Try
Compute 1−1/2+1/3−1/4 one term at a time. The running total moves above and below its eventual limit. Predict whether decreasing step sizes make the two alternating subsequences approach one another, and which assumptions are needed to turn that picture into a proof.
Predict
Compare Σ(−1)^(n+1)/n with Σ1/n. Their term magnitudes are identical, but the signs change the accumulated behavior. Predict why proving convergence of the alternating sum does not prove absolute convergence, and why convergence of the absolute series would be stronger information.
Experiment
Build the first four partial sums and bound the next unknown contribution.
| N | S_N | Next term magnitude |
|---|
| 1 | 1 | 1/2 |
| 2 | 1/2 | 1/3 |
| 3 | 5/6 | 1/4 |
| 4 | 7/12 | 1/5 |
Plot these sums on a number line rather than only against n. The odd sums decrease and the even sums increase, with the limit between them. For this example, the omitted remainder after N terms has magnitude at most 1/(N+1). Use that bound to plan a sufficient term count before computing more terms.
Understand
The alternating-series test applies to Σ(−1)ⁿbₙ when positive magnitudes bₙ eventually do not increase and tend to zero. Pairing terms shows that one partial-sum subsequence is monotone in one direction and the other in the opposite direction, each bounding the other. Their gap tends to zero, so they approach the same limit. Initial exceptions can be separated off without changing convergence.
Under the same decreasing-tail conditions, the remainder magnitude after N terms is at most the next magnitude b_N+1, once N is in that controlled tail. The remainder has the sign of the first omitted term. This locates the total on one side of the partial sum as well as bounding its distance. Alternating signs alone, without shrinking and suitably ordered magnitudes, do not supply this guarantee.
For the alternating harmonic series, error at most 1/100 is guaranteed when 1/(N+1)≤1/100, so 99 terms suffice. This does not prove that 98 terms have actual error above the target; an upper bound is not necessarily attained. Distinguish a sufficient certificate from a minimal computational requirement.
A series is absolutely convergent if Σ|aₙ| converges. Absolute convergence implies ordinary convergence because the magnitude of any signed tail is at most the corresponding absolute tail. A convergent series whose absolute series diverges is conditionally convergent. The alternating harmonic series is conditional; Σ(−1)ⁿ/n² is absolute by the p-series test.
Conditional convergence requires care with order. Rearranging a conditionally convergent series can alter its sum or destroy convergence; infinite sums do not inherit unrestricted finite rearrangement rules. Absolutely convergent series are stable under rearrangement. The absolute case is proved below; no general theorem prescribing sums of conditional rearrangements is assumed. Keep the given order when evaluating a conditional sum or claiming an error bound.
Apply
Classify ordinary and absolute convergence separately in the practice bank. State why the magnitudes decrease and tend to zero before using the next-term bound. For the proof task, explain how two subsequences bracket a common limit, not merely that the plotted points alternate.
Why absolute convergence protects the sum
Suppose Σ|aₙ| converges and S=Σaₙ. Let p be a permutation of the positive integers: it uses each original index exactly once. Given ε>0, choose N so Σ beyond N of |aₙ| is less than ε. There is an M such that p(1),…,p(M) includes every index from 1 through N. For m≥M, subtract the finite rearranged prefix from S. Every remaining term has index greater than N, so the absolute value of the difference is at most that absolute tail, hence less than ε. This inequality follows by first comparing finite sums and then taking their convergent limits. Therefore the rearranged partial sums tend to S. The argument uses exhaustion of all indices as well as the absolute tail bound; deleting or duplicating terms is not a permutation.
Consecutive regrouping also preserves S whenever group endpoints tend to infinity, because grouped partial sums are a subsequence of the original partial sums. After an arbitrary permutation, absolute convergence permits this regrouping as well. Group sizes may grow, but each group must be finite and the groups must exhaust the indices.
Worked decision. For aₙ=(−1)^(n+1)/2ⁿ, read indices in the order 2,1,4,3,6,5,… and then group consecutive pairs. Absolute convergence follows from Σ2⁻ⁿ=1, so both operations preserve S=1/3. The first m pairs contain exactly the first 2m original terms; their error is at most Σ beyond 2m of 2⁻ⁿ=2⁻²ᵐ. This absolute bound is valid regardless of sign cancellation.
Transfer and error repair. (1) For bₙ=(−1)ⁿ/n², read each block of three indices backwards: 3,2,1,6,5,4,… . Prove its sum is unchanged without knowing the sum. (2) A learner reads the alternating harmonic series in the order 1,3,2,5,7,4,… (two new positive terms followed by one new negative term), and claims the same theorem proves equality. Locate the missing hypothesis. (3) Explain why reading only even indices is not protected even for bₙ.
Review the proof obligations
For (1), the p-series Σ1/n² converges, and the block reversal is a bijection of all indices. Apply the finite-core argument; after m complete blocks the prefix even equals the original prefix through 3m. For (2), Σ1/n diverges, so the crucial small absolute tail is unavailable; a bijection alone does not justify equality. No calculation of a new conditional sum is required to identify this logical failure. For (3), the omitted odd indices never enter any prefix, violating exhaustion; the new series is a subseries, not a rearrangement.
Credit requires naming the absolute comparison, checking bijectivity or exhaustion, and distinguishing a failed justification from an asserted value of the rearranged sum. These written responses are practice for review, not independently authenticated scores.
Continue
Next use ratio and root tests to recognize geometric decay hidden inside factorials or nth powers. Their inconclusive boundary at limit 1 will make these alternative tests essential.
