Tail extrema and subsequential limits

Try

For positive integers n, examine xₙ=(−1)ⁿ+1/n. Its first terms are 0, 3/2, −2/3, 5/4, −4/5, 7/6. The even terms approach 1 from above; the odd terms approach −1 from above. The whole sequence does not converge, but its two persistent extremes have precise meanings.

For each starting index N, look at the entire infinite tail n≥N. Define U(N) as its supremum and L(N) as its infimum. These are not necessarily values achieved by terms. Predict U(3) and L(3), then compare the table.

Each displayed supremum is attained at the first eligible even term. The infimum −1 is never attained: every odd term is greater than −1, but odd terms come arbitrarily close. A finite table of inspected values would not by itself establish that infimum; the formula and its limiting behavior do.

Predict

Removing early terms cannot increase the supremum and cannot decrease the infimum. Therefore U(N) decreases weakly with N, while L(N) increases weakly. Predict their limiting values for the worked sequence. They are 1 and −1 respectively, called the limit superior and limit inferior, or limsup and liminf.

Now imagine only the first 32 terms of a sequence have been shown, all zero. Could the sequence converge to 1? Yes: define every term from index 64 onward to be 1. Could it be unbounded? Yes: continue with value n at index n after the displayed prefix. A finite sample can be accurate about every observed term while giving no certificate about the unseen tail.

An infinite tail always contains terms because N is a positive integer and the sequence is defined for all positive integers. Nevertheless it can be unbounded above or below. We use extended-real values: the supremum of a tail unbounded above is +∞, and its infimum when unbounded below is −∞. These values describe bounds; they are not real-valued terms of the sequence.

Experiment

The explorer separates a finite window from an independently justified analytic statement about its prescribed infinite sequence. In the default delayed example, the window ends at 32 while the change occurs at 64. Predict what the finite maximum reports and what the actual limit is before revealing the infinite-tail result. Then change the window endpoint to 64. The newly visible jump changes the sample, not the definition of the sequence.

What can a finite tail window show?

Choose a sequence, then predict the smallest and largest terms from index k through N. This is a finite list, not the infinite tail beginning at k. All indices start at 1.

Prescribed sequenceAlternating: (−1)^n + 1/n1 at powers of two; 0 otherwiseDelayed step: 0 before M, 1 from M Window endpoint N8163264 Tail start k1234567891011121314151617181920212223242526272829303132 Delayed cutoff M163264128 Reset

Selected finite range: k=1 through N=32, containing 32 terms. Its minimum is 0 and maximum is 0. These are not infinite-tail infimum and supremum.

Circles mark the selected finite tail; small squares mark earlier displayed terms. Blank space after N means not displayed, not zero. Fractions such as 1/n are evaluated numerically; displayed decimals are rounded.

Reveal the prescribed infinite-tail result

Complete displayed window; finite tail selection is explicit.

Index n	Value	In k through N?
1	0	Yes
2	0	Yes
3	0	Yes
4	0	Yes
5	0	Yes
6	0	Yes
7	0	Yes
8	0	Yes
9	0	Yes
10	0	Yes
11	0	Yes
12	0	Yes
13	0	Yes
14	0	Yes
15	0	Yes
16	0	Yes
17	0	Yes
18	0	Yes
19	0	Yes
20	0	Yes
21	0	Yes
22	0	Yes
23	0	Yes
24	0	Yes
25	0	Yes
26	0	Yes
27	0	Yes
28	0	Yes
29	0	Yes
30	0	Yes
31	0	Yes
32	0	Yes

Try the delayed sequence with N=32 and M=64, then extend N to 64. Next choose power-of-two spikes and set k=N=16. Explain why a singleton finite minimum of 1 does not prove an infinite-tail lower bound of 1. Controls are temporary and award no completion.

For the power-of-two spike example, choose a window endpoint of 16 and tail start 16. This finite window contains only the spike at 16, so its minimum is 1. At non-power-of-two indices, the prescribed sequence is zero. There are infinitely many such indices, so the infinite-tail infimum is 0. Explain why replacing an infinite tail with a finite window loses precisely the terms needed for this conclusion.

Paper action: a changed alternating sequence

For yₙ=3+(−1)ⁿ/2, every tail contains both 5/2 and 7/2. Its tail infimum and supremum are therefore 5/2 and 7/2 for every N, both attained infinitely often. Contrast this with the worked sequence, whose infimum is unattained and whose supremum changes as early even terms are removed. The same endpoint vocabulary applies to both situations.

Understand

For any real sequence define U(N)=sup of the terms with n≥N and L(N)=inf of those terms, allowing infinite bounds. The tail sets decrease by inclusion, so U is nonincreasing and L nondecreasing. Define β=limsup xₙ=inf over N of U(N) and α=liminf xₙ=sup over N of L(N). Monotonicity means these are also the extended-real limits of the two bound sequences. For any starting indices N and M, choose a term at index at least both; it lies above L(N) and below U(M). Hence L(N)≤U(M), and taking the appropriate supremum and infimum gives α≤β. No expression such as ∞−∞ is needed.

Finite endpoints. If β is real, then for every ε>0, eventually xₙ<β+ε, and infinitely often xₙ>β−ε. For the first assertion, the definition of infimum supplies a tail with U(N)<β+ε. For the second, every tail has supremum at least β, so it must contain some term greater than β−ε. Otherwise β−ε would be an upper bound below that supremum. Having such a term in every tail means infinitely many indices satisfy the inequality.

Conversely, these two properties force limsup to equal β. The eventual upper bounds put its value at most β+ε for every ε, and the infinitely-often lower excursions put every tail supremum at least β−ε. Letting ε shrink gives equality. For a finite α the reversed characterization is: eventually xₙ>α−ε, and infinitely often xₙ<α+ε. These are statements about all sufficiently late terms and about arbitrarily late witnesses, respectively; neither can be replaced by a few displayed samples.

Construct an endpoint subsequence. Suppose β is finite. At stage k, choose a tail beginning beyond the previously chosen index and beyond a threshold where every term is below β+1/k. This tail still has supremum at least β, so choose within it a term greater than β−1/k. Its index nₖ is strictly greater than the preceding index and its value lies within 1/k of β. Thus x at those indices converges to β. This construction uses approximate witnesses, not a largest term. Reverse inequalities to construct a subsequence approaching finite α.

Infinite endpoints. If β=+∞, every tail is unbounded above. Recursively choose nₖ greater than its predecessor with x at nₖ greater than k. This subsequence tends to +∞. Equivalently, for every real threshold M, terms exceed M infinitely often. If α=−∞, choose arbitrarily late terms below −k; equivalently every real lower threshold is crossed infinitely often. If β=−∞, some tail supremum lies below each prescribed real M, so the whole sequence eventually lies below M and tends to −∞. If α=+∞, the whole sequence similarly tends to +∞. These last two cases also provide subsequences to the respective endpoint.

Extremal characterization. Every convergent subsequence, including one converging in the extended real sense, has limit between α and β. For example, eventual bounds below β+ε rule out any subsequential limit above finite β. When β=−∞ the whole sequence tends to −∞; when β=+∞ the upper bound is automatic. The lower cases are symmetric. Since we have constructed subsequences approaching each endpoint, α is the smallest and β the largest extended-real subsequential limit. For a bounded sequence they are finite and attained as subsequential limits, even when no tail extremum is attained by an actual term. Interior subsequential limits may or may not occur.

Apply

Equal endpoints characterize convergence. If α=β=ℓ is finite, the eventual lower bound ℓ−ε and upper bound ℓ+ε hold together, giving ordinary convergence to ℓ. If both endpoints are +∞, the eventual lower-threshold property gives convergence to +∞; if both are −∞, use upper thresholds. Conversely, convergence to a finite or infinite ℓ forces both tail endpoints to be ℓ by the same threshold definitions.

Consider xₙ=(−1)ⁿn. Every tail contains arbitrarily large positive even terms and arbitrarily negative odd terms, so its endpoints are −∞ and +∞. It has subsequences to both endpoints but does not converge in the extended real line. Now consider a sequence equal to n only at square indices and zero elsewhere. Its endpoints are 0 and +∞: the large spikes do not force the whole sequence to tend to +∞ because arbitrarily late zeros remain.

What do equal endpoints fail to tell us about a finite prefix?

They impose no bound on the finitely many early terms. Replacing a finite prefix leaves every sufficiently late tail unchanged and therefore leaves liminf and limsup unchanged. Likewise, a long but finite observed prefix cannot determine the endpoints without additional information about the generating rule. The practice asks you to construct several incompatible continuations of exactly the same finite observations.

Continue

When using limsup or liminf in a later convergence test, translate the endpoint inequality into an eventual bound or an infinitely-often witness before applying another theorem. For example, if limsup xₙ<q for a real q, all sufficiently late terms are below q. If q<limsup xₙ, terms exceed q infinitely often. These implications also cover infinite endpoints with the threshold interpretations above.

The practice includes exact tails, unattained bounds, unbounded sequences and a block sequence whose subsequential limits fill an interval. State the domain of each limit and keep the subsequence indices increasing. A finite-window calculation is useful evidence about that window; the general tail proof is what controls the infinite sequence.