Stopping a series at every possible place produces one number for each stopping place, and those numbers form a sequence of their own. Subtracting consecutive members of it returns the original terms, and a series whose terms are themselves differences collapses to the two ends.
Core ideas
One shift of attention and one technique. The shift is from a single sum to the list of all of them.
“Arithmetic Sequences and Series” was emphatic that S₂₀ = 860 is one number and not a list, and that is still true. But there is an S₁, an S₂, an S₃, one for each n, and a rule assigning one number to each positive integer is a sequence by “Sequences and Recursion”'s definition — recursion and all.
Sₙ contains Sₙ₋₁ plus one extra term, so aₙ = Sₙ - Sₙ₋₁. A formula for the sums therefore determines the sequence, and the two directions are an addition and a subtraction. The catch is n = 1, where there is no Sₙ₋₁ to subtract.
If every term can be written as bₖ - bₖ₊₁ for some sequence b, the sum of n of them is b₁ - bₙ₊₁ and everything between has cancelled in pairs. “Geometric Sequences and Series”'s Sₙ - rSₙ argument was exactly this. Finding the bₖ is the hard part, and there is no procedure for it.
The material
One number for each stopping place, the subtraction that undoes the adding, and series whose middle cancels.
Given a sequence a₁, a₂, a₃, … , its partial sums are S₁ = a₁, S₂ = a₁ + a₂, and in general Sₙ = Σ aₖ with k running from 1 to n. The list S₁, S₂, S₃, … is the sequence of partial sums of the series.
A series and its sequence of partial sums are two names for one situation. Every question about the series — how big does it get, does it settle — is a question about this sequence.
The partial sums have a recursion, and it is the cheapest way to build a table of them: S₁ = a₁ and Sₙ = Sₙ₋₁ + aₙ. Each new partial sum costs one addition rather than n of them, which is what makes computing twenty of them by hand a reasonable thing to do.
a₁ = S₁, and aₙ = Sₙ - Sₙ₋₁ for every n ≥ 2.
The proof is one line: Sₙ = (a₁ + … + aₙ₋₁) + aₙ = Sₙ₋₁ + aₙ, and subtracting Sₙ₋₁ from both sides gives the statement. The restriction to n ≥ 2 is not fussiness — S₀ was never given a value.
So a formula for Sₙ determines the terms, and the first one has to be handled on its own. Take Sₙ = n² + 5. For n ≥ 2, aₙ = (n² + 5) - ((n - 1)² + 5) = n² - (n - 1)² = 2n - 1. But a₁ = S₁ = 6, and 2n - 1 at n = 1 is 1, so the sequence is 6, 3, 5, 7, 9, … with a genuinely exceptional first term. With Sₙ = n² the same computation gives 2n - 1 again and a₁ = S₁ = 1 does match it. Which of the two happens has to be checked.
A sum whose terms are the differences of consecutive members of one sequence: aₖ = bₖ - bₖ₊₁ for every k in the range. The name is the instrument — the sum closes up on itself and only the two ends are left showing.
For any sequence b whatever, Σ (bₖ - bₖ₊₁) = b₁ - bₙ₊₁ with k running from 1 to n. In the other orientation, Σ (bₖ₊₁ - bₖ) = bₙ₊₁ - b₁; it is the same statement with every sign reversed.
No condition is placed on b. It need not be arithmetic, geometric, positive or increasing.
Write the sum as n brackets: (b₁ - b₂) + (b₂ - b₃) + … + (bₙ - bₙ₊₁). Take any bₘ with 2 ≤ m ≤ n. It occurs twice: as the subtracted half of bracket m - 1, and as the added half of bracket m. Those two occurrences cancel.
What has no partner is the added b₁ in the first bracket and the subtracted bₙ₊₁ in the last. So the total is b₁ - bₙ₊₁.
The only property used is that a finite sum may be regrouped freely, which is the associative law applied finitely many times. There are n brackets and 2n terms, and the finiteness is what makes the argument legal.
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First check the split rather than asserting it: 1/k - 1/(k + 1) = ((k + 1) - k)/(k(k + 1)) = 1/(k(k + 1)). So with bₖ = 1/k, the term 1/(k(k + 1)) is exactly bₖ - bₖ₊₁, and the sum from k = 1 to n is b₁ - bₙ₊₁ = 1 - 1/(n + 1) = n/(n + 1). For n = 1, 2, 3, 4 that predicts 1/2, 2/3, 3/4, 4/5, and adding the terms by hand gives the same four numbers.
The geometric sum formula was a telescoping in disguise. Put bₖ = a₁rᵏ⁻¹, the k-th term of the sequence. Then bₖ - bₖ₊₁ = a₁rᵏ⁻¹ - a₁rᵏ = a₁rᵏ⁻¹(1 - r), so summing from 1 to n gives (1 - r)Sₙ on one side and b₁ - bₙ₊₁ = a₁ - a₁rⁿ on the other. That is Sₙ(1 - r) = a₁(1 - rⁿ), which is where “Geometric Sequences and Series” arrived by multiplying the sum by r and subtracting. The two routes are the same cancellation described differently.
The difficulty is always finding bₖ, and there is no method. Single fractions with a factorable denominator are the usual candidates, and the test is whether the split recombines: if bₖ - bₖ₊₁ does not simplify back to aₖ, the split is wrong. One more that works: with bₖ = (k - 1)², bₖ₊₁ - bₖ is k² - (k - 1)² = 2k - 1, so the first n odd numbers sum to bₙ₊₁ - b₁ = n² - 0 = n² — a second derivation of a result “Arithmetic Sequences and Series” reached by pairing. A useful telescope has to reveal the sum from a bₖ that can be found without already knowing the partial sums; manufacturing one from the unknown answer would explain nothing.
Interactive
S1, S2, S3, ... beside the terms, and a telescope collapsing bracket by bracket
Method
Four steps. The third has no procedure behind it, and the fourth is what tells you whether it worked.
S₁ = a₁, then add one term at a time. Four numbers cost four additions and show at once whether the sums are climbing, falling, settling or oscillating.
aₙ = Sₙ - Sₙ₋₁ for n ≥ 2. Then compute a₁ = S₁ on its own and check whether your formula also produces it. Sometimes it does; when it does not, say so, because the first term really is exceptional.
Split the term, then combine the split back over a common denominator. If it does not return the term you started with, the split is wrong. A denominator that factors into consecutive things is the case that usually yields; nothing else is guaranteed to.
The result is b₁ - bₙ₊₁, and the commonest error puts bₙ there instead. Evaluating your closed form at n = 1 and n = 2 against the sums added by hand catches an index error at either end.
The terms are 1/2, 1/6, 1/12, 1/20, … and nothing about them suggests a formula. Splitting each one into a difference is what turns the sum into two numbers.
The four checks establish the formula at n = 1, 2, 3, 4 and nothing beyond that. What makes it true for every n is the telescoping argument: written out, the sum is (1/1 - 1/2) + (1/2 - 1/3) + … + (1/n - 1/(n + 1)), every interior fraction appears once with each sign, and 1 and 1/(n + 1) are left.
The last survivor is 1/(n + 1) and not 1/n. There are n brackets, the last is 1/n - 1/(n + 1), and it is the second half of that bracket that has no partner. Writing 1 - 1/n gives (n - 1)/n, wrong by 1/(n(n + 1)) and entirely plausible on the page.
The partial sums 1/2, 2/3, 3/4, 4/5, … climb and every one is below 1. Whether they add up to 1 is the question of “Infinite Geometric Series”, and having n/(n + 1) in closed form is what makes it answerable.
For a faded telescope, sum 2/((2k - 1)(2k + 1)) from k = 1 to k = 5. The supplied split is 1/(2k - 1) - 1/(2k + 1). Recombine it first, write all five brackets, identify the two survivors, and compare the collapsed value with the five fractions added directly before opening the quiz.
You should be able to build four partial sums of any sequence, recover aₙ from a given Sₙ and check a₁ against S₁ separately, and cancel a telescoping sum to two surviving values with the right index on each. The Σ 1/(k(k+1)) derivation should follow from the split alone.
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