alpha <0$, use $f(x)=x^{\alpha }$, which is decreasing and tends to zero. Its truncated improper integral isThis has a finite limit exactly when $\alpha <-1$. If $\alpha \ge 0$, the terms $n^{\alpha }$ do not tend to zero, so the term test for divergence applies. Thus the full p-series criterion isFor every $\alpha \ge -1$ its nonnegative partial sums tend to $+\infty$.

The alternating series test states that if $b_n\ge 0$ decreases to zero, then $\sum (-1{)}^n{b}_n$ converges. Indeed, its even and odd partial sums are monotone in opposite directions and bound each other, while their difference tends to zero. Here $b_n=1/(1+\sqrt{n})$ decreases to zero, so the series converges.

It fails absolute convergence, since for $n\ge 1$,The p-series $\sum n^{-1/2}$ diverges. This also follows by comparison with the divergent harmonic series, because $n^{-1/2}\ge n^{-1}$. The comparison test for series therefore shows that $\sum b_n$ diverges. Thus

For $s=\pm 1$, the identity $s/(2+s)=-1/3+(2/3)s$ givesThe alternating series test applies to $\sum (-1{)}^n/n$ because $1/n$ decreases to zero; its partial sums consequently have a finite limit. The harmonic series $\sum 1/n$ diverges to $+\infty$. Applying the displayed identity to each finite partial sum yieldsTherefore $\boxed{\text{the series diverges to }-\infty }$. Alternating signs alone do not satisfy the alternating series test: the magnitudes jump from $1/(3n)$ at even $n$ to $1/(n+1)$ at the next odd index.

Let $s_N={\sum }_{n=1}^N(-1{)}^{n-1}{b}_n$. Pairing consecutive terms shows $s_{2m}\ge 0$ andThe even partial sums increase and the odd partial sums decrease. They satisfy $s_{2m}\le s_{2m+1}\le b_1$, whileBoth are monotone bounded sequences and hence have limits, and the displayed difference makes their limits equal. Every partial sum belongs to one subsequence, so the alternating series converges. This is a proof of the alternating series test; the monotonicity and zero-limit hypotheses play separate roles.

The summand is positive, so it suffices to show that its partial sums are unbounded. On the block $2^k\le n<2^{k+1}$, for $k\ge 1$,There are $2^k$ terms in that block, so its sum is at least $1/[2(k+1)\log 2]$. Adding blocks gives a constant multiple of the divergent harmonic series. HenceEquivalently the integral test for convergence gives ${\int }_2^{R}dx/(x\log x)=\log \log R-\log \log 2\to \infty$. The block proof supplies the divergence without assuming the integral test.

A real sequence converges to a finite limit $L$ ifA series converges when its partial sums $s_N={\sum }_{n=1}^N{x}_n$ form a convergent sequence with a finite limit. These are different conditions: a series concerns accumulated terms, whereas sequence convergence concerns the terms themselves.

If $s_N\to S$, then $s_{N-1}\to S$ as well and $x_N=s_N-s_{N-1}\to 0$. Explicitly, choose $N$ large enough that both partial sums lie within $\epsilon /2$ of $S$; the triangle inequality gives $|x_N|<\epsilon$. Hence convergence of a series forces its terms to tend to zero.

For the converse, take $x_n=1/n$. This is a null sequence, but its harmonic series diverges. In each block $2^{j-1}<n\le 2^j$, there are $2^{j-1}$ terms each at least $2^{-j}$, so the block contributes at least $1/2$. ThusA zero limit for the terms is necessary, but not sufficient, for convergent series.

A real sequence $(x_n)$ has limit of a sequence $x$ whenA series ${\sum }_{n\ge 1}{x}_n$ is a convergent series with sum $s$ when its partial sums $S_N={\sum }_{n=1}^N{x}_n$ have limit of a sequence $s$.

For the first comparison test for series, either of the two allowed bounds implies $0<x_n\le a_n+b_n$. HenceThe partial sums form a monotone bounded sequence, so they converge. The series $\sum x_n$ is convergent.

For the exponent $2$, use $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. The resulting telescoping series bounds every partial sum by $2$, proving that the series converges by the monotone bounded sequence property. To prove divergence of the harmonic series, group the terms with $2^j<n\le 2^{j+1}$: their sum is at least $2^j/2^{j+1}=1/2$. Thus its partial sums are unbounded. If $\alpha \le 1$, then $n^{-\alpha }\ge n^{-1}$, so the comparison test for series proves

For the weighted square assumption, the finite-dimensional Cauchy-Schwarz inequality gives, for every $N$,Again the positive partial sums form a monotone bounded sequence. This is the weighted square roots of a summable sequence argument, with $a_n=n^2{x}_n^2$. The weighted square condition implies $\sum x_n<\infty$.

The converse is false. Take $x_n=n^{-3/2}$. Each block $2^j\le n<2^{j+1}$ contributes at most $2^j{2}^{-3j/2}=2^{-j/2}$, and the resulting geometric series converges. But $n^2{x}_n^2=1/n$, whose harmonic series diverges. This proves the counterexample without assuming the full P-series criterion.

A sequence $(a_k)$ converges to $a\in \mathbb{R}$ when for every $\epsilon >0$ there is $N$ such that $|a_k-a|<\epsilon$ for all $k\ge N$. A series ${\sum }_{k=1}^{\infty }{a}_k$ converges to a finite $S$ when its partial sums $s_n={\sum }_{k=1}^n{a}_k$ converge to $S$.

If the series converges, then $a_n=s_n-s_{n-1}\to S-S=0$, so its terms tend to zero. The converse is false: the harmonic series has terms tending to zero but diverges.

For the arithmetic averages, suppose $a_k\to a$. Fix $\epsilon >0$ and choose $K$ so $|a_k-a|<\epsilon /2$ for $k\ge K$. The finite initial contribution $C={\sum }_{k=1}^{K-1}|a_k-a|$ is fixed, andFor sufficiently large $n$, $C/n<\epsilon /2$, provingThis is the Cesàro theorem for convergent sequences. Its hypothesis concerns a convergent sequence; the corresponding series need not be a convergent series.

The arithmetic-geometric mean inequality gives $a_k+1/a_k\ge 2$, so $u_n\ge 2n\to +\infty$. Hence $(u_n)$ has no finite limit.

For the example, take $\boxed{a_k=1+2^{-k}}$. Every term is positive and different from $1$, andThe bounding geometric series converges, so the comparison test for series proves that the partial sums $v_n$ converge.

In general, if $v_n$ converges, the necessary condition for a convergent series gives $d_k=a_k-1/a_k=v_k-v_{k-1}\to 0$. Since $a_k>0$,Apply the Cesàro theorem for convergent sequences to these summands:Equivalently, $a_k=(d_k+\sqrt{d_k^2+4})/2\to 1$. The series of the differences may converge while the series of the positive sums still diverges linearly.

Write $S_n={\sum }_{j=1}^n{Y}_j$ for the partial sum. Fix an integer $M\ge 1$. The tail-sum formula for expectation, applied to the nonnegative random variable $|Y_1|$, givesbecause $\mathbb{E}|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are independent, so the second Borel--Cantelli lemma shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in \mathbb{N}$ yieldsalmost surely.

Now $Y_n=S_n-S_{n-1}$, and the triangle inequality givesIf the limit superior of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Thereforealmost surely.

For a series of complex numbers, form the partial sums $S_N={\sum }_{n=1}^N{a}_n$. The series converges when these partial sums have a finite limit of a sequence $S\in \mathbb{C}$:The series has absolute convergence when ${\sum }_{n=1}^{\infty }|a_n|$ converges as a series of nonnegative real numbers. A series beginning at $n=0$ uses the same definitions with that starting index.

Write the partial sums as $S_N(x)={\sum }_{n=1}^N{f}_n(x)$. The series has uniform convergence to $s:E\to \mathbb{R}$ ifThe same threshold $N_0$ must control every point of $E$. Equivalently, the supremum error ${\sup }_{x\in E}|S_N(x)-s(x)|$ tends to zero, for nonempty $E$.

For real-valued functions, the Uniform Cauchy criterion applied to partial sums says that $\sum u_n$ has uniform convergence exactly whenSufficiency uses completeness of the real numbers.

Let $M$ be a common bound for the uniformly bounded family of functions $(g_n)$; increase it if necessary so that $M>0$. The assumed uniform absolute convergence of $\sum f_n$ gives $N$ such thatThe triangle inequality now yieldsThe Uniform Cauchy criterion therefore proves uniform convergence of $\sum f_n{g}_n$. The same estimate with $|f_n{g}_n|$ proves uniform absolute convergence as well.