Sum of infinite GP — when it converges, proof: Worked examples

This is a companion drill-page for the parent proof note. There we proved that an infinite geometric progression (GP) sums to $\frac{a}{1 - r}$ only when $∣ r ∣ < 1$ . Here we hunt down every kind of situation that a question can throw at you, so you never meet a surprise in an exam.

Intuition Read this first — what "

$a$ ", " $r$ ", " $S_{N}$ " and " $S_{\infty}$ " mean Every GP looks like $a, a r, a r^{2}, a r^{3}, \dots$

$a$ = the first number you actually see (the "first term").
$r$ = the common ratio = $\frac{next term}{current term}$ (later divided by earlier).
$S_{N}$ = the partial sum: the total of just the first $N$ terms, e.g. $S_{3} = a + a r + a r^{2}$ . It is an ordinary finite addition — always a real number.
$S_{\infty}$ = the infinite sum: the single number the partial sums $S_{1}, S_{2}, S_{3}, \dots$ approach as $N$ grows without limit. We write $S_{N} \to S_{\infty}$ to mean " $S_{N}$ homes in on $S_{\infty}$ ". If the $S_{N}$ never home in on one value, $S_{\infty}$ does not exist and we say the series diverges.

The whole game of "does $S_{\infty}$ exist?" is decided by the size of $r$ , written $∣ r ∣$ (the "magnitude" — the value with any minus sign stripped off, so $∣ - \frac{1}{3} ∣ = \frac{1}{3}$ ). If $∣ r ∣ < 1$ the terms shrink to nothing and the partial sums settle; otherwise they never settle.

The scenario matrix

Before working anything, here is the full map of cases a question can hit. Each case class is listed with what makes it special and which worked example covers it — every cell below gets its own fully-worked example later on the page.

Definition The twelve case classes

C1 — $0 < r < 1$ (positive, shrinking). Cleanest case; all terms positive. → Ex 1
C2 — $- 1 < r < 0$ (negative, shrinking). Signs alternate $+, -, +, -$ . → Ex 2
C3 — $r = 0$ (degenerate). Every term after the first is zero. → Ex 3
C4 — $∣ r ∣$ just under $1$ (near-boundary). Converges but slowly; big sum. → Ex 4
C5 — $r > 1$ (positive, diverges). Terms grow, total runs to $+ \infty$ . → Ex 5
C6 — $r = 1$ (boundary, diverges). Every term equals $a$ ; total unbounded. → Ex 6
C7 — $r = - 1$ (boundary, oscillates). Partial sums flip, never settle. → Ex 7
C8 — $r < - 1$ (negative, $∣ r ∣ > 1$ , diverges). Oscillates and blows up. → Ex 8
C9 — real-world word problem. Bouncing ball / money — build $a, r$ yourself. → Ex 9
C10 — recurring decimal. Turn $0. \overline{ab}$ into a fraction. → Ex 10
C11 — exam twist: solve backwards. Given the sum, find $a$ or $r$ . → Ex 11
C12 — exam twist: GP that starts mid-list. First visible term is not the $a$ of the "nice" form. → Ex 12

Case C1 — positive shrinking ratio

Worked example Example 1 (cell C1) — the friendly case

Find $S_{\infty} = 9 + 3 + 1 + \frac{1}{3} + \frac{1}{9} + \dots$

Forecast: each term is one-third of the last and stays positive. Guess: a bit more than $13$ ? Write your guess before reading on.

Find $r$ . $r = \frac{3}{9} = \frac{1}{3}$ . Why this step? $r$ is next-over-current; pick any adjacent pair.
Check convergence. $∣ r ∣ = \frac{1}{3} < 1$ ✅. Why this step? The formula $\frac{a}{1 - r}$ is only legal when $∣ r ∣ < 1$ ; skipping this is the #1 exam mistake.
Read off $a$ . $a = 9$ (the first number you see). Why this step? $a$ must be the very first term — beginners wrongly grab the second term ( $3$ ).
Apply the formula. $S_{\infty} = \frac{a}{1 - r} = \frac{9}{1 - \frac{1}{3}} = \frac{9}{\frac{2}{3}} = \frac{27}{2} = 13.5.$ Why this step? With $∣ r ∣ < 1$ confirmed, the parent formula $S_{\infty} = \frac{a}{1 - r}$ is now legal to use.

Verify: add the first five terms: $9 + 3 + 1 + 0.333 \dots + 0.111 \dots = 13.444 \dots$ , already creeping up toward $13.5$ . ✅ Matches our forecast.

Figure — Sum of infinite GP — when it converges, proof

In the figure the bars are laid end to end. Look at how each coloured bar is exactly $\frac{1}{3}$ the width of the one to its left — the lavender $9$ -bar, then the coral $3$ -bar, then the mint $1$ -bar, and so on. Even though the bars go on forever, their total width never crosses the dashed line at $13.5$ : that dashed line is $S_{\infty}$ . This is the whole reason a shrinking GP has a finite sum.

Case C2 — negative shrinking ratio (alternating signs)

Worked example Example 2 (cell C2) — the see-saw case

Find $S_{\infty} = 8 - 6 + \frac{9}{2} - \frac{27}{8} + \dots$

Forecast: signs alternate, so the total should land somewhere between the first two partial sums, $8$ and $2$ . Guess before reading.

Find $r$ . $r = \frac{- 6}{8} = - \frac{3}{4}$ . Why this step? The minus sign in the second term tells us $r < 0$ ; compute it exactly, don't just guess "negative".
Check convergence. $∣ r ∣ = \frac{3}{4} < 1$ ✅. Why this step? The magnitude is what matters — the sign does not affect convergence, only how the total is approached.
Read $a$ . $a = 8$ . Why this step? Same rule as always: $a$ is the first visible term.
Apply the formula. $S_{\infty} = \frac{8}{1 - ( - \frac{3}{4} )} = \frac{8}{1 + \frac{3}{4}} = \frac{8}{\frac{7}{4}} = \frac{32}{7} \approx 4.571.$ Why this step? Subtracting a negative $r$ turns into adding, so the denominator grows to $\frac{7}{4}$ .

Verify: partial sums are $8, 2, 6.5, 3.125, 5.16 \dots$ — they zig-zag inward, squeezing toward $\frac{32}{7} \approx 4.57$ . ✅

In the figure, follow the coral dots left to right — the running total $S_{N}$ . Notice it lands above the mint line (the limit $\frac{32}{7}$ ), then below it, then above, each time by a smaller gap. That overshoot-then-undershoot zig-zag is the fingerprint of a negative ratio, and the fact that the gaps keep shrinking is exactly why $∣ r ∣ < 1$ still corrals the total into a finite home.

Case C3 — the degenerate ratio $r = 0$

In the figure, only the first lavender bar has any width — the coral tick marks where the second, third, fourth terms should be are pinned flat at zero. There is nothing to add after the first term, so the dashed total sits exactly at $5$ . This is the visual meaning of " $r = 0$ kills every later term."

Case C4 — ratio just below 1 (slow convergence)

Worked example Example 4 (cell C4) — near the cliff edge

Find $S_{\infty} = 1 + 0.99 + 0.9 9^{2} + 0.9 9^{3} + \dots$

Forecast: $r = 0.99$ is just under $1$ , so terms shrink very slowly. Do you expect a small total or a big one?

Read $a$ and $r$ . $a = 1$ , $r = 0.99$ . Why this step? $a$ is the first term; $r = \frac{0.99}{1} = 0.99$ is next-over-current.
Check convergence. $∣0.99∣ < 1$ ✅, so it does converge — but only barely. Why this step? We must confirm $∣ r ∣ < 1$ before daring to use $\frac{a}{1 - r}$ ; here it passes by a whisker.
Apply the formula. $S_{\infty} = \frac{1}{1 - 0.99} = \frac{1}{0.01} = 100.$ Why this step? The denominator $1 - r = 0.01$ is tiny, so dividing by it produces a large total.

Verify: because $r$ is close to $1$ , the denominator $1 - r = 0.01$ is tiny, so dividing by it makes a huge sum ( $100$ ). Sanity check: $\frac{1}{1 - r} \to \infty$ as $r \to 1^{-}$ , which matches "near-boundary GPs have large sums." ✅

Lesson for the matrix: convergence is yes/no (decided by $∣ r ∣ < 1$ ), but the size of the answer blows up as $r$ nears the boundary.

In the figure, trace the lavender curve $\frac{1}{1 - r}$ from left to right. Near $r = 0$ it is calm and low, but as it approaches the coral dashed line at $r = 1$ it shoots up like a wall. The coral dot marks Ex 4 at $r = 0.99$ , already up at height $100$ — sitting far up that steep wall. That is why a ratio just below $1$ still converges but produces a giant sum.

Case C5 — positive ratio $> 1$ (diverges to $+ \infty$ )

Case C6 — the boundary $r = 1$ (diverges)

Case C7 — the oscillating boundary $r = - 1$

Worked example Example 7 (cell C7) — never settles

Find $S_{\infty} = 7 - 7 + 7 - 7 + \dots$

Forecast: the partial totals will be $7$ , then $0$ , then $7$ , then $0$ … does that "approach" any single number?

Find $r$ . $r = \frac{- 7}{7} = - 1$ . Why this step? Next-over-current; the alternating sign makes it negative.
Check convergence. $∣ r ∣ = 1 \neq < 1$ ❌. Why this step? The boundary is excluded; $∣ r ∣ < 1$ is a strict inequality.
Look at partial sums directly. $S_{1} = 7, S_{2} = 0, S_{3} = 7, S_{4} = 0, \dots$ They oscillate forever and never home in on one value → diverges. Why this step? With the formula illegal, we test the definition: do the $S_{N}$ approach one number? They don't.

Verify: since the sequence of partial sums has no single limit, no finite $S_{\infty}$ exists. (Plugging into the formula gives $\frac{7}{1 - ( - 1 )} = \frac{7}{2}$ , but that is meaningless here.) ✅

Compare the two colours in the figure. The green dots ( $r = \frac{1}{3}$ ) climb once and then pile up on a single height — that piling-up is convergence. The coral squares ( $r = - 1$ ) bounce forever between two heights, $7$ and $0$ , never choosing one. That visible refusal to settle on a single line is exactly what "diverges by oscillation" looks like.

Case C8 — negative ratio with $∣ r ∣ > 1$ (oscillates and blows up)

Worked example Example 8 (cell C8) — the wild one

Find $S_{\infty} = 1 - 2 + 4 - 8 + 16 - \dots$

Forecast: signs alternate and the sizes keep doubling. Will the running total settle, oscillate calmly, or explode?

Find $r$ . $r = \frac{- 2}{1} = - 2$ . Why this step? Next-over-current; the sign flip and the doubling together give $- 2$ .
Check convergence. $∣ r ∣ = 2 \geq 1$ ❌ → diverges. Why this step? Only the magnitude $∣ r ∣$ decides convergence; $2 \geq 1$ fails, regardless of the minus sign.
Look at partial sums directly. $S_{1} = 1, S_{2} = - 1, S_{3} = 3, S_{4} = - 5, S_{5} = 11, \dots$ They flip sign and grow in size, swinging out to $\pm \infty$ . Why this step? This shows the failure is worse than Ex 7: not only no single limit, but unbounded swings.

Verify: the partial sums $1, - 1, 3, - 5, 11$ neither approach a value nor stay bounded, so no finite $S_{\infty}$ exists. (The formula would give $\frac{1}{1 - ( - 2 )} = \frac{1}{3}$ , pure nonsense.) ✅

In the figure, watch the coral dots swing across the zero line. Unlike Ex 7 — where the bounces stayed pinned between two fixed heights — here each swing is taller than the one before: $1$ , then $- 1$ , then $3$ , then $- 5$ , then $11$ , launching further out every step. That growing, sign-flipping spread is the visual signature of a negative ratio with $∣ r ∣ > 1$ .

Case C9 — a real-world word problem

Worked example Example 9 (cell C9) — the bouncing ball

A ball is dropped from a height of $10 m$ . After each bounce it returns to $\frac{3}{5}$ of its previous height. What is the total vertical distance it travels before coming to rest?

Forecast: it falls $10$ , bounces up-and-down repeatedly. Rough guess: more than $10$ , but finite?

Split the journey. First fall $= 10 m$ . Then every bounce is a round trip — up then down. Why this step? Each bounce height is travelled twice (once up, once down), so we must count it twice later.
Build the up-heights GP. Bounce heights: $10 \cdot \frac{3}{5}, 10 \cdot \frac{3}{5}^{2}, \dots$ so the up-distances form a GP with first term $a = 10 \cdot \frac{3}{5} = 6$ and ratio $r = \frac{3}{5}$ . Why this step? "Returns to $\frac{3}{5}$ of previous height" is literally a common ratio of $\frac{3}{5}$ .
Sum the up-distances. $∣ r ∣ = \frac{3}{5} < 1$ ✅, so $S_{up} = \frac{6}{1 - \frac{3}{5}} = \frac{6}{\frac{2}{5}} = 15 m$ . Why this step? Convergence confirmed, apply $\frac{a}{1 - r}$ to add all the (infinitely many) up-climbs.
Total distance. Each up-climb is matched by an equal down-fall, so double the up-total, then add the very first drop: $Total = 10 + 2 \times 15 = 10 + 30 = 40 m .$ Why this step? The $2 \times$ pairs each up-climb with its matching down-fall (same distance travelled twice), and the leading $10$ is the initial drop, which had no "up" before it and so is not doubled.

Verify (units + magnitude): all in metres ✅. Total $40 m$ is finite and larger than the $10 m$ drop, exactly as forecast; if $r$ were $\geq 1$ the ball would bounce forever higher — physically impossible, and indeed the formula would diverge. ✅

Case C10 — recurring decimal to fraction

Worked example Example 10 (cell C10) — decode

$0. \overline{45}$ Express $0.454545 \dots$ as a fraction. See also Recurring Decimals as Fractions.

Forecast: a repeating decimal is a hidden GP. Guess the fraction before solving.

Write as a sum of blocks. $0. \overline{45} = 0.45 + 0.0045 + 0.000045 + \dots$ Why this step? Each repeat of "45" sits two decimal places further right, so the number is a sum of shrinking blocks — a GP in disguise.
Identify $a, r$ . $a = 0.45 = \frac{45}{100}$ , and each block is $\frac{1}{100}$ of the previous, so $r = 0.01 = \frac{1}{100}$ . Why this step? Shifting two decimal places = multiplying by $\frac{1}{100}$ , which is exactly the common ratio.
Check convergence. $∣ r ∣ = 0.01 < 1$ ✅. Why this step? Confirm before summing (it comfortably passes).
Sum. $S_{\infty} = \frac{0.45}{1 - 0.01} = \frac{0.45}{0.99} = \frac{45}{99} = \frac{5}{11} .$ Why this step? Apply $\frac{a}{1 - r}$ and simplify the fraction.

Verify: long-divide $5 \div 11 = 0.4545 \dots$ ✅ — the fraction reproduces the decimal exactly.

Case C11 — exam twist: solve backwards

Worked example Example 11 (cell C11) — given the sum, find the first term

An infinite GP has common ratio $r = \frac{1}{4}$ and total $S_{\infty} = 8$ . Find its first term $a$ and its third term.

Forecast: we run the formula in reverse. Guess whether $a$ is bigger or smaller than $8$ .

Confirm it converges. $∣ r ∣ = \frac{1}{4} < 1$ ✅. Why this step? Only if $∣ r ∣ < 1$ is the relation $S_{\infty} = \frac{a}{1 - r}$ true, so we may rely on it to work backwards.
Rearrange for $a$ . From $8 = \frac{a}{1 - \frac{1}{4}} = \frac{a}{\frac{3}{4}}$ , multiply both sides by $\frac{3}{4}$ : $a = 8 \times \frac{3}{4} = 6.$ Why this step? We know $S_{\infty}$ and $r$ and want $a$ — isolate $a$ by clearing the denominator.
Third term. Term $_{3} = a r^{2} = 6 \times (\frac{1}{4})^{2} = 6 \times \frac{1}{16} = \frac{3}{8} .$ Why this step? The $n$ th term is $a r^{n - 1}$ , so the 3rd term uses $r^{3 - 1} = r^{2}$ — see Geometric Progression — nth term.

Verify: rebuild the sum: $\frac{6}{1 - \frac{1}{4}} = \frac{6}{\frac{3}{4}} = 8$ ✅. And $a = 6 < 8$ as expected (the tail adds the extra $2$ ).

Case C12 — exam twist: GP starting mid-list

Worked example Example 12 (cell C12) — the sum from the

second term onward For the GP $12 + 4 + \frac{4}{3} + \frac{4}{9} + \dots$ , find (a) the full infinite sum, and (b) the sum of all terms after the first.

Forecast: part (b) is just part (a) minus the first term — guess how they relate.

Set up. $a = 12$ , $r = \frac{4}{12} = \frac{1}{3}$ . Why this step? $a$ is the first visible term; $r$ is next-over-current from the first pair.
Check convergence. $∣ r ∣ = \frac{1}{3} < 1$ ✅. Why this step? Needed before either sub-sum can use $\frac{a}{1 - r}$ .
(a) Full sum. $S_{\infty} = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 18.$ Why this step? Straight application of the formula with $a = 12$ .
(b) Sum after the first term — subtract. $18 - 12 = 6.$ Why this step? "After the first" means the whole sum minus that leading term $12$ .
(b) Sanity via re-anchoring. The tail $4 + \frac{4}{3} + \dots$ is itself a GP with a new first term $a^{'} = 4$ and the same $r = \frac{1}{3}$ : $\frac{4}{1 - \frac{1}{3}} = \frac{4}{\frac{2}{3}} = 6.$ Same answer. Why this step? It cements that "first term" always means the first term of the series you are summing — a classic trap.

Verify: $12 + 6 = 18$ ✅, and both methods for the tail give $6$ ✅.

Active recall

Recall Which cell, which check? (hide answers)

Every example ran the same two-step reflex: find $r$ , then check $∣ r ∣ < 1$ . Test yourself:

Sum of $9 + 3 + 1 + \dots$ (C1) ::: $13.5$ Sum of $8 - 6 + \frac{9}{2} - \dots$ (C2) ::: $\frac{32}{7}$ Sum of $5 + 0 + 0 + \dots$ (C3) ::: $5$ Sum of $1 + 0.99 + 0.9 9^{2} + \dots$ (C4) ::: $100$ Does $4 + 6 + 9 + \dots$ converge? (C5) ::: No, $r = \frac{3}{2} \geq 1$ , diverges to $+ \infty$ Does $5 + 5 + 5 + \dots$ converge? (C6) ::: No, $r = 1$ , partial sums $5 N \to \infty$ Does $7 - 7 + 7 - \dots$ converge? (C7) ::: No, $r = - 1$ , partial sums oscillate $7, 0, 7, 0$ Does $1 - 2 + 4 - 8 + \dots$ converge? (C8) ::: No, $r = - 2$ , swings out to $\pm \infty$ Bouncing ball total distance (C9) ::: $40 m$ $0. \overline{45}$ as a fraction (C10) ::: $\frac{5}{11}$ First term when $r = \frac{1}{4}, S_{\infty} = 8$ (C11) ::: $a = 6$ Tail sum $4 + \frac{4}{3} + \dots$ of Ex 12 (C12) ::: $6$

Connections

Parent: the proof
Geometric Progression — nth term (the $a r^{n - 1}$ used in Ex 11)
Sum of finite GP
Limits of Sequences (why $r^{N} \to 0$ , and what $S_{N} \to S_{\infty}$ means)
Convergence of Series
Recurring Decimals as Fractions (Ex 10)
Present Value & Perpetuities (backward-solving like Ex 11)