Follow terms toward a limit

Try

List the first five values of aₙ=1−1/n, starting at n=1. They increase while remaining below 1. Now list the first five values of bₙ=(−1)ⁿ. They stay between −1 and 1 but do not settle. Predict which sequence has a limit and identify the extra information that distinguishes them.

Predict

A sequence is a function whose inputs here are positive integers. Predict whether a smooth curve drawn through its points is uniquely determined. It is not: many continuous extensions agree at the integers. A continuous-variable limit can establish a sequence limit when applicable, but the sequence itself only requires statements about its integer indices.

Experiment

For the recurrence a₁=1 and aₙ₊₁=(aₙ+3)/2, calculate terms and their distances below 3.

n	aₙ	3−aₙ
1	1	2
2	2	1
3	5/2	1/2
4	11/4	1/4

Plot the terms as isolated points with a horizontal reference at 3. The error halves each time. Derive that relationship algebraically before extending the table. The difference-equation investigation lets you compare shrinking and growing errors; its bounded trials are examples, not a proof about infinitely many updates.

Understand

The statement aₙ→L means that for every positive tolerance ε there is an index N such that every n≥N satisfies |aₙ−L|<ε. The index can depend on ε. A thousand close terms do not by themselves control all later terms. For 1−1/n, the error is exactly 1/n, so choosing N greater than 1/ε establishes the definition.

For (3n+2)/(2n−1), divide numerator and denominator by n. The reciprocal terms vanish, yielding limit 3/2. This reasoning concerns the behavior of a whole formula, not an apparent trend in a plot. The denominator is nonzero for positive integer n, so the sequence is defined throughout its stated domain.

A real sequence that is monotone and bounded converges. For an increasing bounded sequence, its supremum supplies the limit: eventually a term must lie within any positive tolerance below the least upper bound, and monotonicity keeps all later terms at least that high. A decreasing sequence uses an infimum. This result proves existence but may not identify the value without another argument.

For the recurrence, set eₙ=3−aₙ. Then eₙ₊₁=eₙ/2 and eₙ=2(1/2)^(n−1), which tends to zero. Hence the limit is 3. Solving L=(L+3)/2 only finds a possible limit; it does not establish convergence. Unstable recurrences can have fixed points that nearby trajectories never approach.

Every subsequence of a convergent sequence has the same limit. The even terms of (−1)ⁿ equal 1 and the odd terms equal −1, proving nonconvergence. Bounded oscillation is therefore a concrete counterexample to “bounded implies convergent.” A sequence can also converge without being monotone, as (−1)ⁿ/n does.

Apply

For each practice argument, identify the quantified tail behavior or theorem that supports the conclusion. Distinguish calculating a₃ from finding a limit. For the recurrence, show that the error actually shrinks, and for oscillation identify two conflicting subsequences rather than simply saying the graph looks unsettled.

Continue

An infinite series introduces another sequence: the running totals of its terms. Next distinguish the limit of those totals from the limit of the individual terms being added.