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.
What a learner can do afterwards
- 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
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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.
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.
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.
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.
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.
Check whether terms approach a finite limit. Prove convergence before using a recursion's limit equation. A series is tested through its partial sums.
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Checking a sequence limit
The index tells you which term to use. For a_n = 7 - 4/n, term 1 is 3 and term 2 is 5. Its distance from 7 is 4/n. Given any positive distance d, choosing n > 4/d makes the distance smaller than d. It stays smaller for every later index, so the sequence converges to 7.
A finite prefix does not determine a limit. If a sequence is 12 for its first four terms and 9 thereafter, it converges to 9. Every sufficiently late term equals 9 exactly.
A sequence uses integer indices. If a function has a finite limit as its real input grows, sampling it at positive integers gives the same limit. The converse need not hold because the function can behave differently between integers.
For polynomial quotients, divide numerator and denominator by the highest denominator power. Use the quotient of the resulting limits only when the denominator limit is nonzero.
Proving a recursive sequence converges
Several calculated terms can suggest a pattern, but a proof must cover every later term. To prove an upper bound, check the first term. Then show that a term satisfying the bound produces a next term satisfying it too. This is induction.
To prove increasing behavior, calculate a_(n+1) - a_n and show it is nonnegative. An increasing real sequence with an upper bound converges. A decreasing real sequence with a lower bound also converges. A bound alone is insufficient: the sequence 0, 1, 0, 1, ... never approaches one number.
Completeness gives the terms of an increasing bounded real sequence a least upper bound L. For any d > 0, some term is greater than L - d, since otherwise L - d would be a smaller upper bound. Every later term lies between that term and L. Thus all sufficiently late terms stay within d of L.
After proving convergence, a recurrence a_(n+1) = F(a_n) with F continuous at L gives L = F(L). This equation identifies possible limits. Starting at 0, the rule a_(n+1) = 1 - a_n alternates forever even though L = 1 - L has solution 1/2.
Terms and running totals
For a sequence beginning 6, 3, 3/2, the first three terms are those individual numbers. The first three partial sums are 6, 9 and 21/2. A partial sum includes every term from the start through its index.
An infinite series converges when its partial sums approach a finite number. That number is its sum. If a formula for S_n is supplied, take its limit to find the series sum.
Convergence of a series forces its terms to tend to 0. For n >= 2, a_n = S_n - S_(n-1), and the two partial sums then approach the same number. The converse fails.
Consider 1 + 1/2 + 1/3 + ... . For each integer k >= 1, the block from index 2^(k-1) + 1 through 2^k has 2^(k-1) terms. Each is at least 1/2^k, so every block adds at least 1/2. There are arbitrarily many blocks. The partial sums are unbounded even though the individual terms tend to 0.
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