Pre-Calc 11 · Lesson 5

Infinite Series.

Add up infinitely many shrinking pieces and, astonishingly, the total can settle on a single finite number you can find with one formula: Sᵢnfinity = t₁ / (1 - r).

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By the end, you can

1I can decide whether an infinite geometric series converges or diverges by checking whether the common ratio r satisfies -1 < r < 1.
2I can find the sum of a convergent infinite geometric series using Sᵢnfinity = t₁ / (1 - r).
3I can work backward from a known sum to find a missing first term, common ratio, or the value of a variable x.
4I can model real situations (bouncing balls, pendulums, repeating decimals, shaded areas) as infinite geometric series and find their totals.
5I can explain in plain language WHY an infinite sum can be finite by watching the sequence of partial sums approach a fixed value.

01 / CONCEPTS

The core ideas.

Partial Sums and What 'The Sum' Even Means

You cannot literally add infinitely many numbers one at a time. Instead, you watch the sequence of partial sums S₁, S₂, S₃, ... where each Sₙ is the total of the first n terms. If those running totals home in on a single fixed number as n grows, that number is defined to be the sum of the infinite series.

S_n = t_1 + t_2 + \cdots + t_n

Why it works

Think of the series 4 + 2 + 1 + 0.5 + 0.25 + .... The partial sums go 4, 6, 7, 7.5, 7.75, 7.875, ... creeping toward 8 but never overshooting. The gap to 8 keeps halving, so it shrinks past any tiny amount you name. 'The sum is 8' is shorthand for 'the partial sums get and stay as close to 8 as you like.'

Convergent vs Divergent

An infinite series is convergent if its partial sums approach a finite number; that number is the sum. It is divergent if the partial sums never settle, usually because they grow without bound. For a geometric series, the deciding factor is the common ratio r.

\text{converges} \iff -1 < r < 1

Why it works

If |r| < 1 each new term is a fraction of the one before, so terms shrink toward zero and the leftover tail becomes negligible. If |r| >= 1 the terms stay the same size or grow (2 + 4 + 8 + 16 + ...), so the total keeps climbing and there is nothing finite to land on. The terms must vanish for the sum to settle, and only |r| < 1 makes geometric terms vanish.

Why rⁿ Disappears

The finite geometric sum is Sₙ = t₁(1 - rⁿ)/(1 - r). The only part that depends on n is rⁿ. When -1 < r < 1, raising r to higher and higher powers drives rⁿ toward 0, so the whole formula collapses to t₁/(1 - r).

S_n = \dfrac{t_1(1 - r^n)}{1 - r} \;\xrightarrow[n\to\infty]{}\; \dfrac{t_1}{1 - r}

Why it works

Multiplying a number between -1 and 1 by itself repeatedly is like repeatedly taking a fraction of a fraction: 0.5, 0.25, 0.125, ... races to zero. So (1 - rⁿ) becomes (1 - 0) = 1, and the n-dependence simply switches off. The infinite formula is just the finite formula with its one moving part frozen at zero.

The Infinite Geometric Sum Formula

For a geometric series with first term t₁ and common ratio r where -1 < r < 1, the entire infinite sum equals t₁ divided by (1 - r). Always confirm convergence first; if |r| >= 1 the formula does not apply and no sum exists.

S_\infty = \dfrac{t_1}{1 - r}, \quad -1 < r < 1

Why it works

Here is a one-line derivation. Let S = t₁ + t₁ r + t₁ r² + .... Multiply by r: rS = t₁ r + t₁ r² + .... Subtract: S - rS = t₁ because every other term cancels. So S(1 - r) = t₁, giving S = t₁/(1 - r). The clean cancellation is exactly why the formula looks so simple.

Reading Off t₁ and r

To use the formula you need the true first term and the true common ratio. Find r by dividing any term by the one before it (r = t₂/t₁). A negative r produces an alternating series, and you must keep the sign.

r = \dfrac{t_2}{t_1} = \dfrac{t_3}{t_2}

Why it works

The common ratio is the single 'shrink-and-flip' factor that turns one term into the next, so it lives in the quotient of consecutive terms. Forgetting the sign on an alternating series like 1 - 1/3 + 1/9 - ... (where r = -1/3) is the most common slip, and it changes 1 - r from 4/3 to 2/3, doubling your error.

Working Backward From a Known Sum

Because Sᵢnfinity, t₁, and r are tied together by one equation, knowing any two lets you solve for the third. Rearrange Sᵢnfinity = t₁/(1 - r) into t₁ = Sᵢnfinity(1 - r) or r = 1 - t₁/Sᵢnfinity.

t_1 = S_\infty(1 - r), \qquad r = 1 - \dfrac{t_1}{S_\infty}

Why it works

The formula is just an equation with three quantities. Algebra does not care which one is unknown. This is what powers puzzle problems like 'a series sums to 81 with r = 2/3, find t₁' and the convergence ranges for series built with a variable x inside r.

02 / WORKED EXAMPLES

From easy to tricky.

EX 1easy

Decide whether 1 - 1/3 + 1/9 - ... is convergent or divergent, and find its sum if it exists.

Find r

r = t₂/t₁ = (-1/3)/1 = -1/3.

Check convergence

Since -1 < -1/3 < 1, the series converges.

Apply formula

S_∞ = t₁/(1 - r) = 1/(1 - (-1/3)) = 1/(4/3).

Simplify

1 ÷ 4/3 = 3/4.

EX 2easy

Decide whether 2 - 4 + 8 - ... is convergent or divergent. Find its sum if it exists.

Find r

r = (-4)/2 = -2.

Check convergence

|r| = 2, which is not less than 1, so r is outside -1 < r < 1.

Conclude

The terms grow in size (2, 4, 8, ...), so the partial sums never settle.

EX 3medium

The shaded regions of a square form the series 1/4 + 1/16 + 1/64 + .... How much of the largest square is shaded? (MHR Example 2)

Identify t₁ and r

t₁ = 1/4. Each term is 1/4 of the previous: r = (1/16)/(1/4) = 1/4.

Check convergence

-1 < 1/4 < 1, so the series converges.

Apply formula

S_∞ = (1/4)/(1 - 1/4) = (1/4)/(3/4).

Simplify

(1/4)(4/3) = 1/3.

EX 4medium

Express the repeating decimal 0.584584584... as an infinite geometric series and find its value as a fraction. (MHR Example 2 Your Turn)

Break into pieces

0.584 + 0.000584 + 0.000000584 + ...

Identify t₁ and r

t₁ = 0.584. Each block shifts three more decimal places, so r = 0.001.

Apply formula

S_∞ = 0.584/(1 - 0.001) = 0.584/0.999.

Write as fraction

0.584/0.999 = 584/999.

EX 5hard

The first term of an infinite geometric series is -8 and its sum is -40/3. Find r and list the first four terms. (MHR Q7)

Set up the equation

S_∞ = t₁/(1 - r) gives -40/3 = -8/(1 - r).

Solve for (1 - r)

1 - r = -8 ÷ (-40/3) = -8 · (-3/40) = 24/40 = 3/5.

Solve for r

r = 1 - 3/5 = 2/5.

List terms

t₁ = -8, t₂ = -8(2/5) = -16/5, t₃ = -32/25, t₄ = -64/125.

EX 6hard

A ball is dropped from 16 m and rebounds to one half of its previous height each bounce. Find the total vertical distance it travels. (MHR Q15)

Account for the first drop

The initial fall is 16 m (down only).

Model the bounces

After the first drop the ball goes up 8 m and back down 8 m, then 4 and 4, then 2 and 2, .... Each up-down pair counts twice.

Sum the rebounds

Up-distances form 8 + 4 + 2 + ... with t₁ = 8, r = 1/2, so their sum is 8/(1 - 1/2) = 16. The ball travels up 16 m total and back down 16 m, a total of 32 m of bouncing.

Add the first drop

Total = 16 (first drop) + 32 (bounces) = 48 m.

EX 7challenge

The series 1 + 3x + 9x² + 27x³ + ... has a sum of 4. Find x and list the first four terms. (MHR Q9)

Identify the ratio

Each term multiplies the previous by 3x, so t₁ = 1 and r = 3x.

Set up the sum equation

S_∞ = t₁/(1 - r) gives 4 = 1/(1 - 3x).

Solve

1 - 3x = 1/4, so 3x = 3/4, giving x = 1/4.

Check convergence

r = 3x = 3/4, and -1 < 3/4 < 1, so the series truly converges. Valid.

List terms

Substitute x = 1/4: 1, 3(1/4) = 3/4, 9(1/16) = 9/16, 27(1/64) = 27/64.

03 / COMMON TRAPS

Where students slip.

Trap

Any infinite sum must be infinite, so adding forever can never give a finite answer.

Fix

When terms shrink fast enough (|r| < 1 for geometric series) the leftover tail becomes negligible and the partial sums lock onto a finite value. 4 + 2 + 1 + 0.5 + ... sums to exactly 8, not infinity.

Trap

The formula Sᵢnfinity = t₁/(1 - r) works for every geometric series.

Fix

It only applies when -1 < r < 1. If |r| >= 1 the series diverges and has no sum. Always check r before plugging in: 2 - 4 + 8 - ... has r = -2, so it diverges and the formula must not be used.

Trap

Dropping the negative sign on r for alternating series because 'a sum should be positive.'

Fix

r keeps its sign. For 1 - 1/3 + 1/9 - ... the ratio is r = -1/3, so 1 - r = 1 - (-1/3) = 4/3 and S = 3/4. Using r = +1/3 gives the wrong denominator 2/3 and the wrong answer 3/2.

Trap

0.999... is just below 1, never quite equal.

Fix

Write 0.999... = 0.9 + 0.09 + 0.009 + ..., a geometric series with t₁ = 0.9 and r = 0.1. Its sum is 0.9/(1 - 0.1) = 0.9/0.9 = 1 exactly. The repeating decimal IS the number 1.

04 / QUIZ

Test yourself.

The Test

33 problems across three tiers. Auto-graded on submit. Hints available before you submit, full solutions after.

0/33Answered

Core

Core (12)

Which infinite geometric series is convergent?

Need a hint?

A geometric series converges only when -1 < r < 1.
Check the common ratio of each option against that range.

Find the sum of the infinite geometric series with t₁ = 8 and r = -1/4.

Need a hint?

Confirm -1 < r < 1, then use Sᵢnfinity = t₁/(1 - r).
1 - (-1/4) = 5/4.

Find the sum of the infinite geometric series 1 + 0.5 + 0.25 + ....

Need a hint?

r = 0.5, t₁ = 1.

Does the infinite geometric series with t₁ = 3 and r = 4/3 have a sum?

Need a hint?

Compare r = 4/3 with the convergence condition -1 < r < 1.

Find the sum of 4 - 12/5 + 36/25 - ....

Need a hint?

Find r by dividing the second term by the first: (-12/5)/4.
r = -3/5, so 1 - r = 8/5.

Find the sum of 27 - 9/5 + 3/25 - ... (geometric).

Need a hint?

r = (-9/5)/27.
r = -1/15, so 1 - r = 16/15.

What is the common ratio of the convergent series 125 + 25 + 5 + ...?

Need a hint?

Divide a term by the one before it: 25/125.

Express 5 + 5(2/3) + 5(2/3)² + ... as a single number.

Need a hint?

t₁ = 5, r = 2/3.

Find the sum 7 + 7(1/2) + 7(1/2)² + ....

Need a hint?

t₁ = 7, r = 1/2.

C10

Fill in the blanks: An infinite geometric series converges when ___ < r < ___, and its sum is Sᵢnfinity = ___.

Need a hint?

The ratio must keep the terms shrinking toward zero.

C11

Which series is divergent?

Need a hint?

Find r for each and test against -1 < r < 1.

C12

Find the sum of the infinite geometric series with t₁ = 6 and r = -2/3.

Need a hint?

Keep the negative sign on r when computing 1 - r.
1 - (-2/3) = 5/3.

Apply

Apply (15)

The sum of an infinite geometric series is 81 and its common ratio is 2/3. Find the first term and write the first three terms.

Need a hint?

Rearrange S = t₁/(1 - r) to t₁ = S(1 - r).
1 - 2/3 = 1/3, so t₁ = 81 * 1/3.

The first term of an infinite geometric series is -8 and its sum is -40/3. Find the common ratio and the first four terms.

Need a hint?

Use r = 1 - t₁/S.
t₁/S = (-8)/(-40/3) = 3/5.

In its first month an oil well produced 24000 barrels, then 94% of the previous month each month after. If this continues, what is the lifetime production?

Need a hint?

t₁ = 24000, r = 0.94.
1 - 0.94 = 0.06.

The series 1 + 3x + 9x² + 27x³ + ... has a sum of 4. Find x and list the first four terms.

Need a hint?

The common ratio is 3x; set 1/(1 - 3x) = 4.
1 - 3x = 1/4.

The sum of an infinite geometric series is twice its first term. Find the common ratio.

Need a hint?

Set S = 2t₁ in the formula.
t₁/(1 - r) = 2t₁ means 1/(1 - r) = 2.

A pendulum's initial swing is 50 cm and each successive swing is 0.8 times the previous. If this continues forever, what total distance does the pendulum swing?

Need a hint?

t₁ = 50, r = 0.8.

A hot air balloon rises 25 m in its first minute, and each succeeding minute rises only 80% as high as the minute before. What is the balloon's maximum altitude?

Need a hint?

t₁ = 25, r = 0.8.

A ball is dropped from 16 m and rebounds to half its previous height each bounce. Find the total vertical distance travelled.

Need a hint?

Add the first drop separately, then count each rebound up and down.
Rebound ups: 8 + 4 + 2 + ... = 16, doubled for up and down = 32.

Andrew uses the infinite sum formula on 1 + 1.1 + 1.21 + 1.331 + ... and gets 10. Is his answer reasonable? Explain.

Need a hint?

Find r and check whether |r| < 1.

A10

Each side of an equilateral triangle is 1 cm. Joining the midpoints forms a smaller triangle, repeated forever. Find the sum of all the triangles' perimeters.

Need a hint?

The first perimeter is 3 cm. Each new triangle has sides half as long, so its perimeter is half.
t₁ = 3, r = 1/2.

A11

Express the repeating decimal 0.87878787... as an infinite geometric series and find its value as a fraction.

Need a hint?

Write it as 0.87 + 0.0087 + 0.000087 + ...
t₁ = 0.87, r = 0.01.

A12

For which values of x is the series 5 + 5x + 5x² + 5x³ + ... convergent?

Need a hint?

Here the common ratio is x itself.

A13

For the series 1 + x/3 + (x/3)² + (x/3)³ + ..., find all x for which it converges.

Need a hint?

The common ratio is x/3.
Require -1 < x/3 < 1, then multiply by 3.

A14

Dominique says -1/3 + 4/9 - 16/27 + ... sums to -1/7. Rita says it diverges. Who is correct?

Need a hint?

Compute r = (4/9)/(-1/3).
|r| compared with 1 decides convergence.

A15

A pile driver moves a post 30 cm on the first impact and 27 cm on the second, the distances forming a geometric sequence. If pounded indefinitely, what total distance is the post driven?

Need a hint?

r = 27/30 = 0.9.

Extension

Extension (6)

Does 0.999... = 1? Support your answer with an infinite geometric series.

Need a hint?

Write 0.999... = 0.9 + 0.09 + 0.009 + ...
t₁ = 0.9, r = 0.1.

A 24 cm square is cut into four 12 cm squares; three are placed side by side and the fourth is cut into four 6 cm squares, three placed in the row, and so on forever. What is the total length of the arrangement?

Need a hint?

Three 12 cm squares make a base length of 36 cm. Each later stage adds a strip half as wide.
Added lengths: 6 + 3 + 1.5 + ... with t₁ = 6, r = 1/2.

Show that if x = 1/(z+1) for an integer z, then x + x² + x³ + ... = 1/z.

Need a hint?

The series x + x² + ... has first term x and ratio x, so its sum is x/(1 - x).
Substitute x = 1/(z+1) and simplify 1 - x = z/(z+1).

A mini-lab cuts a unit-area square into 4 parts, gives one part to each of three group members, then repeats on the leftover quarter forever. What total fraction of the square does each member receive?

Need a hint?

Each member gets 1/4 at stage 1, then 1/4 of the remaining 1/4, and so on.
Each member's series is 1/4 + 1/16 + 1/64 + ... with t₁ = 1/4, r = 1/4.

For the same pile driver (30 cm then 27 cm, r = 0.9), how far is the post driven after exactly 8 impacts? Round to two decimals.

Need a hint?

This is a finite geometric sum: Sₙ = t₁(1 - rⁿ)/(1 - r).
t₁ = 30, r = 0.9, n = 8.

The first two terms of a series are 1 and 1/4. If it is an infinite geometric series, what is its sum?

Need a hint?

Geometric ratio r = (1/4)/1 = 1/4.