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The supremum of an increasing bounded sequence need not be a maximum of its set of terms. For $a_n=1-1/n$, every term is strictly less than one; the supremum and limit of a sequence are one, but there is no maximum. Every candidate term is beaten by the next. The decreasing sequence $1+1/n$ similarly has infimum one without a minimum.

If an increasing sequence with limit of a sequence $L$ attains $a_k=L$, it must remain there: $L=a_k\le a_n\le L$ for $n\ge k$. The decreasing case is identical with inequalities reversed. Thus attainment of the limit of a sequence is equivalent to an eventually constant tail of a sequence for a convergent monotone sequence. The increasing sequence $0,1/2,1,1,\dots$ has both supremum and maximum one. ^[ref]

A Cauchy sequence is a sequence whose sufficiently late terms are all close to one another. In a metric space $(X,d)$, the precise condition isFor real numbers with their usual metric, $d(x_m,x_n)=|x_m-x_n|$. The useful feature is that the definition compares computed terms to other computed terms: it does not require an already known limit of a sequence. The one index $N$ must control every pair of later indices, not merely successive terms. ^[ref].

For an explicit example, take $q_n=1-2^{-n}$ for $n\ge 1$. Its tail of a sequence beginning at $N$ lies in $[1-2^{-N},1)$, and for every $m,n\ge N$,Choose $N$ with $2^{-N}<\epsilon$ to certify the entire future sequence. For $\epsilon =0.1$, $N=4$ works because $2^{-4}=0.0625$. The limit of a sequence is one, but the pairwise bound establishes the Cauchy sequence property without first needing that answer. By contrast, $a_n=(-1{)}^n$ is not a Cauchy sequence: arbitrarily late even integers and odd integers give values separated by two. ^[ref].

Tiny successive changes do not provide this guarantee. For the harmonic numbers $H_n={\sum }_{j=1}^{n}1/j$, the successive difference $H_{n+1}-H_n=1/(n+1)$ tends to zero, butThus, beyond every proposed $N$, choosing $n\ge N$ and $m=2n$ defeats the fixed tolerance $\epsilon =1/2$. Even every fixed positive lag $h$ passes the weaker test: $0<H_{n+h}-H_n\le h/(n+1)\to 0$. The failing lag grows with $n$, as explained more generally by convergence from arbitrary variable-lag increments. ^[ref].

Every convergent sequence in a metric space is a Cauchy sequence: apply the triangle inequality to two late terms and their common limit of a sequence. The reverse implication requires completeness. It holds in the real numbers by completeness of the real numbers, but can fail in the rational numbers: the rational endpoints from the bisection method approaching $\sqrt{2}$ are a Cauchy sequence with no rational limit of a sequence. The existing construction in completeness of the real numbers shows how this happens. A Cauchy sequence is always a bounded sequence; in $\mathbb{R}$, boundedness gives a convergent subsequence by the Bolzano-Weierstrass theorem, and the pairwise condition forces the whole sequence to share its limit. ^[ref].

The limit inferior of a real sequence isallowing a value in the extended real number line. For a finite value $L$, it has two complementary meanings: for every $\epsilon >0$, all sufficiently late terms satisfy $a_n>L-\epsilon$, while arbitrarily late terms satisfy $a_n<L+\epsilon$. The first prevents persistent dips below $L$; the second prevents any larger number from being the true lower edge.

For $a_n=(-1{)}^n(1+1/n)$, the lower bounds in limit inferior and limit superior approach $-1$. Every late term eventually exceeds $-1-\epsilon$, and the subsequence at odd integers comes arbitrarily close to $-1$. For $a_n=1/n$, the lower bound of every infinite tail of a sequence is already zero, although no term equals zero. Thus a limit inferior need not be the minimum of any tail.

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.

Choose $M$ with $|b_n|\le M$. Every tail of a sequence is nonempty and bounded, so its supremum $c_n$ exists and lies in $[-M,M]$. Removing one term cannot increase a supremum, hence $c_{n+1}\le c_n$. The bounded monotone sequence theorem, applied to $-c_n$, provesThus $L$ is the limit superior of $(b_n)$.

For $\epsilon >0$, eventually $c_n<L+\epsilon$, and then $b_n\le c_n<L+\epsilon$. Only finitely many indices can satisfy $b_n\ge L+\epsilon$. If only finitely many satisfied $b_n>L-\epsilon$, some entire tail of a sequence would have $b_n\le L-\epsilon$. Its supremum would also be at most $L-\epsilon$, contradicting $c_n\ge L$.

Inductively choose $n_k>n_{k-1}$ so late that $c_{n_k}<L+1/k$ and $b_{n_k}>L-1/k$. There are infinitely many choices for the latter inequality, so both requirements can be met. Then $|b_{n_k}-L|<1/k$, proving $\boxed{b_{n_k}\to L}$. This constructs a subsequence attaining the limit superior.

Use uniform convergence to choose $N$ with $|f_n(x)-f_N(x)|\le 1$ for all $n\ge N$ and $x\in E$; this follows, for example, from the Uniform Cauchy criterion. Since each $f_j$ is a bounded function, put $M_j={\sup }_E|f_j|<\infty$ for $1\le j\le N$. Thenfor every $n$ and $x$. Thus $(f_n)$ is a uniformly bounded family of functions. The finite initial segment is essential: uniform convergence directly controls only a tail of a sequence.

A sequence assigns a value to each position in an ordered index list. For an infinite sequence in a set $X$, one standard convention is a function $x:{\mathbb{N}}_{>0}\to X$, written $(x_n{)}_{n=1}^{\infty }$ with $x_n=x(n)$. Some applications start the index at zero, and a finite sequence has only finitely many positions. Values may repeat; both order and repetition matter. ^[ref].

For example, $a_n=(-1{)}^n$ begins $-1,1,-1,1,\dots$, whereas $b_n=(-1{)}^{n+1}$ begins $1,-1,1,-1,\dots$. Their sets of values are both $\{-1,1\}$, but these are different sequences. A subsequence keeps selected positions in increasing order; the subsequence of $a_n$ at its even integers is the constant sequence $1,1,1,\dots$. A tail of a sequence retains all positions from one index onward.

Sequences express successive approximations, repeated observations, and the iterates of a recurrence relation. A limit of a sequence describes where its terms eventually concentrate. Finite observations alone cannot establish that behaviour: for any positive integer $K$, the all-zero sequence and the sequence with $u_n=0$ for $n\le K$ and $u_n=1$ for $n>K$ agree through index $K$, yet have different limits of sequences. A formula or bound controlling the entire future, such as the tail diameter of a sequence in the Cauchy sequence condition, supplies information that a finite list cannot.

Define a sequence by $a_n=1$ when $n$ is a power of two, including $1=2^0$, and $a_n=0$ otherwise. Every tail of a sequence contains both a later power of two and a later nonpower. Therefore $\mathrm{liminf}{a}_n=0$ and $\mathrm{limsup}{a}_n=1$, and the sequence does not converge. Its peak indices become increasingly sparse, yet never stop appearing.

Through index $N\ge 1$, the number of peaks is $1+\lfloor {\log }_{2}N\rfloor$, so their fraction isThe set of powers of two has natural density zero. The same fraction is the Cesàro mean of the first $N$ terms, which tends to zero even though the limit superior stays one. At $N=1024$, eleven peaks give mean $11/1024\approx 0.0107422$. An average and a long-run upper edge answer different questions about the same sequence.