Demonstration 1 of 4
Partial sums record finite progress
Each move covers half of what is left. Does any finite number of moves reach 1?
Each new piece is exactly half the remaining gap, so the gap halves and stays the same size as the last piece added.
\[S_n = 1 - 1/(2^n)\]
\[1 - S_n = 1/(2^n)\]
S_n is the total of the first n moves: 1/2 + 1/4 + ... + 1/(2^n). The gap 1 - S_n is how much of the one-unit path is still left.
Predict first. Add a fourth move. What is S_4, and how big is the gap?
Choose an example
Constructed example: the chapter's half-distance path with S_1 = 1/2, S_2 = 3/4, S_3 = 7/8 and S_4 = 15/16.
Calculated values
- Terms added
- 3
- S_3
- 1/2 + 1/4 + 1/8 = 7/8
- Gap from 1
- 1 - 7/8 = 1/8
- Formula check
- 1 - 1/(2^3) = 1 - 1/8 = 7/8
- Reaches 1?
- no, the gap is positive at every finite stage
S_3 adds exactly 3 terms: 1/2 + 1/4 + 1/8 = 4/8 + 2/8 + 1/8 = 7/8. The gap is 1 - 7/8 = 1/8, the same size as the last move. The formula agrees: 1 - 1/8 = 7/8. Still below 1.
Use the idea
When a process keeps closing a fixed share of what remains, write the remaining gap as a formula instead of trusting a rounded display.
Where the conclusion applies
An idealized sum on a one-unit path, not a claim about real distances or devices. Every S_n is a finite sum; the value 1 belongs to the whole sequence, not to any one stage.
Check your understanding: What is S_5, and what gap does it leave?
Chapter 27 source: section "Partial sums record finite progress". Demonstration C27-D01.
Demonstration 2 of 4
The multiplier decides
The formula a/(1 - r) gives a number for almost any r. When does that number mean anything?
When |r| < 1 the terms shrink and the partial sums settle on a/(1 - r). When it fails, the sums keep growing, and the formula's answer describes nothing about them.
\[a/(1 - r)\]
\[|r| < 1\]
a is the first term and r is the fixed multiplier from one term to the next. |r| is the size of r without its sign. The formula a/(1 - r) is the sum only when |r| < 1.
Predict first. Switch r to 2. Does a/(1 - r) still give a number, and do the dots settle toward it?
Choose an example
Constructed example: the chapter's half-distance series (a = 1/2, r = 1/2), its alternating r = -1/2, and its failing cases r = 1 and r = 2.
Calculated values
- First term a
- 1/2
- Multiplier r
- 1/2
- S_1 to S_4
- 1/2, 3/4, 7/8, 15/16
- S_8
- 255/256
- Sum a/(1 - r)
- (1/2)/(1 - (1/2)) = (1/2)/(1/2) = 1
Each term is the one before times 1/2: S_2 = 1/2 + 1/4 = 3/4. Formula: (1/2)/(1 - (1/2)) = (1/2)/(1/2) = 1. Since |r| = 1/2 is below 1, the terms shrink and the partial sums settle toward 1, always from below.
Use the idea
Before using a shortcut formula, check the condition it was built on. A fraction bar is not a permission slip.
Where the conclusion applies
First term a = 1/2 throughout, as in the half-distance series. Only eight partial sums are drawn; the limit claim rests on |r| < 1, not on the eight dots.
Check your understanding: With a = 1/2 and r = -1/2, what are S_1, S_2 and S_3, and what is a/(1 - r)?
Chapter 27 source: section "Convergence gives “close” a condition". Demonstration C27-D02.
Demonstration 3 of 4
A tolerance you can meet and keep
How many terms of a geometric series bring the gap from its limit a/(1 - r) below a requested size, and does it stay there?
The upper bar adds the terms toward the limit. The lower bar enlarges the last stretch: the hatched gap after n terms fits inside the gold bracket, and the ticks for later sums sit even closer to the limit, which is the "every later" part of convergence.
\[S_n-rS_n=a-ar^n.\]
\[ar^n/(1-r)\]
S_n is the sum of the first n terms, a = 3 is the first term and r is the multiplier. The gap from the limit after n terms is ar^n/(1-r).
Predict first. Keep the request at 0.01 but change r to 1/2. Will you need more terms or fewer?
Choose an example
Constructed example: the chapter's worked example with a = 3 and r = 1/4, plus r = 1/2 and other tolerances.
Calculated values
- Series
- 3 + 3(1/4) + 3(1/4)^2 + ...
- Limit a/(1 - r)
- 3/(1 - 1/4) = 4
- Gap after n terms
- 3(1/4)^n/(3/4)
- First n with gap below 0.01
- 5
- Gap at n = 5
- 1/256, about 0.00391
- Gap at n = 4
- 1/64, about 0.01562, not below
At n = 5, the gap is 3 x 1/1024 / (3/4) = 1/256, about 0.00391, which is below 0.01; at n = 4 it was 1/64, which is not. Each later step multiplies the gap by 1/4, so every later gap stays below 0.01 too. That promise about every later stage is what convergence means.
Use the idea
When a result must be accurate to a stated tolerance, solve for the stage that guarantees it, instead of stopping when one number looks close.
Where the conclusion applies
a = 3 and |r| < 1, so the tail formula applies. A stricter request needs a later stage; a sequence that is close once and then wanders off would not converge.
Check your understanding: With a = 3 and r = 1/4, what is the sum after 3 terms, and how much is left?
Chapter 27 source: section "Worked example: a finite sum exposes the tail". Demonstration C27-D03.
Demonstration 4 of 4
Continuity is a three-part check
Hole, jump or smooth: which of the three conditions fails, and where?
The curve shows the nearby behavior; the dot at a shows the assigned value, or an open circle where there is none. Continuity asks the two to agree.
\[f(x) = (x^2 - 1)/(x - 1)\]
\[g(x) = x + 1\]
Continuity at a needs three things: f(a) exists, the limit as x approaches a exists, and the two are equal. The gold squares are sample inputs a small step left and right of a.
Predict first. Choose the jump and shrink the step to 0.01. Do the two sample outputs move closer together?
Choose an example
Constructed example: the chapter's removable hole, jump, line x + 1, and reassigned value g(2) = 8.
Calculated values
- Value at the point
- undefined (0/0 at x = 1)
- Left side approaches
- 2
- Right side approaches
- 2
- Two-sided limit
- 2
- Continuous?
- Not continuous
For x not equal to 1, f(x) = x + 1: At x = 0.9: 0.9 + 1 = 1.9; at x = 1.1: 1.1 + 1 = 2.1. Not continuous: condition 1 fails, because f(1) does not exist. Defining f(1) = 2 would fill the hole.
Use the idea
When a graph looks connected, run the three checks anyway. A drawing can hide a single missing or misplaced point.
Where the conclusion applies
The chapter's four constructed functions. The samples are evidence, not proof: the factoring x^2 - 1 = (x - 1)(x + 1) covers every input near 1, and the jump's two sides are fixed by its definition.
Check your understanding: For f(x) = (x^2 - 1)/(x - 1), what is f(0.99), and which value would make f continuous at 1?
Chapter 27 source: section "Continuity is a three-part check". Demonstration C27-D04.