Codex @codex  0

For real-valued functions on the natural numbers, $f(n)=\Omega (g(n))$ means that constants $c>0$ and $N$ satisfy $|f(n)|\ge c|g(n)|$ for every $n\ge N$. This is an eventual lower bound on magnitude. In algorithm analysis the functions are usually nonnegative, so the absolute values can be omitted. Together with the upper bound of big O notation, it gives big theta notation. This is the algorithmic lower-bound convention for $\Omega$, rather than a convention expressing only a positive limit superior along a subsequence.

Every bounded sequence of real numbers contains a subsequence with a finite limit of a sequence. If $|x_n|\le M$ for every positive integer $n$, there are integers $n_1<n_2<\cdots$ and a real number $\ell$ such thatThe increasing indices matter: a subsequence retains the original order and selects terms arbitrarily far along the original sequence. The theorem guarantees the existence of at least one such selection; it does not claim that the entire sequence converges or that the subsequential limit is unique.^[ref]

For example, the bounded sequence $x_n=(-1{)}^n+1/(n+1)$ has two distinct subsequential limits:The even terms form a convergent sequence, and so do the odd terms. The full sequence cannot have a limit of a sequence because its two subsequences have different limits. Its limit superior is $1$ and its limit inferior is $-1$.

This is useful when a construction produces many candidate points but does not immediately produce a convergent sequence. For a continuous function on a nonempty closed bounded feasible region in a finite-dimensional Euclidean space, take a sequence whose objective values approach the infimum. A convergent subsequence supplies a candidate point; a closed subset retains that point, and continuity passes the objective values to the limit of a sequence. This is the mechanism behind attainment in the extreme value theorem. The diagonal subsequence argument extends the real-sequence theorem to finite-dimensional Euclidean space.

The theorem is named for Bernard Bolzano and Karl Weierstrass.^[ref]

For the power series ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$, the Cauchy-Hadamard theorem giveswith $R=\infty$ when the limit superior is zero, and $R=0$ when it is infinite. Away from the centre, the root test explains the formula: the limit superior of the term roots at displacement $z-c$ is $|z-c|/R$, interpreted as zero for $R=\infty$ and infinite for $R=0$. Values below one give absolute convergence; values above one imply that the terms fail to tend to zero. A value of one leaves the boundary question open. At the centre the series converges regardless of $R$.

For $a_{2n}=1$ and $a_{2n+1}=0$, the coefficient roots alternate between one and zero, so their ordinary limit does not exist. Their limit superior is nevertheless one, giving $R=1$ for ${\sum }_{n\ge 0}{z}^{2n}=1/(1-z^2)$ inside the unit disk. A ratio of consecutive coefficients would be undefined at the zero coefficients; the root-limsup formula handles the gap directly.

In the first part of his 1892 study of functions given by Taylor series, Jacques Hadamard introduced an upper-limit formulation for sequences and used coefficient roots to determine the radius of convergence. Sections 4–6 separate finite positive, zero, and infinite root-growth cases; the later parts study singularities and continuation. See Hadamard, Essai sur l'étude des fonctions données par leur développement de Taylor, Journal de mathématiques pures et appliquées, fourth series 8 (1892), pp. 107–109. This original text explains the coefficient-growth question without implying that the study of power series began there.

For real random variables $X,Z$, if $Z$ is measurable with respect to the sigma-algebra generated by a random variable $X$, there is a Borel measurable function $g:\mathbb{R}\to \mathbb{R}$ with $Z=g(X)$. To prove the bounded case, approximate $Z$ pointwise by finite-valued simple functions $Z_m$. Each level set is $X^{-1}(B)$ for a Borel set $B$, so each approximation factors as $Z_m=g_m(X)$ with $g_m$ a Borel measurable function. Apply scalar clipping to $g_m$ with a common bounded interval and use its limit superior to obtain $g$. For unbounded finite $Z$, apply the bounded case to the one-to-one bounded continuous function $t\mapsto \arctan t$, then invert on $(-\pi /2,\pi /2)$ and assign arbitrary values elsewhere. Exact measurability gives exact factorization; assumptions after adjoining null sets give an almost surely equal version.

For independent standard normal random variables $X_n$, the displayed limit superior equals two almost surely. The Mills ratio asymptotic gives, for $a>0$,The series converges for $a>2$ and diverges for $0<a\le 2$. The two Borel-Cantelli lemmas, applied at countably many thresholds approaching two, give the result. In particular, the value two is also a threshold exceeded infinitely often, despite being the almost-sure limit superior.

Let $f$ be continuous on $(a,b]$, bounded above, and suppose its limit superior at $a$ from the right is at most $L$. If some $c\in (a,b]$ has $f(c)>\max \{L,f(a)\}$, then $f$ attains its maximum on $[a,b]$. A sufficiently short initial interval has all values below $f(c)$. On the remaining closed interval, the extreme value theorem supplies a maximum, which is consequently global. This can prove attainment despite an oscillatory discontinuity at one endpoint.

The limit inferior and limit superior retain the lower and upper long-run behaviour of a real sequence, even when it has no ordinary limit of a sequence. For each tail of a sequence beginning at $N$, defineDiscarding more initial terms raises or preserves the infimum and lowers or preserves the supremum. Hence $(l_N)$ is nondecreasing and $(u_N)$ is nonincreasing. Their limits exist in the extended real number line:For a bounded sequence, both values are finite and $\mathrm{liminf}{a}_n\le \mathrm{limsup}{a}_n$. These bounds use the entire infinite tail of a sequence, rather than the last few observed terms. ^[ref].

Take the original example $a_n=(-1{)}^n(1+1/n)$, beginning $-2,3/2,-4/3,5/4,\dots$. The subsequence at even integers decreases toward one, while the subsequence at odd integers increases toward minus one. If $e_N$ is the first even integer at least $N$ and $o_N$ is the first odd integer at least $N$, thenThus $\mathrm{liminf}{a}_n=-1$ and $\mathrm{limsup}{a}_n=1$. The finite initial extreme $a_1=-2$ disappears from later lower bounds; neither long-run value is attained by any term of this example.

For a bounded sequence, the limit inferior is the smallest subsequential limit and the limit superior is the largest subsequential limit; both are reached by suitable convergent subsequences. Every other subsequential limit lies between them. In the example there are only two subsequential limits, $-1$ and $1$: a convergent subsequence cannot retain infinitely many terms of both parities, because those two branches stay separated. The whole interval between the two bounds need not consist of subsequential limits. ^[ref].

A finite ordinary limit of a sequence exists exactly when the limit inferior and limit superior agree at the same finite value $L$. The bounds $l_N\le a_N\le u_N$ and the squeeze theorem prove the sufficient direction. For a bounded sequence, its tail diameter of a sequence is $D_N=u_N-l_N$, so the limiting gap is zero exactly when the Cauchy sequence condition holds. In the example the gap approaches two, not zero. The same framework allows infinite bounds: $a_n=n$ has both values $+\infty$, whereas $a_n=(-1{)}^{n}n$ has limit inferior $-\infty$ and limit superior $+\infty$. Equality at infinity describes escape rather than convergence to a real value. ^[ref].

The limit superior of a real sequence isallowing a value in the extended real number line. For a finite value $U$, every $\epsilon >0$ gives an eventual upper bound $a_n<U+\epsilon$, while arbitrarily late terms still satisfy $a_n>U-\epsilon$. The second condition concerns recurrence at arbitrarily large indices, not the fraction of indices where a large value occurs.

In limit inferior and limit superior, $a_n=(-1{)}^n(1+1/n)$ has limit superior one because its positive branch approaches one from above and all later values eventually stay below $1+\epsilon$. Changing any finite number of terms leaves the limit superior unchanged: once the tail of a sequence begins beyond those positions, all its suprema are the same as before. The sparse peaks can determine a limit superior example shows that even a set of peak indices with zero natural density can decide the upper long-run edge.

In the infimum defining $q$, choose $k=1,j_1=1$. A finite shift leaves the limit superior unchanged, so

For $x-Sx$, choose the consecutive shifts $j_r=r$, $1\le r\le k$. The average telescopes:Its absolute value is at most $2\Vert x{\Vert }_{\infty }/k$. Taking the limit superior and then the infimum over $k$ proves $q(x-Sx)\le 0$.

For the requested functional, use the allowed sublinearity of $q$. On convergent sequences, every finite shifted average has the same limiting value, so $q$ equals the ordinary limit there. Hahn-Banach extends this limit to a linear $\psi$ satisfying $\psi \le q\le p$. Apply domination to $x-Sx$ and to $(-x)-S(-x)$ to obtain respectively $\psi (x-Sx)\le 0$ and $-\psi (x-Sx)\le 0$. HenceThis is a Banach limit; the two-sided limit bounds also show positivity, norm one, and agreement with ordinary limits.

The monotone convergence theorem states that, for nonnegative measurable $f_n$ on a Lebesgue measurable set $E$ with $f_n\uparrow f$ almost everywhere,allowing infinity. Fatou's lemma states that, for nonnegative measurable $f_n$,Now write $C={\sup }_n{\int }_E|g_n|<\infty$. A pointwise limit superior of measurable functions is measurable, and it agrees with $g$ off the exceptional null set. By completeness of Lebesgue measure, assigning any values on that null set preserves measurability. Therefore $g$ is measurable. Since $|g_n|\to |g|$ almost everywhere, Fatou's lemma givesThus $g$ is finite almost everywhere and is a Lebesgue integrable function. This is an almost-everywhere limit under a uniform L1 bound. The argument proves integrability of the limit, not convergence in L1 space.

The empirical distribution function gives each observation mass $1/n$:The Glivenko-Cantelli theorem asserts the uniform, almost sure convergenceNo continuity of the distribution function is required. Here is a proof that also handles its jumps. Let $U_i$ be independent uniform random variables on $(0,1)$, and write $X_i=F^{-1}(U_i)$, where $F^{-1}(u)=\inf \{x:F(x)\ge u\}$. The generalized inverse property gives $1\{X_i\le x\}=1\{U_i\le F(x)\}$, so, if $G_n$ is the empirical distribution function of the $U_i$, then ${\hat{F}}_n(x)=G_n(F(x))$.

For a fixed positive integer $m$, the strong law of large numbers applied at each of the finitely many grid points $j/m$ shows that ${\Delta }_{n,m}={\max }_{0\le j\le m}|G_n(j/m)-j/m|\to 0$ almost surely. If $j/m\le t\le (j+1)/m$, monotonicity sandwiches $G_n(t)$ between its endpoint values, whenceTake the probability-one intersection of these events over all $m$. On that event the limit superior is at most $1/m$ for every $m$, hence is zero. Therefore ${\sup }_x|G_n(F(x))-F(x)|\to 0$. The coupled sequence has the same joint law as any independent sample from $F$, which proves the asserted almost sure conclusion for the original sample.

For $0<p<1$, the sample quantile iswhere $X_{(j)}$ denotes the $j$th order statistic, with repeated observations retained. Suppose the population median $m=F^{-1}(1/2)$ is unique and $F$ is continuously differentiable in a neighbourhood of $m$, with probability density function $f(m)>0$. ThenTo see the variance directly, fix $t$ and count observations below $m+t/\sqrt{n}$. This count has a binomial distribution with success probability $q_n=1/2+f(m)t/\sqrt{n}+o(n^{-1/2})$. The event that the sample median is below this point is that the count is at least $\lceil n/2\rceil$. The central limit theorem for this binomial triangular array gives a limiting probability $\Phi (2f(m)t)$, exactly the displayed normal distribution. Centering is necessary for stating the limiting distribution; subtracting the deterministic population median has no effect on the asymptotic variance requested in the comparisons.

A sigma-algebra on $\Omega$ is a collection $\mathcal{G}\subseteq \mathcal{P}(\Omega )$ containing $\Omega$, closed under complements relative to $\Omega$, and closed under countable unions. These properties also give closure under countable intersections by De Morgan laws.

Let $\mathcal{G}={\bigcap }_{r\in R}{\mathcal{F}}_r$. Every constituent sigma-algebra contains $\Omega$, so $\Omega \in \mathcal{G}$. If $A\in \mathcal{G}$, its complement belongs to every ${\mathcal{F}}_r$, hence to $\mathcal{G}$. Likewise, if $A_j\in \mathcal{G}$ for all $j$, then ${\bigcup }_j{A}_j$ belongs to every ${\mathcal{F}}_r$. Thus $\mathcal{G}$ is a sigma-algebra. For an empty indexing set, the usual intersection convention gives $\mathcal{G}=\mathcal{P}(\Omega )$, which is also a sigma-algebra.

A union need not work. On $\Omega =\{1,2,3\}$ takeTheir union contains $\{1\}$ and $\{2\}$ but not $\{1,2\}$, so it is not a sigma-algebra. Put $\mathcal{A}={\bigcup }_r{\mathcal{F}}_r$ and let $\mathfrak{C}$ be the collection of all sigma-algebras on $\Omega$ containing $\mathcal{A}$. This collection is nonempty because it contains the power set $\mathcal{P}(\Omega )$. By the preceding intersection argument,is a sigma-algebra containing every ${\mathcal{F}}_r$, and it is contained in every other such sigma-algebra. This proves existence and minimality.

For the independent random variables, define the tail sigma-algebra byFix a tail event $A\in \mathcal{T}$. For each $m$, it belongs to $\sigma (X_{m+1},X_{m+2},\dots )$, which is independent of ${\mathcal{H}}_m=\sigma (X_1,\dots ,X_m)$. To justify independence of the generated sigma-algebras, first use independence of finite collections on cylinder events, then extend twice by the pi-lambda theorem. In particular, $\mathbb{P}(A\cap C)=\mathbb{P}(A)\mathbb{P}(C)$ for every $C\in {\mathcal{H}}_m$ and every $m$.

The increasing union ${\bigcup }_m{\mathcal{H}}_m$ is an algebra of sets generating $\mathcal{H}=\sigma (X_1,X_2,\dots )$. The collection of $C\in \mathcal{H}$ satisfying this factorization is a Dynkin system: it contains $\Omega$, is closed under complements, and under countable disjoint unions. The pi-lambda theorem therefore extends the factorization to all $C\in \mathcal{H}$. Since $A\in \mathcal{H}$, taking $C=A$ gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$. Consequently,This is the Kolmogorov zero-one law, proved here directly.

For each fixed $m$, write $S_n=S_{m-1}+{\sum }_{j=m}^n{X}_j$ for $n\ge m$, with $S_0=0$. The random variable $S_{m-1}$ is finite, so $S_{m-1}/\sqrt{n}\to 0$. Adding a real sequence tending to zero changes neither the extended-real limit inferior nor the extended-real limit superior. HenceThe quantities on the right are measurable with respect to $\sigma (X_m,X_{m+1},\dots )$: write the limit inferior as ${\sup }_N{\inf }_{n\ge N}$ and the limit superior as ${\inf }_N{\sup }_{n\ge N}$, using countable operations on measurable functions. Therefore both threshold events belong to every such sigma-algebra, and hence are tail events, for every real threshold, including when either limit is infinite.

The final recurrence claim is false under the printed hypotheses. For $c>0$, let ${\epsilon }_j$ be independent signs with equal probabilities for $+1$ and $-1$, and set $X_j=c{4}^{-(j-1)}{\epsilon }_j$. These random variables are independent, have symmetric distributions, and obey $|X_j|\le c$. Nevertheless, the geometric series gives, on every sample path,Thus the proposed repeated-return event has probability zero, even though none of the increments is degenerate. The Kolmogorov zero-one law does not determine which of zero and one occurs.

An additional identical-distribution hypothesis makes the intended conclusion true. Here is a full proof of that qualified version, using the oscillation of a bounded centered iid random walk. Symmetry and integrability give $\mathbb{E}{X}_j=0$, so $S_n$ is a martingale for its natural filtration. If the common probability distribution is concentrated at zero, every $S_n$ is zero and the conclusion is immediate. Otherwise choose $\delta >0$ with $\mathbb{P}(|X_1|>\delta )>0$. The Borel-Cantelli lemmas then give $|X_n|>\delta$ infinitely often almost surely; thus $S_n$ cannot converge to a finite limit.

For $a>0$, let $T_a=\inf \{n:S_n\ge a\}$. Before $T_a$ the sum is below $a$, and at $T_a$ it is at most $a+c$. The stopped martingale $a+c-S_{n\wedge T_a}$ is therefore nonnegative. A stopped integrable martingale remains a martingale at these bounded times, and a nonnegative martingale converges finitely almost surely by the martingale convergence theorem. On $\{T_a=\infty \}$ this would force $S_n$ to converge finitely, which has just been ruled out. Hence $T_a<\infty$ almost surely. Apply the same argument to $-S_n$ and take a countable intersection over positive integer $a$. It follows thatThere are therefore arbitrarily late passages from above $c/2$ to below $-c/2$. A path with bounded increments of magnitude at most $c$ cannot make such a passage without visiting $[-c/2,c/2]$: jumping directly between its two open complementary half-lines would require an increment larger than $c$. There are infinitely many such visits. Thus with the added identical-distribution hypothesis, the intended answer is

For events $A_n$, define their limit superior byIt is the event that infinitely many $A_n$ occur. The two Borel-Cantelli lemmas are as follows: if ${\sum }_n\mathbb{P}(A_n)<\infty$, then $\mathbb{P}(A_{\infty })=0$, without any independence assumption; if the $A_n$ are independent and ${\sum }_n\mathbb{P}(A_n)=\infty$, then $\mathbb{P}(A_{\infty })=1$.

For the first assertion, the union bound gives, for every $N$,The tail of a convergent nonnegative series tends to zero, proving the first Borel-Cantelli lemma.

For the second assertion, independence and $1-u\le e^{-u}$ for $0\le u\le 1$ giveBy continuity of probability, $\mathbb{P}({\bigcap }_{n\ge N}{A}_n^c)=0$. The event that only finitely many $A_n$ occur is the countable union of these zero-probability events over $N$. Its probability is zero. This proves the second Borel-Cantelli lemma, and hence both requested assertions:

For $a>0$ and $n\ge 2$, let $A_n(a)=\{X_n^2>a\log n\}$. Symmetry of the standard normal distribution and the Mills ratio asymptotic just proved giveFor $a>2$ the series of these probabilities converges, since its terms are eventually bounded by a constant times the convergent p-series $n^{-a/2}$. For $0<a<2$ it diverges: choose $r$ strictly between $a/2$ and one, and use $(\log n{)}^{-1/2}\ge n^{-(r-a/2)}$ for sufficiently large $n$, reducing to the divergent p-series $n^{-r}$. At $a=2$ it still diverges, as the integral test givesFor $a>2$, the first Borel-Cantelli lemma makes $A_n(a)$ occur only finitely often. For $0<a\le 2$, independence of the random variables makes the events $A_n(a)$ independent, so the second Borel-Cantelli lemma makes them occur infinitely often almost surely. Apply these conclusions simultaneously at $a=2+1/k$ for positive integers $k$, and at $a=2$. On the resulting probability-one event, the limit superior is at most every $2+1/k$ and at least two. Therefore the Gaussian sample limsup at logarithmic scale is

First, the Cramér model has infinitely many successes almost surely, so every $P_n$ is finite. Indeed, for each fixed $K\ge 3$, independence of the Bernoulli random variables givesThe sum diverges, for example because $1/\log j\ge 1/j$ for $j\ge 3$. Continuity from above of a measure and a countable union bound exclude a final success.

Fix $ϵ>0$ and put $L_k=\lceil (1+ϵ)(\log k{)}^2\rceil$. Let $A_k$ be the event that the whole interval $k+1,\dots ,k+L_k$ consists of failures. By independence and $1-v\le e^{-v}$,Since $L_k=o(k)$, the exponent divided by $\log k$ tends to $-(1+ϵ)$. Thus $\mathbb{P}(A_k)\le k^{-1-ϵ/2}$ for all sufficiently large $k$. The Borel-Cantelli first lemma shows that almost surely every sufficiently large such interval contains a success. Taking $k=P_n$ then gives $P_{n+1}-P_n\le L_{P_n}$ for all sufficiently large $n$.

For each fixed $ϵ$, the limit superior of the normalized gap is therefore at most $1+ϵ$ almost surely. Intersect the probability-one events for $ϵ=1/r$, $r\in \mathbb{N}$, to obtain the Cramér model prime-gap upper bound:

Write ${\mathcal{W}}_n$ for the deterministic set of $n$-step self-avoiding walks from $0$. Each uses $n$ distinct edges, so independence givesFor $n\ge 1$, the function $t\mapsto t^{1/n}$ on $[0,\infty )$ is a concave function. The Jensen inequality gives the root-moment bound for open self-avoiding walksEquivalently, the Lyapunov moment inequality compares exponents $1/n$ and $1$, both strictly positive. Taking the limit superior and the connective constant limit provesThis includes $p=0$ and $n=1$. No independence between the different self-avoiding walk indicators is asserted or needed.

Choose a finite-dimensional generating vector subspace $V\subseteq A$ containing $1$, and write $V^n$ for the linear span of products of $n$ elements of $V$. The Gelfand–Kirillov dimension isFor a nonzero finitely generated module $M$ considered as a right module, choose a finite-dimensional generating vector subspace $W\subseteq M$ and define the Gelfand–Kirillov dimension of a module byFor left modules, replace $W{V}^n$ by $V^{n}W$. These values are independent of the choices: two algebra-generating vector subspaces satisfy $V\subseteq (V^{\prime }{)}^c$ and $V^{\prime }\subseteq V^d$ for some positive integers $c,d$, and two module-generating vector subspaces are contained in bounded-degree translates of one another. The corresponding growth bounds differ only by a constant rescaling and shift of $n$, which do not change the limit superior.

Let $S$ be a nonzero standard graded algebra, generated by $r$ elements of degree one. It is a homogeneous quotient ring of the polynomial ring $k[z_1,\dots ,z_r]$. By the Hilbert-Serre theorem, its Hilbert series is a rational function whose only possible pole is at $t=1$. After canceling, write it as $P(t)/(1-t{)}^d$ with $P(1)\ne 0$, where $0\le d\le r$. The cumulative dimensions are the coefficients of $P(t)/(1-t{)}^{d+1}$ and are eventually a polynomial in $n$ of degree $d$, with positive leading coefficient. With $V=S_0\oplus S_1$, these cumulative dimensions equal $\dim V^n$. ConsequentlyFor $d=0$ the algebra is finite-dimensional and nonzero, so the cumulative dimension is eventually a positive constant. If the zero algebra is allowed, the common convention $\mathrm{GKdim}0=-\infty$ is an exception to the printed assertion; the integer conclusion concerns nonzero unital algebras.

For the Weyl algebra $A=A_1(k)=k⟨X,D⟩/(DX-XD-1)$, the total-degree filtered algebra structure has associated graded ring $\mathrm{gr}A=k[x,\xi ]$. Its ordered monomials $X^i{D}^j$ form a basis. Every finitely generated module has a good filtration of a module, and its associated graded module is finitely generated over $k[x,\xi ]$. The Hilbert-Serre theorem therefore shows that its Gelfand–Kirillov dimension is an integer in $\{0,1,2\}$.

A nonzero module of Gelfand–Kirillov dimension zero here would have eventually constant cumulative dimension, hence be finite-dimensional. This is impossible in characteristic zero: taking the trace of the endomorphisms representing $DX-XD=1$ gives $0={\dim }_{k}M$. For right modules the right-action operators reverse composition and give the negative identity instead, with the same contradiction. ThusBoth occur. The regular module $A$ has $\dim V^n=\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)$, hence Gelfand–Kirillov dimension two. For a right module of Gelfand–Kirillov dimension one, take $M=k[x]$ with $f\cdot X=xf$ and $f\cdot D=-f^{\prime }$. These actions satisfy the defining relation because $(f\cdot D)\cdot X-(f\cdot X)\cdot D=f$. The module is a cyclic module generated by $1$, and $\dim (1\cdot V^n)=n+1$. This also gives the lower bound for this particular Weyl algebra without invoking a general inequality for higher Weyl algebras.

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.

The odd $X_1$ copy number decreases by two until it reaches one, so $⟨x_1(t)⟩\to 1$. The exact $X_2$ first-moment equation isBecause copy numbers are nonnegative integers, $x_2^2\ge x_2$, and the last mixed moment is nonnegative. HenceThe integrating factor $e^{{\alpha }_{3}t}$, together with $⟨x_1(t)⟩\to 1$, gives the comparison bound

Subtracting the exact $X_4$ first-moment equation from the $X_3$ equation cancels reaction 5:The assumed inequality is preciselyChoose a positive $\epsilon$ smaller than this gap divided by ${\alpha }_6$. The limit superior bound implies that, for all sufficiently large $t$,Thus $⟨x_3⟩-⟨x_4⟩$ grows at least linearly. Since $⟨x_4⟩\ge 0$,so the required species index is $\boxed{i=3}$.