Try
An infinite series is a limit of finite sums, not an instruction to add all terms simultaneously. For Σ from n=0 of (1/3)^n, the Nth partial sum is (1−(1/3)^(N+1))/(1−1/3), which tends to 3/2. The finite identity and the vanishing remainder establish the sum. A long decimal accumulation may suggest the same answer, but it does not control every remaining tail.
Predict
Predict whether a_n→0 guarantees Σa_n converges. It is necessary because a_n=S_n−S_(n−1) if partial sums converge. It is not sufficient: harmonic terms tend to zero while harmonic partial sums diverge. Grouping terms from 2^(k−1)+1 through 2^k gives at least 1/2 per block, so partial sums become arbitrarily large. This is an argument about all blocks, not a numerical trend.
Experiment
Compare three series.
| Series | Terms tend to zero? | Absolute convergence? |
|---|
| Σ(1/3)^n | Yes | Yes |
| Σ(−1)^(n+1)/n | Yes | No |
| Σ1/n | Yes | No, and the original diverges |
For the alternating series, inspect even and odd partial sums separately. Even sums increase, odd sums decrease, and each even sum lies below each following odd sum. Their gap is one term, tending to zero. Use monotone convergence to conclude both subsequences have the same limit and hence the full partial-sum sequence converges.
Understand
The Cauchy criterion says a real series converges exactly when, for every ε>0, all sufficiently late finite tail sums have absolute value below ε. This follows by applying the sequence Cauchy theorem to S_m−S_n. If Σ|a_n| converges, then |Σ from n+1 to m a_k|≤Σ from n+1 to m |a_k| makes the original partial sums Cauchy. This proves absolute convergence implies convergence.
An absolutely convergent series also has a sum independent of order. Given ε, retain an original finite prefix whose remaining absolute tail is below ε. A sufficiently long permuted prefix contains every index in that original prefix. Its omitted terms all lie in the small absolute tail, so its difference from the full sum is below ε. The permutation must be bijective: dropping or repeating terms changes the problem.
Apply
Repair “addition is commutative, therefore every infinite rearrangement preserves the sum.” Finite commutativity does not justify passage to an infinite limit. For the alternating harmonic series, group two positive terms followed by one negative. After n such groups the sum is H_(4n)−(1/2)H_(2n)−(1/2)H_n, which tends to (3/2)log 2, whereas the original sum is log 2. Establishing those logarithmic limits requires an additional known harmonic asymptotic or integral argument; the displayed grouping alone should not be presented as its proof.
Construct an order with a chosen sum
Can the same terms be ordered to approach a number you choose? For a conditionally convergent real series, the answer is yes for every finite real target T. Here “conditionally” means that Σa_n converges but Σ|a_n| diverges. We must construct a bijection of the original indices and prove convergence of all its partial sums.
First separate positive terms from negative terms. Write p_n=max(a_n,0) and q_n=max(−a_n,0), so a_n=p_n−q_n and |a_n|=p_n+q_n. For prefixes through N, let P_N=Σp_n, Q_N=Σq_n and S_N=Σa_n. The identities P_N=(Σ|a_n|+S_N)/2 and Q_N=(Σ|a_n|−S_N)/2 show that both P_N and Q_N tend to positive infinity: the absolute sums tend to infinity and S_N stays bounded. Thus each sign supplies infinitely many terms, with unlimited total magnitude. Also a_n→0, because the original series converges.
List the strictly positive terms in their original order and separately list the strictly negative terms in their original order. Start with sum zero. Take at least one unused positive term, then keep taking positive terms until the sum is strictly above T. Next take at least one unused negative term and continue until the sum is strictly below T. Repeat these two phases forever. Every phase ends after finitely many terms, since deleting a finite prefix cannot make either divergent sign sum finite. After each pair of phases, insert the next unused zero term if one exists. Zeros do not change the sum.
This really uses each original index exactly once. Terms are removed only from the front of their sign lists, never repeated. Every cycle consumes at least one term from each nonzero sign list and at most the next zero. After enough cycles, any fixed position in any of the three lists has been consumed. An arbitrary enumeration that keeps choosing convenient terms without this exhaustion argument would not prove a rearrangement theorem.
After the first cycle, every upward phase starts below T. Its final overshoot is at most its last positive term: just before the crossing, the sum was at most T. Every downward phase starts above T and has undershoot at most the magnitude of its last negative term. These last terms tend to zero, since their original indices increase without bound. Hence the sums at both kinds of crossing tend to T. Within each phase, partial sums move monotonically between its starting and ending crossings. Once both endpoints are within ε of T, every intermediate sum is too; inserted zeros preserve that bound. This proves convergence of the entire reordered series, not only of crossing endpoints. The exceptional first upward phase has no effect on a limit.
Try a changed target. Use the alternating harmonic terms and T=1/2. The first positive phase gives 1; the negative phase goes through 1/2 to 1/4 because the crossing is strict. The next positive phase gives 7/12, and the next negative phase gives 5/12. Record the original indices used: 1, 2, 4, 3, 6. Predict the next positive term and the next negative term, then check both crossings. Explain why this short trace alone proves neither exhaustion nor convergence.
Check the trace and repair a proof
The next positive term is 1/5, producing 37/60. The next negative term is −1/8, producing 59/120, just below 1/2. Both are the next unused terms of their respective sign lists. A claim that “crossing sums converge, therefore the series converges” omits the intermediate sums; monotonicity inside each phase supplies the missing bound. A claim that “we use infinitely many terms, therefore every term is used” omits surjectivity; consuming each list in order and advancing it every cycle supplies that proof.
For independent written practice, reconstruct the argument for a general convergent but non-absolutely-convergent real series with infinitely many zero terms and target −2. Your proof must establish divergence of both sign magnitudes, termination of each phase, exhaustion of all three lists, vanishing crossing errors and control of every partial sum. The target changes the stopping conditions, not those five obligations. This is public supported practice; revealing this explanation is not independent assessment evidence.
Continue
Optional depth compares rearrangement with regrouping consecutive terms. Grouping finitely many consecutive terms corresponds to a subsequence of partial sums and preserves a convergent sum; reversing that inference may fail if intermediate partial sums behave differently.
Write a convergence proof using a tail estimate and a separate argument about order. Identify whether absolute values are placed around the entire tail or around each term. That distinction controls which theorem applies and will reappear in improper integrals. Continue to neighborhoods on the real line, which organize the domains where limits and continuous functions live.
