Sequences and convergence

An estimate can cross its target and still settle

An exact calculation produces estimates 1, 5/2, 5/3, 9/4, 9/5, 13/6, … . They lie alternately below and above 2. Their distances from 2 are 1, 1/2, 1/3, 1/4, 1/5, 1/6, … . Each estimate has an index: a₁=1 is first, a₂=5/2 is second. The rule is aₙ=2+(−1)ⁿ/n for every positive integer n.

Ask a question stronger than whether the dots seem close: after which index is every remaining estimate less than 1/4 away from 2? Since the error is exactly 1/n, n=4 is on the boundary and does not satisfy a strict less-than requirement. Every n≥5 does. We have a guarantee for the entire infinite tail, not merely the points displayed.

Preparation: be comfortable with fractions, inequalities, functions and absolute value as distance. The earlier limit lessons explain the same kind of error guarantee for real-variable inputs. Here the input takes only integer values. Follow the first error example and the oscillation comparison before returning to the proofs about arbitrary sequences. No differentiation or integration is needed for these arguments.

A sequence has an order and a stated starting index

A real sequence is a function from the positive integers to the real numbers. We write its value at n as aₙ, and the whole sequence as (aₙ). Order and repetition matter: (1,2,1,2,…) is different from a set containing just 1 and 2. A graph of a sequence consists of separate points (n,aₙ); a line joining them is not an extra claim about fractional indices.

An explicit definition gives aₙ directly, as in 2+(−1)ⁿ/n. A recursive definition gives starting data and a rule for obtaining later terms. For example x₁=0 and xₙ₊₁=(xₙ+3)/2 for n≥1 determine 0, 3/2, 9/4, 21/8, … . Without the starting value the same update rule does not specify one particular sequence.

Some sequences start at index 0 or another integer; write that convention explicitly and adjust formulas accordingly. This lesson starts at 1. A finite displayed list does not determine a unique infinite rule. For example, a sequence may agree with 1/n for the first million terms and then equal 7 forever. A conclusion about its tail requires its definition, not just a long prefix.

The term aₙ is one value. The expression a₁+⋯+aₙ is a different value: a partial sum. The next lesson will study the sequence of these partial sums. Convergence of the terms and convergence of their accumulated sums are separate questions.

Choose the tolerance first, then a whole tail

The statement aₙ→L means: for every real ε>0 there is a positive integer N such that |aₙ−L|<ε for every integer n≥N. Epsilon, written ε, is the permitted error. The cutoff N can depend on ε and on the sequence, but once chosen it must work for all later indices. The finite real number L is the limit.

Think of the open band L−ε<aₙ<L+ε. Terms before N may lie anywhere; every term from N onwards must stay inside. One visit to the band is insufficient. A fixed-width band is also insufficient, because the definition asks for every positive width, however small.

For aₙ=2+(−1)ⁿ/n, the error is 1/n. Given ε>0, choose N=⌊1/ε⌋+1, where the floor ⌊x⌋ is the largest integer no greater than x. Then N>1/ε, and n≥N gives 1/n≤1/N<ε. This proves aₙ→2. The rule works even when 1/ε is an integer, so the strict boundary is handled correctly.

N need not be the smallest valid cutoff. Any larger one works. A proof can safely overestimate the needed index. Conversely, finding an N for ε=0.01 alone proves one accuracy guarantee, not convergence for every accuracy.

A finite limit is one kind of long-term behaviour

The sequence bₙ=(−1)ⁿ takes the value −1 at every odd index and +1 at every even index. It is bounded, yet cannot approach one finite number. For any cutoff there are still both an odd and an even index beyond it. A band of radius 1/2 about one number cannot contain both −1 and +1, since their separation is 2 while two such radii total only 1.

The sequence cₙ=n instead escapes above every fixed threshold. Writing cₙ→+∞ means that for every real M there exists N such that cₙ>M for all n≥N. This is not convergence to a real number: +∞ is not a real output value. Define escape to −∞ by reversing the inequality. Divergence also includes bounded oscillation, so it does not always mean escape.

Shrinking oscillation can converge. Our opening estimates alternate around 2 but their oscillation amplitude 1/n tends to zero. Merely alternating signs or crossing a target says nothing by itself about convergence. Inspect how large the deviations remain arbitrarily far along the sequence.

The tail fixes a unique limit

A sequence cannot have two distinct finite limits L and K. If d=|L−K|>0, choose ε=d/3 for both proposed limits and take the larger of their two cutoffs. For any later term, the triangle inequality gives d≤|L−aₙ|+|aₙ−K|<2d/3, a contradiction. This proves uniqueness.

Changing finitely many terms does not change a sequence's limit or whether it converges. Move the cutoff beyond every changed position as well as beyond the original accuracy cutoff. All later terms are exactly the original ones. The same reasoning covers deleting a finite prefix and reindexing the remaining tail; a fixed shift of the index merely shifts the cutoff.

Every convergent real sequence is bounded: take ε=1 to bound its tail by |L|+1 in absolute value, then enlarge that bound to include the finitely many earlier values. Bounded means that one finite B satisfies |aₙ|≤B for all n. The oscillating sequence (−1)ⁿ proves that boundedness alone does not force convergence.

Skipping terms cannot change an existing limit

A subsequence is (aₙₖ), where n₁<n₂<n₃<⋯ are positive integer indices. The indices must strictly increase, so this is an infinite selection in the original order, not a repeated choice of one early term. Since nₖ≥k, its indices eventually pass every cutoff.

If aₙ→L, choose a cutoff N for any ε>0. For k≥N, nₖ≥k≥N, so |aₙₖ−L|<ε. Every subsequence therefore has the same limit L. For (−1)ⁿ, the even-index subsequence is constantly +1 and the odd-index subsequence constantly −1. Their distinct limits provide another proof that the original sequence diverges.

One convergent subsequence does not establish convergence of the whole sequence: the even terms of (−1)ⁿ already give a counterexample. If both the even and odd subsequences converge to the same L, however, the whole sequence does converge to L. Take cutoffs for both selections and pass a sufficiently large original index; its parity places it in one of the two guaranteed tails.

Combine limits only after checking that they exist

If aₙ→A and bₙ→B are finite limits, then aₙ+bₙ→A+B, aₙ−bₙ→A−B and c aₙ→cA for a fixed real c. For the sum, make each input error less than ε/2 and take the larger cutoff. The triangle inequality bounds the total error by their sum. The difference uses the same argument; scalar multiplication uses input tolerance ε/|c| when c≠0, with c=0 immediate.

Products satisfy aₙbₙ→AB. To see why, write aₙbₙ−AB=aₙ(bₙ−B)+B(aₙ−A). Boundedness of (aₙ) supplies a finite C≥1 with |aₙ|≤C. Make |bₙ−B|<ε/(2C) and |aₙ−A|<ε/[2(|B|+1)]. Their two contributions then total less than ε. This controls the changing factor rather than treating it as a fixed constant.

For a quotient, require B≠0 and a defined denominator in the tail. Once |bₙ−B|<|B|/2, the triangle inequality gives |bₙ|>|B|/2, so the denominator stays away from zero. Moreover |1/bₙ−1/B|≤2|bₙ−B|/|B|². This proves reciprocal convergence and, with the product law, aₙ/bₙ→A/B. A zero denominator at an earlier position requires starting the quotient sequence after that position.

For example (5n−2)/(2n+7)=(5−2/n)/(2+7/n)→5/2, because 1/n→0 and the denominator limit is 2. A formal expression 0/0 supplies no quotient limit: (1/n)/(1/n)=1, while (1/n²)/(1/n)=1/n→0. Both numerator and denominator tend to zero in each example, but the actual quotients have different limits.

Use an envelope when individual terms are awkward

Suppose lₙ≤aₙ≤uₙ for all sufficiently large n, and both lₙ and uₙ tend to the same finite L. For a given ε>0, their two guaranteed tails lie between L−ε and L+ε. The middle terms then lie there too. Taking a cutoff that also passes any initial exceptions proves aₙ→L. This is the squeeze theorem for sequences.

For aₙ=(1+(−1)ⁿ)/(2n), the numerator is either 0 or 2, so 0≤aₙ≤1/n. Both bounds tend to zero, hence so does aₙ. The argument covers every index without treating the even and odd cases as unexplained numerical evidence.

More generally, if |vₙ|≤C and wₙ→0, then |vₙwₙ|≤C|wₙ|→0. The bounded factor need not have a limit. When vₙ=(−1)ⁿ and wₙ=1/n, their product tends to zero although vₙ itself diverges. Applying the ordinary product law directly would fail its hypothesis; the envelope supplies the missing argument.

A one-way sequence needs a bound in the direction it moves

A sequence is nondecreasing if aₙ₊₁≥aₙ for every n, and nonincreasing if aₙ₊₁≤aₙ. Either kind is called monotone; equality is allowed. Nondecreasing sequences already have the lower bound a₁, so an upper bound is the additional requirement. Nonincreasing sequences need a lower bound. A sequence that becomes monotone after finitely many terms can be treated by its tail.

A nondecreasing real sequence bounded above has a finite limit. Here is the completeness step: the nonempty set of its values has a least upper bound L, also called its supremum. This is a property of the real numbers. L need not itself be one of the values, and an arbitrary upper bound need not be the limit.

For every ε>0, L−ε is smaller than the least upper bound and therefore cannot be an upper bound. Some term a_N must exceed L−ε. For every n≥N, monotonicity and the upper bound give L−ε<a_N≤aₙ≤L, hence |aₙ−L|<ε. This proves convergence to L. Apply the same argument to (−aₙ) for a nonincreasing sequence bounded below.

Each hypothesis matters. The nondecreasing sequence n has no upper bound and diverges. The bounded sequence (−1)ⁿ is not monotone and diverges. The sequence 2−1/n is nondecreasing and bounded above by 100, but its limit is 2, its least upper bound, not the convenient larger bound 100.

A fixed-point equation is a candidate, not a convergence proof

Return to x₁=0 and xₙ₊₁=(xₙ+3)/2. If 0≤xₙ<3, then 0≤xₙ₊₁<3, so induction keeps every term in [0,3). Also xₙ₊₁−xₙ=(3−xₙ)/2>0. The sequence is increasing and bounded above; the theorem proves that a finite limit exists.

Only now take limits in the recurrence. A shifted convergent sequence has the same limit, and the sum and scalar laws give L=(L+3)/2, so L=3. Solving this equation at the beginning would identify only a possible limit. For y₁=1 and yₙ₊₁=2yₙ, the equation L=2L suggests 0 but the actual values 1,2,4,8,… diverge.

This recursion also has an exact error calculation: 3−xₙ₊₁=(3−xₙ)/2. Starting with error 3 gives 3−xₙ=3/2ⁿ⁻¹ by induction. For |xₙ−3|<1/100, the requirement is 2ⁿ⁻¹>300. Since 2⁸=256 and 2⁹=512, N=10 works for every later index and N=9 does not.

For completeness, 2ᵏ≥k+1 for every nonnegative integer k: it holds at k=0, and doubling k+1 gives at least k+2 at the next step. Thus 3/2ⁿ⁻¹≤3/n→0, proving decay for arbitrary tolerance, not just 1/100. The exact formula and the monotone bounded theorem give two compatible justifications.

Separate the limit from the evidence needed to use it

A finite graph or table helps you predict, compare errors and catch a false formula. The definition or an all-index theorem establishes what happens indefinitely. Keep the exact error bound beside a numerical approximation whenever accuracy matters. A sequence can converge so slowly that a visually stable prefix gives little practical information about the remaining error.

Having |aₙ₊₁−aₙ| small is weaker than being close to one limit. The partial sums 1+1/2+⋯+1/n have successive differences 1/(n+1)→0 but, as the next lesson proves by grouping, they grow without bound. Small individual updates do not guarantee that their accumulated effect is small.

Use this checklist in a calculation: define every term and index; decide whether a finite real limit is claimed; justify a whole tail for every tolerance or check a suitable theorem; and distinguish a proved bound from the displayed samples. These are the foundations for infinite series and, later, sequences of functions.

Worked examples

Try it yourself

1. If aₙ=1/n, are a₃ and a₁+a₂+a₃ the same number?

   Show the answer and reasoning

   No: a₃=1/3, while a₁+a₂+a₃=11/6.

   One is a term; the other is a partial sum. Limits of these two sequences answer different questions.

2. Every term after index 50 is within 0.1 of 2. Does that alone prove convergence to 2?

   Show the answer and reasoning

   No.

   Convergence requires a whole tail for every positive tolerance. A sequence constantly equal to 2.05 satisfies this statement but converges to 2.05.

3. Change the first ten terms of 1/n to 1,000,000. What happens to its limit?

   Show the answer and reasoning

   It remains 0.

   A cutoff beyond ten leaves the original tail. A large finite prefix does not alter an infinite-tail guarantee.

4. A sequence equals 0 at every even index and 1 at every odd index. Is visiting 0 infinitely often enough for convergence to 0?

   Show the answer and reasoning

   No.

   There are still odd-index values equal to 1 beyond every cutoff. A band of radius 1/2 around 0 cannot contain the tail.

5. Find the limit of (1+(−1)ⁿ)/(2n).

   Show the answer and reasoning

   0.

   For every positive index it lies between 0 and 1/n, and both bounding sequences tend to 0.

6. The sequence 3−1/n is increasing and bounded above by 8. Must its limit be 8?

   Show the answer and reasoning

   No; the limit is 3.

   The least upper bound is 3. The monotone bounded theorem guarantees convergence but does not turn every upper bound into the limit.

7. A sequence has a subsequence converging to 5. Must the whole sequence converge to 5?

   Show the answer and reasoning

   No.

   Let its even terms be 5 and its odd terms be −5. The even subsequence converges to 5, while the whole sequence has incompatible parity tails.

8. Both the even and odd subsequences converge to 5. Can the original sequence still diverge?

   Show the answer and reasoning

   No.

   For any tolerance, take the two parity cutoffs. Every sufficiently late original index belongs to one of the two tails and is within that tolerance of 5.

9. May the quotient limit law evaluate (1/n)/(1/n²) by dividing the limiting values?

   Show the answer and reasoning

   No; the denominator limit is zero. The actual quotient is n and diverges to +∞.

   Every finite denominator is nonzero, but the additional nonzero-limit hypothesis still fails. Simplify the exact quotient instead.

10. For y₁=1 and yₙ₊₁=2yₙ, does solving L=2L prove a limit of 0?

    Show the answer and reasoning

    No.

    That equation is necessary only if a finite limit already exists. The actual formula is yₙ=2ⁿ⁻¹, which grows without bound.

11. Can a convergent sequence fail to be monotone?

    Show the answer and reasoning

    Yes: 2+(−1)ⁿ/n converges to 2.

    Its successive terms cross 2, while their exact distances from 2 tend to zero. Monotonicity is a useful sufficient hypothesis when combined with a bound, not a necessary one.

12. Give a valid cutoff for 1/n<ε, for an arbitrary ε>0.

    Show the answer and reasoning

    N=⌊1/ε⌋+1 works for every n≥N.

    This integer is strictly greater than 1/ε, including the case where 1/ε is itself an integer.

What to carry forward

• Convergence controls every term in a sufficiently late tail for every positive error tolerance.
• A sequence's finite real limit is unique and unchanged by finitely many altered terms.
• Convergence implies boundedness and forces every subsequence to share its limit. Boundedness or the existence of one convergent subsequence alone does not guarantee convergence.
• Limit laws need existing finite input limits and a nonzero denominator limit for quotients.
• A squeeze controls a sequence through two bounds with the same limit.
• A real monotone sequence converges when bounded in its direction of motion; completeness supplies the least upper or greatest lower bound.
• A recursive fixed-point equation identifies a possible limit only after convergence is justified; an exact error formula can also give usable accuracy.