The sum of an infinite series is a definition, not a longer addition: it is the number the partial sums approach, and a series with no such number has no sum. For a nonzero geometric series the whole question reduces to the size of rⁿ; the all-zero series is the transparent exception that has sum 0 for any chosen multiplier.
Core ideas
“Geometric Sequences and Series” supplied a formula for every finite stopping place. This lesson asks whether the resulting sequence of finite sums approaches one number.
Nobody adds infinitely many numbers, and no definition asks them to. S∞ is the number the sequence S₁, S₂, S₃, … approaches; if that sequence approaches nothing, the series has no sum and the expression naming it names nothing. Almost everything that sounds paradoxical here comes from reading the + … as an instruction to be carried out.
Write Sₙ = a₁/(1 - r) - [a₁/(1 - r)]rⁿ. The first piece does not depend on n at all, and the second is that same constant times rⁿ. So the entire question is whether rⁿ is driven to 0, and for |r| < 1 it is, at a rate you can compute. If a₁ = 0, every term and partial sum is already 0, so no condition on a chosen multiplier is needed.
For 4 + 2 + 1 + 1/2 + … the gap is exact: 8 - Sₙ = 8(1/2)ⁿ. Getting within 1/1000 of 8 takes 13 terms and no fewer, since 8 - S₁₂ = 1/512 and 8 - S₁₃ = 1/1024. "Approaches 8" means every distance you name has a term count attached.
The material
What an infinite sum is defined to be, why |r| < 1 settles every nonzero geometric series, and the zero exception.
“Partial Sums and Telescoping” made S₁, S₂, S₃, … a sequence in its own right. This lesson asks one question about that sequence, and the answer to it is what the phrase "the sum of an infinite series" is defined to mean.
Given a₁ + a₂ + a₃ + … , form the partial sums Sₙ. If there is a number S such that the partial sums can be brought within any distance you name of S by taking n large enough, and stay within it for every larger n, the series converges and S is its sum, written S∞. Otherwise the series diverges and has no sum.
The definition is about a sequence of numbers, each of which is an ordinary finite sum. At no point does anything infinite get added.
The sum of a convergent series is also written Σ aₖ with k running from 1 to ∞. Writing the symbol does not bring the number into existence: for a divergent series it is notation with nothing behind it. For 2 + 6 + 18 + …, the expression a₁/(1 - r) returns -1; the partial sums run away, so that number is not a sum and the expression was used without a licence.
If a₁ ≠ 0, the series a₁ + a₁r + a₁r² + … converges if and only if |r| < 1, and then S∞ = a₁/(1 - r). If a₁ = 0, every term and every partial sum is 0, so the series converges to 0 for any chosen r.
The case r = 0 needs no exclusion. With a₁ ≠ 0 it gives a₁, 0, 0, …, and the formula returns a₁; with a₁ = 0 it is the all-zero case.
If a₁ = 0, the recurrence multiplies 0 at every step and the partial sums are 0, 0, 0, …, which already proves the stated exception. Now take a₁ ≠ 0.
For r = 0, the partial sums are a₁, a₁, a₁, …, so they converge to a₁ = a₁/(1 - 0). For 0 < |r| < 1, Exponential and Logarithmic Functions's exponential-decay result says the positive sequence |r|ⁿ approaches 0. Since |rⁿ| = |r|ⁿ, the signed power rⁿ does too, alternating when r < 0. In Sₙ = a₁/(1 - r) - [a₁/(1 - r)]rⁿ, the second piece therefore approaches 0, leaving a₁/(1 - r).
Suppose instead |r| > 1. Then |aₙ| = |a₁||r|ⁿ⁻¹ grows without bound, and by “Partial Sums and Telescoping” the gap between consecutive partial sums is aₙ. Partial sums moving further and further apart cannot be settling on a number. For r = 1, Sₙ = na₁, which runs away; for r = -1, Sₙ is a₁, 0, a₁, 0, … , visiting two values for ever and approaching neither. Those three exhaust |r| ≥ 1.
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The gap 8(1/2)ⁿ is positive for every n, so no partial sum ever equals 8 and none exceeds it. S∞ = 8 holds anyway, because "sum" was defined as the number approached and not a number reached. This is where 0.999… = 1 stops being surprising: that no truncation of it equals 1 is a fact about truncations.
1 - 1 + 1 - 1 + … is geometric with a₁ = 1 and r = -1, so |r| = 1 and it has no sum; its partial sums are 1, 0, 1, 0, … . Bracketing it as (1 - 1) + (1 - 1) + … suggests 0. Bracketing it as 1 - (1 - 1) - (1 - 1) - … suggests 1. The formula a₁/(1 - r) returns 1/2. Three manipulations, three answers, and none of the three is licensed: brackets may be moved around a finite sum, and this is not one.
If a series converges then aₙ = Sₙ - Sₙ₋₁ approaches 0, since both partial sums approach the same number. The converse is false. In the harmonic series 1 + 1/2 + 1/3 + 1/4 + … the terms shrink to 0; but group the finite partial sum S₁₆ as 1 + 1/2 + (1/3 + 1/4) + (1/5 + … + 1/8) + (1/9 + … + 1/16) and each bracket exceeds 1/2, so S₁₆ > 3, and every doubling of n adds more than another 1/2. The partial sums pass every bound. Each step there regroups one finite sum, which is what the previous example was not entitled to do.
For a geometric series the question is now closed: find r, compare |r| with 1, and either write a₁/(1 - r) or say there is no sum. The general question — hand me a series, does it converge — has no one-line answer of that kind, and it is the business of calculus.
Interactive
The partial sums approaching a limit, or refused and shown running away
Method
Check the transparent zero case first; every other geometric series is decided by the size of its multiplier.
If a₁ = 0 and every later term is 0, every partial sum is already 0 and the sum is 0. Otherwise divide consecutive nonzero terms at least twice, or verify the multiplier directly when a zero appears. Write r as an exact fraction, not a rounded decimal.
If |r| ≥ 1 there is no sum and a₁/(1 - r) must not be written down as one: it returns a number whenever r ≠ 1, and that number has no meaning here. Only |r| < 1 licenses the next step.
If the series starts at k = 0, or the first few terms were peeled off and handled separately, a₁ is the first term of what remains. The formula sums from its own first term onward and knows nothing about what came before it.
With positive terms the sum must exceed every partial sum. With r negative the partial sums alternate around it, so the sum lies between S₁ and S₂. Either check catches a sign lost in 1 - r.
A negative ratio, so the partial sums come at the answer from both sides. The point of the example is the third column: the exact distance from each partial sum to the sum.
The gap column is the lesson. It is never 0, so no partial sum equals 1/3; its size halves at every step, so naming any distance names a term beyond which every partial sum is closer than that. 1/3 is the sum of this series in that sense and in no other.
Adding six terms by hand and computing S₆ from the closed form agree at 21/64, which checks this one instance — a₁ = 1/2, r = -1/2, n = 6. “Geometric Sequences and Series”'s cancellation is what makes Sₙ = a₁(1 - rⁿ)/(1 - r) true in general, and (-1/2)ⁿ being driven to 0 is what makes S∞ = 1/3 true.
Because r is negative, S∞ is not larger than every partial sum: S₁ = 1/2 overshoots, S₂ = 1/4 undershoots, and 1/3 stays caught between consecutive partial sums for ever. Step 4's check has to be read in the form matching the sign of r.
For a faded convergence pass, use 3 - 1 + 1/3 - 1/9 + …. The supplied ratio is r = -1/3, so |r| < 1 licenses a sum. Compute S∞, add the first four terms to obtain S₄, and derive the exact signed gap S∞ - S₄ from the closed form. Check that its sign matches which side of the limit S₄ lies on before opening the quiz.
You should be able to say what S∞ means without using the word "add", dispatch the all-zero exception, decide any other geometric series in a line, compute how many terms bring a partial sum within a stated distance of the limit, and name what goes wrong in each of the three divergent cases |r| > 1, r = 1 and r = -1.
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