--- title: "Sequences and Their Limits" description: "Study real sequences indexed by positive integers and decide whether their terms approach a finite limit. Increasing sequences bounded above converge, as do decreasing sequences bounded below." canonical: https://lightmysky.com/learn/mathematics/sequences-and-their-limits-mt_PQl3Q6n5dc source: https://lightmysky.com/learn/mathematics/sequences-and-their-limits-mt_PQl3Q6n5dc.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Sequences and Their Limits Study real sequences indexed by positive integers and decide whether their terms approach a finite limit. Increasing sequences bounded above converge, as do decreasing sequences bounded below. Subject: Mathematics · Area: Calculus & Analysis · Ages 19 to 20 Page: https://lightmysky.com/learn/mathematics/sequences-and-their-limits-mt_PQl3Q6n5dc ## Ready when they can - Decide convergence for a sequence given by a formula in n - Show a recursively defined sequence is increasing and bounded above - Distinguish a sequence from the series formed by summing it ## Lesson: Where sequences settle down A sequence gives a real number a_n for each positive integer n; a_n means term n. It converges to a finite number L if, for every chosen positive distance, all sufficiently late terms stay within that distance of L. Otherwise it diverges. Only the tail matters: changing finitely many early terms leaves convergence and the limit unchanged. A sequence that equals one constant from some index onward converges to that constant. For polynomial quotients, divide numerator and denominator by the highest denominator power. For 3n/(2n + 5), this gives 3/(2 + 5/n), with limit 3/2. For (n + 2)/n^2, it gives 1/n + 2/n^2, with limit 0. For any positive integer k, 1/n^k tends to 0. Sums of convergent sequences have the sum of their limits. Quotients have the quotient of their limits when the denominator limit is nonzero. *(drawing: These are the first four terms of 3n/(2n + 5). The algebra establishes the limit; a finite sample only illustrates the terms.)* The factor (-1)^n is -1 for odd n and 1 for even n. Thus (-1)^n has no limit: its odd and even terms stay at different values. But (-1)^n/(n + 2) lies between -1/(n + 2) and 1/(n + 2), which both tend to 0. The squeeze theorem gives limit 0. More generally, if a sequence lies between two sequences with the same limit from some index onward, it has that limit too. Increasing means a_(n+1) >= a_n; decreasing means a_(n+1) <= a_n. An upper bound M satisfies a_n <= M for every n. An increasing real sequence bounded above converges, as does a decreasing one bounded below. A bound alone does not give convergence. Completeness says every nonempty set of real numbers bounded above has a least upper bound, its smallest upper bound. For an increasing sequence, this is its limit. **Example.** A recursive rule uses earlier terms. Let c_1 = 2 and c_(n+1) = (c_n + 8)/3. The first term lies between 0 and 4. If 0 <= c_n <= 4, the rule gives 0 <= c_(n+1) <= 4. Starting with the first term, this implication proves the bound for every term by induction. Also, c_(n+1) - c_n = 2(4 - c_n)/3 >= 0. The sequence is increasing and bounded above, so it converges. If a_n converges to L, its tail a_(n+1) has the same limit. For a_(n+1) = F(a_n), continuity at L lets us take limits: L = F(L). Continuity means F(x) approaches F(L) as x approaches L. For the preceding rule, L = (L + 8)/3 gives L = 4. This fixed-point equation identifies possible limits; it does not prove convergence. For a quotient rule, check that its limiting denominator is nonzero. **Example.** Suppose an established limit satisfies L^2 = 3L + 4 and L > 0. Move all terms to one side: L^2 - 3L - 4 = 0. Factor: (L - 4)(L + 1) = 0. Either factor must be zero, giving 4 or -1. The positive condition selects 4. A series is the expression a_1 + a_2 + ... . Its nth partial sum S_n adds terms 1 through n. The series converges if S_n approaches a finite number; that number is its sum. For terms 2, 1, 1/2, ... , the first three partial sums are 2, 3 and 7/2. *(drawing: This row lists individual terms.)* *(drawing: This row lists running totals: 2, then 2 + 1, then 2 + 1 + 1/2.)* To find a series sum from a supplied formula for S_n, take the limit of S_n. For example, if S_n = 6 - 2/(n + 1), the series sum is 6. The terms 1/n tend to 0, but that alone does not establish convergence of 1 + 1/2 + 1/3 + ... . Individual terms and running totals require separate limit checks. **Recap.** Check whether terms approach a finite limit. Prove convergence before using a recursion's limit equation. A series is tested through its partial sums. *Lesson, questions and flashcards written for this learning dataset.* ## Practice 14 questions on this page, each with its working shown. ## Optional connections - [Geometric Sequences and Common Ratios](https://lightmysky.com/learn/mathematics/geometric-sequences-and-common-ratios-mt_NDnbJm2hTY) ## Opens up - [Infinite Series and the Geometric Series](https://lightmysky.com/learn/mathematics/infinite-series-and-the-geometric-series-mt_EPE-ERFFQt) - [The Completeness Axiom: Suprema and Infima](https://lightmysky.com/learn/mathematics/the-completeness-axiom-suprema-and-infima-mt_xjI-pIfh95)