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The limit of a sequence of real numbers is $a$ if, for every $\epsilon >0$, there is an integer $N$ such that $n\ge N$ implies $|a_n-a|<\epsilon$. It tends to positive infinity if, for every real $M$, there is an integer $N$ such that $n\ge N$ implies $a_n>M$.

If $a_n\to +\infty$, choose the latter threshold with $M=1/\epsilon$. Then $a_n>1/\epsilon >0$ eventually, so $|1/a_n|<\epsilon$. Thus $1/a_n\to 0$. The converse is false: $a_n=-n$ has reciprocal tending to zero and tends to negative infinity. The precise sign-free statement is given by reciprocal limits and escape in absolute value: $1/a_n\to 0$ is equivalent to $|a_n|\to \infty$.

If $a_n\to a\ne 0$, eventually $|a_n-a|<|a|/2$, and the triangle inequality gives $|a_n|>|a|/2$. ConsequentlyGiven $\epsilon >0$, take $N$ large enough that $|a_n-a|<\min (|a|/2,\epsilon |a{|}^2/2)$ for all $n\ge N$. This proves $1/a_n\to 1/a$ directly from the definition of the limit of a sequence.

Here is a nested-interval proof of the Bolzano-Weierstrass theorem. Put every term of the bounded real sequence in a closed interval $I_0=[-M,M]$, with $M>0$. Bisect this interval and choose a closed half containing infinitely many terms. Repeating gives nested closed intervals $I_j=[l_j,r_j]$, each containing infinitely many sequence terms, with length $2M/2^j$.

By the least-upper-bound property, $x_{\ast }={\sup }_j{l}_j$ exists. For each $j$, every lower endpoint lies at most $r_j$, while $l_j\le x_{\ast }$; hence $x_{\ast }\in I_j$. Choose increasing indices $n_j$ with $a_{n_j}\in I_j$, possible because each interval contains infinitely many terms. ThenThus every bounded real sequence has a convergent subsequence.

For a sequence with no convergent subsequence, take $a_n=n$: any subsequence has $n_j\ge j$, so its terms tend to positive infinity rather than to a finite limit of a sequence. For an unbounded sequence with a convergent subsequence, take $a_{2n}=0$ and $a_{2n-1}=n$. Its odd terms are unbounded, while its even subsequence converges to zero.

Group three consecutive terms. Their block sum isSince $2n-1\ge n$ and $4n-1\ge 3n$ for $n\ge 1$, we have $|b_n|\le 1/(2n^2)$. The comparison test for series and the preceding proof of convergence of $\sum n^{-2}$ show that $\sum b_n$ converges absolutely.

This establishes convergence of the original partial sums at indices $3N$. Each intermediate partial sum differs from one of these by at most two of the original terms; those terms tend to zero. Thus all partial sums have the same limit of a sequence, which is the mechanism of convergence from fixed-length series blocks. The original series is not absolutely convergent, since its absolute-value series contains ${\sum }_n(2n-1{)}^{-1}\ge \frac{1}{2}{\sum }_n{n}^{-1}$. Therefore the original series converges conditionally.

Use the least-upper-bound property as the least upper bound axiom: every nonempty set of real numbers bounded above has a least upper bound in $\mathbb{R}$. It implies convergence of a bounded increasing sequence: if $L$ is the supremum of its terms, then for every $\epsilon >0$ some term exceeds $L-\epsilon$, and all subsequent terms lie between that term and $L$. A bounded decreasing sequence converges by applying this argument to its negative.

The alternating series test states that if $b_n\ge 0$, $b_{n+1}\le b_n$, and $b_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n-1}{b}_n$ converges. To prove it, let $S_N$ denote its partial sums. The even partial sums are increasing because $S_{2m+2}-S_{2m}=b_{2m+1}-b_{2m+2}\ge 0$. The odd partial sums are decreasing because $S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\le 0$. Moreover,Completeness of the real numbers therefore gives a limit of a sequence for each subsequence, denoted $L_e,L_o$ respectively for the even and odd subsequences. Their difference is $S_{2m+1}-S_{2m}=b_{2m+1}\to 0$, so $L_e=L_o$. Every sufficiently late partial sum belongs to one of these two subsequences, proving convergence of the full sequence of partial sums. Multiplying the terms by $-1$ leaves convergence unchanged.

Use $T_y=\inf \{n\ge 0:X_n=y\}$ and the positive return time $T_x^{+}=\inf \{n\ge 1:X_n=x\}$. The walk has transitions $p_{uv}=c_{uv}/d(u)$, $d(u)={\sum }_v{c}_{uv}$; its transition matrix defines a reversible Markov chain, although the transition probabilities need not be symmetric.

The voltage $h(u)=P_u(T_x<T_y)$ equals one at $x$, zero at $y$, and is a harmonic function on a graph away from the terminals: $h(u)={\sum }_v{p}_{uv}h(v)$. For this unit electric potential difference the electric current isThe first-step analysis therefore proves the return probability from effective resistance identity:Finite irreducibility ensures eventual hitting of one of the terminals, so no missing escape event appears in this finite-network calculation.

For an infinite locally finite connected graph, exhaust it by finite sets and wire their exteriors to one sink. The event of exiting before the first return decreases to the event of never returning. Thuswith $1/\infty =0$. In particular recurrence is equivalent to infinite effective resistance to infinity.

After deleting an edge, consider any infinite component and choose a root $o$ away from that edge's endpoints. Its vertex degree $d(o)$ is unchanged. On every finite wired exhaustion the Rayleigh monotonicity principle makes the resistance no smaller; the limit of a sequence does too. A recurrent original network consequently leaves a recurrent component. Finite components give recurrent walks as well, with an isolated graph vertex viewed as an absorbing state. This proves that edge deletion preserves recurrence. The phrase “no less recurrent” refers to this recurrence comparison: at a deleted-edge endpoint the vertex degree factor also changes, so monotonicity of its first-return probability alone should not be inferred.

A self-avoiding walk of length $n$ is a sequence of $n+1$ distinct graph vertices joined consecutively by edges. Split a rooted self-avoiding walk of length $m+n$ after $m$ steps. There are ${\kappa }_m$ possible prefixes; each remaining segment is a self-avoiding $n$-step walk from its current graph vertex, and forbidding intersections with the prefix can only reduce its count below ${\kappa }_n$. Translation invariance therefore givesApply the Fekete lemma, proved in Question 4(a), to $\log {\kappa }_n$. It gives $\log \mu ={\inf }_n{n}^{-1}\log {\kappa }_n$ and hence existence of the connective constant. Walks using only the $d$ positive coordinate directions never revisit a graph vertex, so there are at least $d^n$. A walk has $2d$ choices for its first step and at most $2d-1$ subsequently because immediate reversal is forbidden. Thus

For independent bond percolation, let $C(0)$ be the percolation cluster of the origin, $\theta (p)=P_p(|C(0)|=\infty )$, and $p_c=\inf \{p:\theta (p)>0\}$. An infinite percolation cluster in a locally finite graph contains an open self-avoiding path of every length. The union bound givesWhen $p\mu <1$, the right-hand side tends to zero, since its $n$th root tends to $p\mu$. Hence the connective-constant lower bound for percolation is

For the upper bound in two dimensions, use the square planar dual graph. Call a dual edge available when its crossing primal edge is closed. Each dual edge is available with probability $q=1-p$, independently. The planar graph boundary lemma says that a finite primal percolation cluster is enclosed by a graph cycle consisting of such edges: its external edge boundary gives this graph cycle. We show that $q\mu <1$ guarantees positive percolation probability.

Let $N_n$ count enclosing graph cycles of length $n$. Because the origin lies inside their coordinate bounding box and their diameter is at most $n$, every such graph cycle has a graph vertex within a box of side $2n+3$. From each possible root, the first $n-1$ edges give a self-avoiding walk; the final closing edge is then determined. Dropping the closure restriction gives the convenient boundChoose $\rho$ with $q\mu <\rho <1$. The connective-constant limit of a sequence implies ${\kappa }_{n-1}{q}^n\le C{\rho }^n$. Consequently the large-circuit tailis summable.

Condition all primal edges inside a large square $[-L,L{]}^2$ to be open; this event has positive probability. The origin percolation cluster then contains that square. If it is finite, its enclosing graph cycle has length at least $L$. A proposed graph cycle using a pinned-open edge now has conditional probability zero; every other proposed graph cycle still has probability $q^n$ by independence of random variables. For $L$ large enough the union bound for the remaining graph cycles is less than one. There is therefore positive conditional probability that the origin percolation cluster is infinite. This proves the connective-constant Peierls bound without needing the exact value of $p_c$:

Set $L={\inf }_{k\ge 1}{x}_k/k$. It may be $-\infty$, but cannot be $+\infty$ because $x_1$ is finite. Fix $k$ and write $n=qk+r$, $0\le r<k$. Repeated use of the subadditive sequence inequality giveswhere $C_1=0$. Dividing by $n$ and letting $n\to \infty$ gives $\mathrm{limsup}{x}_n/n\le x_k/k$. Since this holds for every $k$, the upper limit of a sequence is at most $L$, while every ratio is at least $L$. If $L$ is finite these bounds agree. If $L=-\infty$, choose $k$ with $x_k/k$ below any prescribed real number; the same upper-limit bound proves convergence to $-\infty$. This establishes the extended-real Fekete lemma:A finite real limit of a sequence requires a lower linear bound. For example $x_n=-n^2$ satisfies the premise but has normalized limit of a sequence $-\infty$, so that qualification cannot be dropped.

In bond percolation on ${\mathbb{Z}}^d$, edges are independently open with probability $p$ and closed otherwise. An open percolation cluster is a connected component using only open edges. In site percolation it is the graph vertices that are independently declared open. The percolation probability $\theta (p)=P_p(0\leftrightarrow \infty )$ measures the chance that the origin lies in an infinite percolation cluster, and the percolation critical probability is $p_c=\inf \{p:\theta (p)>0\}$. Monotone coupling of Bernoulli percolation by independent uniform distribution makes $\theta$ nondecreasing.

The sharpness of the percolation transition says that below $p_c$ connection probabilities on ${\mathbb{Z}}^d$ decay exponentially with distance, while above $p_c$ the percolation probability is positive. There is almost surely a unique infinite percolation cluster in the supercritical regime. On the square lattice, the Harris-Kesten theorem gives bond threshold $1/2$, and there is no infinite percolation cluster at percolation critical probability. The emergence of connections on arbitrarily large scales is the geometric meaning of the phase transition.

Several observables describe this singular behavior. The percolation susceptibility is $\chi (p)=E_p|C(0)|$ for $p<p_c$. A directional correlation length can be defined through exponential two-point decay, $\xi (p{)}^{-1}=-{\lim }_n{n}^{-1}\log P_p(0\leftrightarrow n{e}_1)$. Percolation critical exponents describe leading powers, conveniently defined logarithmically rather than requiring exact asymptotic amplitudes:The first limit of a sequence approaches from above, the other two from below. At percolation critical probability, percolation two-point connection probability is expected to have leading power $|x{|}^{-(d-2+\eta )}$, and the number per site of percolation clusters of size $s$ has leading power $s^{-\tau }$. This cluster-count probability distribution differs from the size-biased probability distribution seen by choosing a site. Scaling relations for critical exponents such as $\gamma =(2-\eta )\nu$ and $2\beta +\gamma =d\nu$ connect the exponents in their applicable scaling regime; hyperscaling relation is not valid above the upper critical dimension of percolation in ordinary mean-field regimes.

The percolation universality hypothesis predicts that dimension and large-scale symmetry, rather than microscopic graph details, determine these exponents and scaling functions. Thus square-lattice bond and triangular-lattice site models are expected to share the planar graph universality class despite different microscopic definitions. Universality does not mean identical critical probabilities, amplitudes or finite-lattice distributions. Exact values and existence of exponents require model-specific theorems, not merely this prediction.

Two dimensions have an additional structure: conformal maps preserve angles, and critical crossing probabilities are expected to be unchanged under conformal transport of the domain and its marked boundary arcs. A crossing probability then depends only on the conformal shape of that marked quadrilateral. The Cardy boundary crossing formula gives this dependence in terms of its boundary cross ratio. For critical triangular-lattice site percolation, conformal invariance and the resulting exponents are established; examples are $\beta =5/36$, $\gamma =43/18$, $\nu =4/3$, with the rigorous statements in leading-power form. These values are summarized in Smirnov and Werner's original paper. Their universality extension to arbitrary planar lattices should not be presented as the same theorem.

An exploration path separating the two colors, with opposite boundary colors on two arcs, has a domain Markov property of a percolation exploration: conditional on the explored portion, the unexplored colors retain their independent laws with the newly exposed boundary colors. Conformal invariance, this domain Markov property and suitable regularity lead to Schramm–Loewner evolution. In the upper half-plane its chordal equation iswhere $B_t$ is standard Brownian motion and time is normalized by half-plane capacity. Critical percolation corresponds to $\kappa =6$. Thus the complicated discrete interface becomes a conformally natural random curve driven by one real Brownian motion. Its crossing and arm probabilities supply critical exponents and, through scaling relations for critical exponents, near-critical singularities. This makes conformal methods predictive tools for the transition, rather than only descriptions of an interface's appearance.

The series ${\sum }_{n\ge 1}{a}_n$ converges to $A$ when its partial sums $s_n$ have limit of a sequence $A$: for every $\epsilon >0$, there is $N$ such that $|s_n-A|<\epsilon$ for all $n\ge N$.

Its averages are the Cesàro means of the sequence of partial sums. To prove their convergence, choose $N$ with $|s_k-A|<\epsilon /2$ for $k\ge N$. ThenThe finite first sum divided by $n$ tends to zero. This proves the Cesàro theorem for convergent sequences in this setting, andThe converse fails. Take $a_n=(-1{)}^{n+1}$, so $s_n=1$ for odd $n$ and $s_n=0$ for even $n$. The partial sums do not converge, butThus the alternating series is Cesàro summable without having an ordinary sum.

The sequence is at least $1$. Fix $\epsilon >0$. By the binomial theorem, for $n\ge 2$,once $(n-1){\epsilon }^2>2$. Taking positive $n$th roots gives $1\le n^{1/n}<1+\epsilon$ for all sufficiently large $n$. This proves directly from the definition of a limit of a sequence that

Pairing the first and last terms of the arithmetic progression gives ${\sum }_{k=1}^{n}k=n(n+1)/2$. Therefore the limit of a sequence is

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.

Use integer graph vertices in the boxes: ${\Lambda }_n=[-n,n{]}^2\cap {\mathbb{Z}}^2$, and set ${\Lambda }_{-1}=\varnothing$. Thus $\partial {\Lambda }_0=\{0\}$, $g_0=1$, and $|\partial {\Lambda }_m|=8m$ for $m\ge 1$.

If $0\leftrightarrow \partial {\Lambda }_{m+n}$, erase loops from an open graph path to obtain a self-avoiding walk. Let $x$ be its first graph vertex on $\partial {\Lambda }_m$. Split at $x$. The prefix certifies $A_x=\{0\leftrightarrow x\}$, while a segment of the suffix certifies $B_x=\{x\leftrightarrow x+\partial {\Lambda }_n\}$: the final graph vertex is at supremum norm distance at least $n$ from $x$, so the suffix first hits that translated boundary. The two certificates use disjoint edges. Therefore the BK boundary-splitting estimate, the union bound, the Van den Berg-Kesten inequality, and translation invariance giveThe cases $m=0$ or $n=0$ follow directly from $g_0=1$, so this proves the bound including the endpoints.

Here is an explicit way to remove the polynomial boundary factor. For $p>0$ put $b_n=32n^2{g}_n$, $n\ge 1$. By symmetry in $m,n$, assume $m\le n$. ThenConsequently $a_n=\log b_n$ is a subadditive sequence. The Fekete lemma states that any real subadditive sequence satisfies $\lim a_n/n={\inf }_{n\ge 1}{a}_n/n$, possibly $-\infty$. In this case the direct horizontal open graph path gives $g_n\ge p^n$, so this limit of a sequence is bounded below by $\log p$. Since $g_n\le 1$, its upper bound is zero. Finally,has the same limit of a sequence. Thus the percolation one-arm decay rate exists andFor $p=0$, every $g_n$ with $n\ge 1$ is zero and $\gamma =0$. No exponential-decay theorem or assumption that $p$ is below the percolation critical probability is needed.

The union bound gives $g_k\le {\sum }_{x\in \partial {\Lambda }_k}{\mathbb{P}}_p(0\leftrightarrow x)$. Hence some boundary graph vertex has percolation two-point connection probability at least $g_k/(8k)$. The rotations and reflections of the square lattice let us choose such a graph vertex as $x=(k,j)$, $-k\le j\le k$. Reflection in the vertical line through $x$ sends $0$ to $e_{2k}=(2k,0)$ and fixes $x$. ThereforeThe Harris-FKG inequality applied to these two increasing events proves the reflection lower bound for two-point percolation:Every graph path from $0$ to $e_{2k}$ reaches $\partial {\Lambda }_{2k}$, so $h_{2k}\le g_{2k}$. Taking roots givesBoth outside expressions have limit of a sequence $\gamma$, proving the even case. For $p>0$, the Harris-FKG inequality with the last horizontal edge gives $h_{2k+1}\ge p{h}_{2k}$, and also $h_{2k+1}\le g_{2k+1}$. The lower bound has rootThe upper bound has the same limit of a sequence. At $p=0$ all positive-distance connection probabilities vanish. Thus ${\lim }_{n\to \infty }{h}_n^{1/n}=\gamma$ for every $p\in [0,1]$.

Let $\alpha ={\inf }_{k\ge 1}{x}_k/k$. The relevant version of the Fekete lemma is the extended-real Fekete lemma: $\alpha$ may be $-\infty$, but cannot be $+\infty$ because $x_1$ is real. We show that this is exactly the limit of a sequence.

Fix $k\ge 1$ and, for $n\ge k$, write $n=qk+r$, where $q\ge 1$ and $0\le r<k$. Iterating the defining inequality gives $x_{qk}\le q{x}_k$. If $r>0$, it also gives $x_n\le q{x}_k+x_r$; if $r=0$, there is no remainder term. Thus, with $C_k=\max (0,x_1,\dots ,x_{k-1})$ and $C_1=0$,As $n\to \infty$, $q/n\to 1/k$, so ${\mathrm{limsup}}_n{x}_n/n\le x_k/k$. This holds for every positive $k$. On the other hand every $x_n/n\ge \alpha$. If $\alpha$ is finite, these two bounds give the desired equality. If $\alpha =-\infty$, for every real $M$ choose $k$ with $x_k/k<M-1$; the same upper bound makes $x_n/n<M$ for all sufficiently large $n$. ThereforeThe index $k=0$ is excluded because its ratio is undefined, and $x_0$ is irrelevant since the defining inequality was required only at positive indices. If the printed word “limit of a sequence” is interpreted as a finite real limit of a sequence, an additional lower bound is necessary: $x_n=-n^2$ defines a subadditive sequence but $x_n/n=-n\to -\infty$. The extended-real statement is the correct unrestricted conclusion.

For each integer $j$, consider the cut of horizontal edges joining column $j$ to column $j+1$ inside the strip. It contains exactly $2k+1$ edges. Its being completely closed has probabilityDifferent cuts use disjoint edge sets, so these closed-cut events are independent. Any graph path from column zero to column $n>0$, even one that wanders to other columns first, must cross each cut with $j=0,\dots ,n-1$. None of those $n$ cuts may therefore be completely closed. This gives the closed-cut bound for percolation in a strip:Taking negative logarithms, dividing by $n$ and taking the established limit of a sequence yieldsThe strict inequality uses both finite width and $p<1$. It is consistent with $f_k(p)\le -\log p$, since $1-(1-p{)}^{2k+1}\ge p$. At $p=1$ all strip connections occur and the decay rate is zero, so that excluded endpoint would invalidate strict positivity. The mechanism is a positive probability of an impermeable finite cut, rather than any assumption that the unrestricted lattice is subcritical.

The strips are nested. Every connection permitted inside $T_k$ is also permitted inside $T_{k+1}$, so $q_k(n)\le q_{k+1}(n)$ for every fixed separation. ConsequentlyThe preceding nonnegative bound makes this a decreasing sequence bounded below. By monotone convergence of real sequences,There is no assertion that this limit of a sequence is strictly positive: positivity at each fixed width need not survive an increasing-width limit of a sequence.

One can also identify the limit of a sequence. Let $q(n)$ be the percolation two-point connection probability in the whole square lattice. Every finite connecting graph path has a bounded vertical extent, so $q_k(n)\uparrow q(n)$. Commuting infima gives the strip approximation to the planar connection decay rate:The same positive-association argument identifies the final infimum with the whole-plane normalized logarithmic limit of a sequence. This is an interchange of infima justified by monotonicity at fixed $n$. For example, when $p>1/2$, the threshold proved in question 2 and uniqueness give an infinite percolation cluster with root probability $\theta (p)>0$. The Harris-FKG inequality makes the probability that both endpoints belong to it at least $\theta (p{)}^2$, so $q(n)\ge \theta (p{)}^2$. The whole-plane rate, and hence ${\lim }_k{f}_k(p)$, is then zero, despite the strict positivity of every finite-strip rate.

If $a_n\to L$, every subsequence also has limit of a sequence $L$: its increasing indices eventually exceed any prescribed threshold. This applies in particular to every monotone subsequence.

Conversely, suppose $(a_n)$ is a bounded sequence and all its monotone subsequences converge to the same $L$. At least one exists by the monotone subsequence theorem. If $a_n$ did not tend to $L$, there would be ${\epsilon }_0>0$ and a subsequence $(a_{n_j})$ such thatApply the monotone subsequence theorem to $(a_{n_j})$. The resulting further subsequence is also a monotone subsequence of $(a_n)$, so it tends to $L$, contradicting the displayed bound. ThereforeThe bounded monotone sequence theorem ensures that these subsequences have finite limits of sequences; the hypothesis requires those values to agree.

Enumerate both sets of peaks of a sequence in increasing order of index. If $a_n\to L$, the two selected subsequences both tend to $L$.

For the converse, suppose $a_{k_r}\to L$ and $a_{m_r}\to L$. Given $\epsilon >0$, choose $r$ large enough that both selected terms differ from $L$ by less than $\epsilon$. For every $n>\max \{k_r,m_r\}$, the two peak of a sequence conditions implyThe lower bound comes from the peak of a sequence of $(-a_n)$ at $m_r$. This is precisely the definition of $a_n\to L$. Thus convergence is equivalent to a common finite limit of a sequence for the upper and lower peak subsequences.

Choose $N$ beyond every peak of a sequence of both $(a_n)$ and $(-a_n)$. Every term with index at least $N$ has a strictly larger later term and a strictly smaller later term.

Suppose for a contradiction that $a_n\to L$. Starting at $a_N$, recursively select a strictly increasing subsequence and a strictly decreasing subsequence. Both must have limit of a sequence $L$. If their second terms are $u>a_N$ and $v<a_N$, respectively, passage to their limits of sequences givesa contradiction. Hence $(a_n)$ does not converge. This argument includes the possibility that either set of peaks of a sequence is empty.