Geometric series — convergence condition, proof: Exercises

This page is a self-test ladder built on top of the parent proof. Each problem states cleanly what to find. The full worked solution lives in a collapsible [!recall]- callout — try the problem first, THEN unfold. Levels climb from recognising a geometric series to synthesising several ideas at once.

Everything here rests on two facts from the parent note, restated so you never have to leave this page:

Two skills decide every problem below:

Find $a$ : plug the starting index into the term and compute the first actual number.
Find $r$ : divide any term by the one before it — $r = \frac{term}{previous term}$ .

Level 1 — Recognition

(Can you spot $a$ , $r$ , and whether it converges?)

L1.1 — Is $5 + 10 + 20 + 40 + \dots$ geometric? If so, give $a$ and $r$ , and state converge/diverge.

Recall Solution L1.1

Test for geometric: divide consecutive terms. $\frac{10}{5} = 2$ , $\frac{20}{10} = 2$ , $\frac{40}{20} = 2$ . The ratio is the same number every time, so yes, it is geometric.

$a = 5$ (the first term written down).
$r = 2$ . Converge? $∣ r ∣ = 2 \geq 1$ , so the terms grow and the series diverges.

L1.2 — For $n = 0 \sum \infty 4 (\frac{1}{3})^{n}$ , write out the first three terms, and give $a$ and $r$ .

Recall Solution L1.2

Plug in $n = 0, 1, 2$ : $4 \cdot (\frac{1}{3})^{0} = 4, 4 \cdot (\frac{1}{3})^{1} = \frac{4}{3}, 4 \cdot (\frac{1}{3})^{2} = \frac{4}{9} .$ So the series is $4 + \frac{4}{3} + \frac{4}{9} + \dots$ .

$a = 4$ (the $n = 0$ term).
$r = \frac{1}{3}$ . Since $∣ r ∣ = \frac{1}{3} < 1$ , it converges.

L1.3 — Which of these converge? (a) $n = 0 \sum \infty (0.9)^{n}$ , (b) $n = 0 \sum \infty (- 0.9)^{n}$ , (c) $n = 0 \sum \infty (1.1)^{n}$ , (d) $n = 0 \sum \infty (- 1)^{n}$ .

Recall Solution L1.3

Convergence depends only on $∣ r ∣ < 1$ .

(a) $∣ r ∣ = 0.9 < 1$ → converges.
(b) $∣ - 0.9∣ = 0.9 < 1$ → converges (sign of $r$ does not matter for whether it converges, only for the value).
(c) $∣ r ∣ = 1.1 \geq 1$ → diverges (terms grow).
(d) $∣ r ∣ = 1$ → diverges ( $r^{N}$ oscillates $1, - 1, 1, - 1, \dots$ and never settles).

Level 2 — Application

(Plug into the formula correctly — mind the starting index.)

L2.1 — Evaluate $n = 0 \sum \infty 6 (\frac{1}{4})^{n}$ .

Recall Solution L2.1

$a = 6$ ( $n = 0$ term), $r = \frac{1}{4}$ , $∣ r ∣ < 1$ ✓. $S = \frac{a}{1 - r} = \frac{6}{1 - \frac{1}{4}} = \frac{6}{\frac{3}{4}} = 6 \cdot \frac{4}{3} = 8.$

L2.2 — Evaluate $n = 1 \sum \infty 6 (\frac{1}{4})^{n}$ . (Same series as L2.1 but starting at $n = 1$ .)

Recall Solution L2.2

WHAT changed: the index now starts at $n = 1$ , so the $n = 0$ term ( $6$ ) is missing. The first written term is the $n = 1$ term: $a = 6 \cdot (\frac{1}{4})^{1} = \frac{6}{4} = \frac{3}{2} .$ The ratio is still $r = \frac{1}{4}$ . $S = \frac{a}{1 - r} = \frac{\frac{3}{2}}{\frac{3}{4}} = \frac{3}{2} \cdot \frac{4}{3} = 2.$ Sanity check: L2.1 minus its $n = 0$ term should equal L2.2: $8 - 6 = 2$ ✓.

L2.3 — Evaluate $n = 0 \sum \infty 3 (- \frac{1}{2})^{n}$ .

Recall Solution L2.3

$a = 3$ , $r = - \frac{1}{2}$ , $∣ r ∣ = \frac{1}{2} < 1$ ✓. The denominator is $1 - r = 1 - (- \frac{1}{2}) = 1 + \frac{1}{2} = \frac{3}{2}$ . Negative $r$ makes the denominator bigger, so the sum lands below $a$ : $S = \frac{3}{\frac{3}{2}} = 2 < 3. ✓$

L2.4 — Write $0. \overline{45}$ (i.e. $0.454545 \dots$ ) as a fraction using a geometric series. See Repeating decimals as fractions.

Recall Solution L2.4

Group the repeating block " $45$ " as chunks of two decimal places: $0. \overline{45} = \frac{45}{100} + \frac{45}{10000} + \frac{45}{1000000} + \dots = \sum_{n = 1}^{\infty} 45 (\frac{1}{100})^{n} .$ First written term $a = \frac{45}{100}$ , ratio $r = \frac{1}{100}$ . $S = \frac{45/100}{1 - 1/100} = \frac{45/100}{99/100} = \frac{45}{99} = \frac{5}{11} .$ Check: $5 \div 11 = 0.4545 \dots$ ✓.

Level 3 — Analysis

(Pull apart, re-index, or reason about parameters.)

L3.1 — For which real $x$ does $n = 0 \sum \infty (x - 2)^{n}$ converge, and what is its sum there? (This is the model behind Power series and radius of convergence.)

Recall Solution L3.1

Here the ratio is $r = x - 2$ and $a = (x - 2)^{0} = 1$ . Convergence needs $∣ r ∣ < 1$ : $∣ x - 2∣ < 1 ⟺ - 1 < x - 2 < 1 ⟺ 1 < x < 3.$ On that interval: $S = \frac{a}{1 - r} = \frac{1}{1 - ( x - 2 )} = \frac{1}{3 - x} .$ The centre is $x = 2$ and the radius is $1$ — see the figure below. It shows a horizontal number line: the accent-red open segment from $x = 1$ to $x = 3$ is exactly where the series converges, with the black centre mark at $x = 2$ and the two regions outside labelled "diverges."

Figure — Geometric series — convergence condition, proof

Figure (s01): number line showing convergence only on the open interval $1 < x < 3$ , centred at $x = 2$ with radius $1$ ; outside the red segment the series diverges.

L3.2 — A geometric series has $n = 0 \sum \infty a r^{n} = 9$ and second term $a r = 2$ . Find $a$ and $r$ .

Recall Solution L3.2

Two equations: $\frac{a}{1 - r} = 9 (1), a r = 2 (2) .$ From (1): $a = 9 (1 - r)$ . Substitute into (2): $9 (1 - r) r = 2 ⟹ 9 r - 9 r^{2} = 2 ⟹ 9 r^{2} - 9 r + 2 = 0.$ Factor: $(3 r - 1) (3 r - 2) = 0$ , so $r = \frac{1}{3}$ or $r = \frac{2}{3}$ . Both satisfy $∣ r ∣ < 1$ , so two valid solutions:

$r = \frac{1}{3} \Rightarrow a = 9 (1 - \frac{1}{3}) = 9 \cdot \frac{2}{3} = 6.$
$r = \frac{2}{3} \Rightarrow a = 9 (1 - \frac{2}{3}) = 9 \cdot \frac{1}{3} = 3.$ Check: $(6, \frac{1}{3})$ : $a r = 2$ ✓, sum $= 6/ (2/3) = 9$ ✓. $(3, \frac{2}{3})$ : $a r = 2$ ✓, sum $= 3/ (1/3) = 9$ ✓.

L3.3 — Split $n = 0 \sum \infty \frac{2 ^{n} + 3 ^{n}}{6 ^{n}}$ into two geometric series and evaluate.

Recall Solution L3.3

WHY split: $\frac{2 ^{n} + 3 ^{n}}{6 ^{n}} = \frac{2 ^{n}}{6 ^{n}} + \frac{3 ^{n}}{6 ^{n}} = (\frac{2}{6})^{n} + (\frac{3}{6})^{n} = (\frac{1}{3})^{n} + (\frac{1}{2})^{n}$ . A finite sum of convergent series may be added term by term. $\sum_{n = 0}^{\infty} (\frac{1}{3})^{n} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}, \sum_{n = 0}^{\infty} (\frac{1}{2})^{n} = \frac{1}{1 - \frac{1}{2}} = 2.$ Total: $\frac{3}{2} + 2 = \frac{7}{2}$ .

Level 4 — Synthesis

(Combine geometric series with another idea.)

L4.1 — A ball dropped from height $h = 10$ m rebounds to $\frac{3}{5}$ of its previous height each bounce. Find the total vertical distance it travels before coming to rest.

Recall Solution L4.1

Picture it: the first drop is $10$ m (down only). After that, every bounce goes up to a peak and back down — so each subsequent height is counted twice.

Figure (s02): the ball's path — one accent-red vertical "first drop" of $10$ m counted once, then black rebound arcs of heights $6, 3.6, \dots$ each travelled up and down (twice); the total works out to $40$ m.

Initial drop: $10$ .
Peaks: $10 \cdot \frac{3}{5}, 10 \cdot (\frac{3}{5})^{2}, \dots$ , each travelled up and down. $D = 10 + 2 \sum_{n = 1}^{\infty} 10 (\frac{3}{5})^{n} .$ The tail is geometric with first written term $a = 10 \cdot \frac{3}{5} = 6$ , ratio $r = \frac{3}{5}$ : $\sum_{n = 1}^{\infty} 10 (\frac{3}{5})^{n} = \frac{6}{1 - \frac{3}{5}} = \frac{6}{\frac{2}{5}} = 15.$ $D = 10 + 2 (15) = 40 m .$

L4.2 — Solve for $x$ : $n = 0 \sum \infty (\frac{x}{x + 3})^{n} = 4$ , and confirm the ratio actually satisfies $∣ r ∣ < 1$ .

Recall Solution L4.2

Ratio $r = \frac{x}{x + 3}$ , $a = 1$ . Sum condition: $\frac{1}{1 - \frac{x}{x + 3}} = 4.$ Simplify the denominator: $1 - \frac{x}{x + 3} = \frac{( x + 3 ) - x}{x + 3} = \frac{3}{x + 3}$ . So $\frac{1}{\frac{3}{x + 3}} = \frac{x + 3}{3} = 4 ⟹ x + 3 = 12 ⟹ x = 9.$ Check ratio: $r = \frac{9}{12} = \frac{3}{4}$ , and $∣ r ∣ = \frac{3}{4} < 1$ ✓ — the sum is genuinely valid.

Level 5 — Mastery

(Re-derive, or handle a subtle/degenerate case.)

L5.1 — Re-derive the partial sum $S_{N} = a + a r + \dots + a r^{N - 1}$ from scratch using the telescoping subtraction, then explain in one line why the limit needs $∣ r ∣ < 1$ .

Recall Solution L5.1

Write $S_{N}$ and $r S_{N}$ aligned: $S_{N} = a + a r + a r^{2} + \dots + a r^{N - 1},$ $r S_{N} = a r + a r^{2} + \dots + a r^{N - 1} + a r^{N} .$ Subtract — every middle term $a r^{k}$ appears in both and cancels, leaving only the first term of $S_{N}$ and the last of $r S_{N}$ : $S_{N} - r S_{N} = a - a r^{N} ⟹ S_{N} (1 - r) = a (1 - r^{N}) ⟹ S_{N} = \frac{a ( 1 - r ^{N} )}{1 - r} (r \neq = 1) .$ Why $∣ r ∣ < 1$ for the limit: as $N \to \infty$ , $r^{N} \to 0$ only when $∣ r ∣ < 1$ ; then $S_{N} \to \frac{a ( 1 - 0 )}{1 - r} = \frac{a}{1 - r}$ . If $∣ r ∣ \geq 1$ , $r^{N}$ does not go to $0$ (it grows or oscillates), so the limit fails. See Sequences — limits and convergence.

L5.2 — Evaluate the shifted series $n = 2 \sum \infty 5 (\frac{2}{5})^{n}$ two ways: (a) directly via the correct first term, (b) as "full sum minus the first two terms."

Recall Solution L5.2

$r = \frac{2}{5}$ , $∣ r ∣ < 1$ ✓. (a) Direct. First written term is at $n = 2$ : $a = 5 (\frac{2}{5})^{2} = 5 \cdot \frac{4}{25} = \frac{4}{5}$ . $S = \frac{\frac{4}{5}}{1 - \frac{2}{5}} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{5} \cdot \frac{5}{3} = \frac{4}{3} .$ (b) Full minus early terms. Full sum from $n = 0$ : $\frac{5}{1 - \frac{2}{5}} = \frac{5}{\frac{3}{5}} = \frac{25}{3}$ . Subtract $n = 0$ and $n = 1$ terms: $5 + 5 \cdot \frac{2}{5} = 5 + 2 = 7$ . $S = \frac{25}{3} - 7 = \frac{25 - 21}{3} = \frac{4}{3} . ✓$ Both routes agree — a clean check on re-indexing.

L5.3 (Degenerate case) — What is $n = 0 \sum \infty a r^{n}$ when $a = 0$ ? Discuss for all $r$ , including $∣ r ∣ \geq 1$ .

Recall Solution L5.3

If $a = 0$ , every term is $0$ regardless of $r$ : $0 + 0 + 0 + \dots = 0$ . The partial sums are all $0$ , so they trivially converge to $0$ even when $∣ r ∣ \geq 1$ . This is why the divergence statement in the parent note carries the caveat "with $a \neq = 0$ ." The convergence gate $∣ r ∣ < 1$ matters only when there is actually something nonzero to sum.

Connections

Geometric series — convergence condition, proof (parent — the formula and proof)
Partial sums and series convergence (L5.1 re-derivation)
Sequences — limits and convergence (why $r^{N} \to 0$ needs $∣ r ∣ < 1$ )
Ratio Test (generalises the constant-ratio idea)
Power series and radius of convergence (L3.1's interval of convergence)
Repeating decimals as fractions (L2.4)
Telescoping series (the cancellation in L5.1)

Skill Map

The diagram below is the decision path you run on any candidate series. In words: start from a given series; divide consecutive terms to test whether the ratio is constant (is it geometric?). If yes, find $r$ and $a$ . Then check the gate $∣ r ∣ < 1$ . If the gate fails and $a \neq = 0$ , the series diverges; if the gate fails but $a = 0$ (the degenerate case), the sum is simply $0$ . If the gate passes, the sum is $\frac{a}{1 - r}$ — but you must still (i) read $a$ off the actual starting index, and (ii) for parameter problems, substitute your answer back to confirm $∣ r ∣ < 1$ .

Recall One-line self-check before you submit any answer

Did I (1) confirm the ratio is constant, (2) compute $a$ from the actual starting index, (3) check $∣ r ∣ < 1$ , and (4) — for parameters — substitute back to confirm $∣ r ∣ < 1$ ? Answer ::: If any box is unticked, the answer is unverified.