Mathematics · Sequences and Series

Geometric Series

Q: How do you compute the sum of the first n terms of a geometric series?

With sₙ = a₁·(qⁿ − 1)/(q − 1) for $q \ne{} 1$; equivalent is $\frac{a_{1}\cdot{}(1 - q^{n})}{(1 - q)}$. You need only three inputs: first term a₁, ratio q (each term divided by the previous one) and count n. Example: 3 + 6 + 12 + ... with 8 terms: $a_{1} = 3$, $q = 2$, so $s_{8} =\frac{ 3\cdot{}(2^{8} - 1)}{(2 - 1)} = 3\cdot{}255 = 765$. Watch n: the exponent holds the number of terms, not the last exponent (the eighth term is 3·2⁷). For $q = 1$ the formula fails (division by 0); then all terms are equal and simply sₙ = n·a₁.

Q: When does an infinite geometric series have a limit?

Exactly when $|q| < 1$, i.e. the ratio lies strictly between −1 and 1. Then the terms shrink fast enough that the partial sums approach $s\infty{} =\frac{ a_{1}}{(1 - q)}$; in the finite formula the term $q^{n}$ dies out as n → ∞. Example: 1 + 1/2 + 1/4 + ... has $q =\frac{ 1}{2}$ and $s\infty{} =\frac{ 1}{(1 - 0.5)} = 2$. For $|q| \ge{} 1$ the series diverges: at $q = 2$ the sums explode, at $q = 1$ they grow linearly, at $q = -1$ they jump forever between a₁ and 0. The most common mistake is thoughtlessly applying $s\infty{} =\frac{ a_{1}}{(1 - q)}$ to q = 1.05 (an interest factor!); savings plan sums are finite series.

Q: Why is 0.999... exactly equal to 1?

Because 0.999... is nothing other than an infinite geometric series: 0.9 + 0.09 + 0.009 + ... with $a_{1} = 0.9$ and $q = 0.1$. Since $|q| < 1$ the limit exists, and it is $s\infty{} =\frac{ 0.9}{(1 - 0.1)} =\frac{ 0.9}{0.9} = 1$. Exactly, not approximately: 0.999... denotes the LIMIT of the partial sums, not any single partial sum, and this limit is the number 1. Every finite partial sum (0.9; 0.99; 0.999; ...) stays below 1, but the gap $10^{-n}$ becomes arbitrarily small and is 0 in the limit. Two numbers at distance 0 are the same number. The same argument shows 0.333... = 1/3, which times 3 confirms 0.999... = 1 again.

Q: What is the difference between a geometric sequence and a geometric series?

The sequence is the list of individual terms: aₙ = a₁·qⁿ⁻¹, e.g. 3, 6, 12, 24, ... The series is the sum of these terms: sₙ = a₁ + a₁q + ... $+ a_{1}q^{n-1}$. Each question has its matching formula: "How large is the 8th term?" requires $a_{8} = 3\cdot{}2^{7} = 384$; "How large is the sum of the first 8 terms?" requires $s_{8} = 765$. Mix-ups are caught immediately by checking magnitudes: the sum must exceed the largest term. Their limit behaviour also differs: for $|q| < 1$ the SEQUENCE tends to 0, while the SERIES tends to the positive value $\frac{a_{1}}{(1 - q)}$.

Q: How do you use the geometric series for compound interest and savings plans?

If you deposit a fixed instalment R at the end of every year, which then earns interest with factor $q = 1 +\frac{ p}{100}$, the final capital after n years is a geometric sum: the first instalment earns interest (n−1) times, the last one not at all, together $K =\frac{ R\cdot{}(q^{n} - 1)}{(q - 1)}$. Example: 1000 € per year, 3 % interest, 10 years: $K =\frac{ 1000\cdot{}(1.03^{10} - 1)}{0.03} =\frac{ 1000\cdot{}0.3439}{0.03} \approx{} 11\,464$ €, i.e. 1464 € of interest on 10,000 € of deposits. The same structure sits inside loans (annuities), pension payouts and drug levels under regular dosing. Careful: here $q > 1$, the series grows; the s∞ formula is off limits.

The geometric series sums terms with a constant factor q; for $|q| < 1$ even the infinite series has a finite value.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Formula

s_{n} = a_{1} \cdot \frac{q ^{n} - 1}{q - 1} (q \neq = 1), s_{\infty} = \frac{a _{1}}{1 - q} (∣ q ∣ < 1)

Dimensionless · in applied contexts unit of the terms (e.g. euros)

Variables & units

Variables & units
Symbol	Meaning	Unit
$a₁$	First term of the series	dimensionless
$q$	Constant ratio of consecutive terms	dimensionless
$n$	Number of summed terms	dimensionless
s_n, s∞	Sum of the first n terms and limit value	dimensionless

LaTeX

s_{n} = a_{1} \cdot \frac{q^{n}-1}{q-1} \quad (q \neq 1), \qquad s_{\infty} = \frac{a_{1}}{1-q} \quad (|q| < 1)

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A square, repeatedly halved into areas 1/2, 1/4, 1/8, 1/16 and ever smaller pieces that together fill the whole square. — With a₁ = 1/2 and q = 1/2 the terms 1/2 + 1/4 + 1/8 + … fill the unit square: s∞ = a₁/(1 − q) = 1.

Derivation & background

Euclid already treated the finite geometric sum (Elements IX, 35). Derivation as a telescoping trick: $s_n-qs_n$ makes almost all terms cancel. For $|q| < 1$ the power $q^{n}$ tends to 0, hence $s_\infty = \frac{a_1}{1-q}$ ; for $|q| \ge{} 1$ the infinite series diverges. Classic: $0.999\ldots = 0.9\cdot\frac{1}{1-0.1} = 1$ . In financial mathematics the final value of a savings plan is a geometric sum with q = interest factor.

Exam blueprint

Validity range

The finite sum formula holds for q ≠ 1 (for $q = 1$ , s_n = n·a₁); the infinite series converges only for $|q| < 1$ , otherwise it diverges.

Derivation steps

Telescoping trick: subtract s_n and q·s_n, almost everything cancels.

1s_n = a₁ + a₁q + ... $+ a_{1}q^{n-1}$ and q·s_n = a₁q + ... $+ a_{1}q^{n}$ ; subtraction leaves only $a_{1} - a_{1}q^{n}$ .
2So s_n(1 − q) = a₁(1 − qⁿ), and for $|q| < 1 q^{n} \to{} 0$ , hence $s\infty{} =\frac{ a_{1}}{(1 - q)}$ .

Rearrangements

Infinite sum

s_{\infty} = \frac{a_{1}}{1-q} \quad (|q| < 1)

Only for $|q| < 1$ ; otherwise no limit exists.

First term

a_{1} = s_{\infty} \cdot (1 - q)

Backwards from limit and ratio.

n-th term of the sequence

a_{n} = a_{1} \cdot q^{\,n-1}

Keep sequence (single term) and series (sum) apart.

Task variant

Compute the sum of the first 8 terms of 3 + 6 + 12 + ...

$a_{1} = 3$ , $q = 2$ : $s_{8} =\frac{ 3\cdot{}(2^{8} - 1)}{(2 - 1)} = 3\cdot{}255 = 765$ .

An infinite geometric series has $s\infty{} = 20$ and $q = 0.6$ . Determine a₁.

$a_{1} = s\infty{}\cdot{}(1 - q) = 20\cdot{}0.4 = 8$ .

Check: $\frac{8}{(1 - 0.6)} =\frac{ 8}{0.4} = 20$ ✓.

Common mistakes

Applying the infinite sum formula for $|q| \ge{} 1$ .

For $|q| \ge{} 1$ the series diverges; s∞ exists only for $|q| < 1$ .

Confusing sequence and series.

a_n = a₁qⁿ⁻¹ is one term, s_n the sum of all terms up to n.

Miscounting exponents (qⁿ⁻¹ vs. qⁿ).

In the sum of n terms the highest exponent is n − 1; the formula contains $q^{n}$ .

Exam context

Compound interest and savings plan tasks, limits, growth and decay chains, justification of 0.999... = 1.

Formula cluster

Sequences and series

Multiplicative counterpart of the arithmetic series; foundation of financial mathematics.

s_n = n·(a₁ + a_n)/2

Arithmetic Series (Gauss Sum Formula)

N(t) = N₀·e^(r·t)

Exponential Growth

log(x·y) = log x + log y

Laws of Logarithms

Worked example

s₈ of 3 + 6 + 12 + ...: $a_{1} = 3$ , $q = 2$ : $s_{8} =\frac{ 3\cdot{}(2^{8} - 1)}{(2 - 1)} = 3\cdot{}255 = 765$ . Infinite: $a_{1} = 1$ , $q =\frac{ 1}{2}$ : $s\infty{} =\frac{ 1}{(1 - 0.5)} = 2$ .

Applications

Compound interest and annuity calculations (savings plans, loans), drug levels under regular dosing, limit tasks, computer science (runtime estimates)

How do you calculate with Geometric Series?

Here is how to work through a typical Geometric Series (sₙ = a₁·(qⁿ−1)/(q−1)) task step by step:

1
Task
Compute the sum of the first 8 terms of 3 + 6 + 12 + ...
Solution path
$a_{1} = 3$ , $q = 2$ : $s_{8} =\frac{ 3\cdot{}(2^{8} - 1)}{(2 - 1)} = 3\cdot{}255 = 765$ .
2
Task
An infinite geometric series has $s\infty{} = 20$ and $q = 0.6$ . Determine a₁.
Solution path
1. $a_{1} = s\infty{}\cdot{}(1 - q) = 20\cdot{}0.4 = 8$ .
Check: $\frac{8}{(1 - 0.6)} =\frac{ 8}{0.4} = 20$ ✓.

Frequently asked questions about Geometric Series

How do you compute the sum of the first n terms of a geometric series?+

With s_n = a₁·(qⁿ − 1)/(q − 1) for $q \ne{} 1$ ; equivalent is $\frac{a_{1}\cdot{}(1 - q^{n})}{(1 - q)}$ . You need only three inputs: first term a₁, ratio q (each term divided by the previous one) and count n. Example: 3 + 6 + 12 + ... with 8 terms: $a_{1} = 3$ , $q = 2$ , so $s_{8} =\frac{ 3\cdot{}(2^{8} - 1)}{(2 - 1)} = 3\cdot{}255 = 765$ . Watch n: the exponent holds the number of terms, not the last exponent (the eighth term is 3·2⁷). For $q = 1$ the formula fails (division by 0); then all terms are equal and simply s_n = n·a₁.

When does an infinite geometric series have a limit?+

Exactly when $|q| < 1$ , i.e. the ratio lies strictly between −1 and 1. Then the terms shrink fast enough that the partial sums approach $s\infty{} =\frac{ a_{1}}{(1 - q)}$ ; in the finite formula the term $q^{n}$ dies out as n → ∞. Example: 1 + 1/2 + 1/4 + ... has $q =\frac{ 1}{2}$ and $s\infty{} =\frac{ 1}{(1 - 0.5)} = 2$ . For $|q| \ge{} 1$ the series diverges: at $q = 2$ the sums explode, at $q = 1$ they grow linearly, at $q = -1$ they jump forever between a₁ and 0. The most common mistake is thoughtlessly applying $s\infty{} =\frac{ a_{1}}{(1 - q)}$ to q = 1.05 (an interest factor!); savings plan sums are finite series.

Why is 0.999... exactly equal to 1?+

Because 0.999... is nothing other than an infinite geometric series: 0.9 + 0.09 + 0.009 + ... with $a_{1} = 0.9$ and $q = 0.1$ . Since $|q| < 1$ the limit exists, and it is $s\infty{} =\frac{ 0.9}{(1 - 0.1)} =\frac{ 0.9}{0.9} = 1$ . Exactly, not approximately: 0.999... denotes the LIMIT of the partial sums, not any single partial sum, and this limit is the number 1. Every finite partial sum (0.9; 0.99; 0.999; ...) stays below 1, but the gap $10^{-n}$ becomes arbitrarily small and is 0 in the limit. Two numbers at distance 0 are the same number. The same argument shows 0.333... = 1/3, which times 3 confirms 0.999... = 1 again.

What is the difference between a geometric sequence and a geometric series?+

The sequence is the list of individual terms: a_n = a₁·qⁿ⁻¹, e.g. 3, 6, 12, 24, ... The series is the sum of these terms: s_n = a₁ + a₁q + ... $+ a_{1}q^{n-1}$ . Each question has its matching formula: "How large is the 8th term?" requires $a_{8} = 3\cdot{}2^{7} = 384$ ; "How large is the sum of the first 8 terms?" requires $s_{8} = 765$ . Mix-ups are caught immediately by checking magnitudes: the sum must exceed the largest term. Their limit behaviour also differs: for $|q| < 1$ the SEQUENCE tends to 0, while the SERIES tends to the positive value $\frac{a_{1}}{(1 - q)}$ .

How do you use the geometric series for compound interest and savings plans?+

If you deposit a fixed instalment R at the end of every year, which then earns interest with factor $q = 1 +\frac{ p}{100}$ , the final capital after n years is a geometric sum: the first instalment earns interest (n−1) times, the last one not at all, together $K =\frac{ R\cdot{}(q^{n} - 1)}{(q - 1)}$ . Example: 1000 € per year, 3 % interest, 10 years: $K =\frac{ 1000\cdot{}(1.03^{10} - 1)}{0.03} =\frac{ 1000\cdot{}0.3439}{0.03} \approx{} 11\,464$ €, i.e. 1464 € of interest on 10,000 € of deposits. The same structure sits inside loans (annuities), pension payouts and drug levels under regular dosing. Careful: here $q > 1$ , the series grows; the s∞ formula is off limits.

Quanta exam set

Curated exam set for "Geometric Series":

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Which formula describes Geometric Series?

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Question (front)

How do you rearrange Geometric Series for Infinite sum?

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Question (front)

Which common mistake happens with Geometric Series?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Algebra and Trigonometry 2e, §13.4: Series and Their Notations

Related formulas

L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|;\quad R=\lim_{n\to\infty}\sqrt[n]{|a_n|}

Convergence Tests for Infinite Series

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Vertex Form of the Parabola

f(x)\sim\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n\cos nx+b_n\sin nx\right)

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Direct Proportion

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Geometric Series

The geometric series sums terms with a constant factor q; for

|q| < 1

even the infinite series has a finite value.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

a₁

First term of the series

dimensionless

q

Constant ratio of consecutive terms

dimensionless

n

Number of summed terms

dimensionless

s_n, s∞

Sum of the first n terms and limit value

dimensionless

Derivation & background

Exam blueprint

Validity range

The finite sum formula holds for q ≠ 1 (for $q = 1$ , s_n = n·a₁); the infinite series converges only for $|q| < 1$ , otherwise it diverges.

Derivation steps

Telescoping trick: subtract s_n and q·s_n, almost everything cancels.

1s_n = a₁ + a₁q + ... $+ a_{1}q^{n-1}$ and q·s_n = a₁q + ... $+ a_{1}q^{n}$ ; subtraction leaves only $a_{1} - a_{1}q^{n}$ .
2So s_n(1 − q) = a₁(1 − qⁿ), and for $|q| < 1 q^{n} \to{} 0$ , hence $s\infty{} =\frac{ a_{1}}{(1 - q)}$ .

How do you calculate with Geometric Series?

Here is how to work through a typical Geometric Series (sₙ = a₁·(qⁿ−1)/(q−1)) task step by step:

Task

Compute the sum of the first 8 terms of 3 + 6 + 12 + ...

Solution path

$a_{1} = 3$ , $q = 2$ : $s_{8} =\frac{ 3\cdot{}(2^{8} - 1)}{(2 - 1)} = 3\cdot{}255 = 765$ .

Task

An infinite geometric series has $s\infty{} = 20$ and $q = 0.6$ . Determine a₁.

Solution path

$a_{1} = s\infty{}\cdot{}(1 - q) = 20\cdot{}0.4 = 8$ .

Check: $\frac{8}{(1 - 0.6)} =\frac{ 8}{0.4} = 20$ ✓.

Frequently asked questions about Geometric Series

How do you compute the sum of the first n terms of a geometric series?+

When does an infinite geometric series have a limit?+

Why is 0.999... exactly equal to 1?+

What is the difference between a geometric sequence and a geometric series?+

How do you use the geometric series for compound interest and savings plans?+

Retain Geometric Series for exams

Create a curated FSRS exam set for sₙ = a₁·(qⁿ−1)/(q−1): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential