Geometric Series

9 min read

Textbook Reference

The partial-sum definition of a series from the previous lesson is the starting point here.

Key idea

The geometric series is the one infinite sum you can add up exactly, with a single formula.

A geometric series is one where each term is a fixed multiple of the one before it. Start with a number $a$, multiply by a ratio $r$ to get the next term, multiply by $r$ again, forever: \[ a + ar + ar^2 + ar^3 + \cdots. \] The question is whether that adds up to something finite. The answer is beautifully simple. If the ratio $r$ is small enough in size that $|r| < 1$, each term is a real fraction of the last, the terms shrink fast, and the total settles down to \[ \frac{a}{1 - r}. \] If $|r| \ge 1$ the terms do not shrink (they hold steady or grow), so the sum runs off and the series diverges.

You can almost feel why the formula looks the way it does. The room you cross in Zeno’s puzzle is a geometric series with $a = \tfrac{1}{2}$ and $r = \tfrac{1}{2}$, and it sums to $\dfrac{1/2}{1 - 1/2} = 1$, the full width of the room. Once you trust the single formula and learn to read $a$ and $r$ off a series, an entire class of infinite sums becomes a one-line calculation.

Prerequisite Check

• Know that the sum of a series is the limit of its partial sums (previous lesson)
• Recall that $r^n \to 0$ when $-1 < r < 1$ (the sequence lessons)
• Work with exponent rules, for example $3^{1-n} = 3 \cdot 3^{-n}$
• Add and simplify fractions
• Read a repeating decimal such as $0.4444\dots$

If these are solid, you are ready.

Quick Reference

Definition. A geometric series has the form \[ \sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + ar^3 + \cdots, \qquad a \neq 0, \] where each term is the previous term times the common ratio $r$.

Convergence and sum (the one formula to memorize). \[ \sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1 - r} \quad \text{if } |r| < 1, \] and the series diverges if $|r| \ge 1$.

How to read $a$ and $r$. The first term is $a$. The ratio is $r = \dfrac{\text{any term}}{\text{the term before it}}$.

Procedure.

1. Confirm the series is geometric: divide each term by the one before it and check the ratio is constant.
2. Identify $a$ (the first term as written) and $r$ (the common ratio).
3. If $|r| \ge 1$, the series diverges. Stop.
4. If $|r| < 1$, the sum is $\dfrac{a}{1 - r}$.

Note on the starting index. If a series starts at $n = 0$, as in $\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots$, the first term is still $a$ (here $a = 1$) and the sum formula is unchanged. Always read $a$ as the actual first term written out.

Key Concepts

1. Where the Sum Formula Comes From

A short derivation makes the formula memorable instead of magic. Write the $n$th partial sum and a shifted copy: \[ s_n = a + ar + ar^2 + \cdots + ar^{n-1}, \] \[ r\,s_n = ar + ar^2 + \cdots + ar^{n-1} + ar^{n}. \] Subtract the second line from the first. Every middle term cancels, leaving \[ s_n - r\,s_n = a - ar^{n} \quad\Longrightarrow\quad s_n = \frac{a(1 - r^{n})}{1 - r} \quad (r \neq 1). \] Now take the limit. When $|r| < 1$, the sequence lessons tell us $r^{n} \to 0$, so \[ \lim_{n\to\infty} s_n = \frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r}. \] When $|r| \ge 1$, the power $r^{n}$ does not go to $0$ (it holds at $1$, grows, or oscillates), so the partial sums have no finite limit and the series diverges.

The partial sum has a clean closed form, which is exactly why the geometric series can be summed exactly.

2. Summing a Geometric Series

Example 1 (alternating ratio). Find the sum of $5 - \dfrac{10}{3} + \dfrac{20}{9} - \dfrac{40}{27} + \cdots$.

Goal. Read off $a$ and $r$, check $|r| < 1$, apply the formula.

Work. The first term is $a = 5$. The ratio is the second term over the first: \[ r = \frac{-10/3}{5} = -\frac{2}{3}. \] Since $|r| = \dfrac{2}{3} < 1$, the series converges, and \[ \frac{a}{1 - r} = \frac{5}{1 - \left(-\frac{2}{3}\right)} = \frac{5}{\frac{5}{3}} = 3. \]

Answer: $3$

Recap. A negative ratio makes the terms alternate in sign, but the formula does not change. What matters for convergence is $|r| < 1$, which holds here.

(Common error: using the second term as $a$. The first term is $a = 5$; the ratio $r$ is found by dividing the second term by the first, not the other way around.)

Example 2 (rewrite into standard form first). Is $\displaystyle\sum_{n=1}^{\infty} 2 \cdot 3^{1-n}$ convergent or divergent? If convergent, find its sum.

Goal. Rewrite the $n$th term in the shape $ar^{n-1}$, then read $a$ and $r$.

Work. Use exponent rules: \[ 2 \cdot 3^{1-n} = 2 \cdot 3 \cdot 3^{-n} = 6 \cdot \frac{1}{3^{n}} = 2\left(\frac{1}{3}\right)^{n-1}. \] So $a = 2$ and $r = \dfrac{1}{3}$. Since $|r| = \dfrac{1}{3} < 1$, the series converges, and \[ \frac{a}{1 - r} = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3. \]

Answer: the series converges and its sum is $3$.

Recap. When the term is written with exponents, massage it into $ar^{n-1}$ first. The ratio $r$ is whatever is raised to the moving power.

3. Repeating Decimals Are Geometric Series

A repeating decimal is secretly a geometric series, which is how you convert one into an exact fraction.

Example 3 (repeating decimal to a fraction). Write $2.3\overline{17} = 2.3171717\dots$ as a ratio of integers.

Goal. Split off the non-repeating part, then sum the repeating tail as a geometric series.

Work. The repeating block “$17$” first appears in the thousandths place: \[ 2.3171717\dots = 2.3 + \frac{17}{10^3} + \frac{17}{10^5} + \frac{17}{10^7} + \cdots. \] The tail is geometric with first term $a = \dfrac{17}{10^3}$ and ratio $r = \dfrac{1}{10^2}$ (each block is two decimal places further right). So the tail sums to \[ \frac{a}{1 - r} = \frac{17/10^3}{1 - 1/10^2} = \frac{17}{990}. \] Add the front part: \[ 2.3\overline{17} = \frac{23}{10} + \frac{17}{990} = \frac{2277 + 17}{990} = \frac{2294}{990} = \frac{1147}{495}. \]

Answer: $\dfrac{1147}{495}$

Recap. Peel off the non-repeating digits, sum the repeating tail with the geometric formula, then add. This is why every repeating decimal is a rational number.

(Common error: choosing the ratio as $\dfrac{1}{10}$ instead of $\dfrac{1}{10^2}$. The ratio is set by how many digits are in the repeating block: a two-digit block shifts by two places, so $r = \dfrac{1}{10^2}$.)

Inline Self-Check

Question. Does $\displaystyle\sum_{n=1}^{\infty} 3 \cdot 2^{n-1}$ converge? If so, find its sum.

Common Errors Summary

Common Misconceptions

Leveled Practice

Attempt each problem before opening the answer.

Level 1: Direct Application

Problem 1. Determine whether $10 - 2 + 0.4 - 0.08 + \cdots$ converges, and find the sum if it does.

Problem 2. Find the sum of $\displaystyle\sum_{n=1}^{\infty} 12(0.73)^{n-1}$.

Problem 3. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(-3)^{n-1}}{4^n}$ converges, and find the sum if it does.

Level 2: Rewriting and Decimals

Problem 4. Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{3 \cdot 2^{n+1}}{3^n}$ converges, and find the sum if it does.

Problem 5. Write $0.\overline{46} = 0.46464646\dots$ as a ratio of integers.

Problem 6. For which values of $x$ does $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{3^n}$ converge, and what is the sum?

Level 3: Applications and Reasoning

Problem 7. A ball is dropped from a height of $H$ meters. Each bounce returns it to $r$ times its previous height, with $0 < r < 1$. Find the total vertical distance the ball travels before coming to rest.

Problem 8. A drug raises a patient’s blood concentration by $0.2$ mg/mL each dose, and only $30\%$ of the drug remains by the next dose. Show the long-run concentration just after a dose approaches a limit, and find it.

Mastery Checklist

You have mastered this skill when you can do all of the following without notes:

• Recognize a geometric series and find its common ratio $r$
• State that it converges exactly when $|r| < 1$
• Apply the sum formula $\dfrac{a}{1 - r}$ correctly, including with negative $r$
• Rewrite a term such as $2 \cdot 3^{1-n}$ into the standard form $ar^{n-1}$
• Convert a repeating decimal to a fraction by summing a geometric tail
• Derive the sum formula from the partial-sum subtraction trick
• Solve a bouncing-ball or repeated-dose application

Mental Model

Picture a strip of paper of length $a$, then a piece $r$ times as long, then a piece $r$ times as long as that, laid end to end. If each new piece is a genuine fraction of the last ($|r| < 1$), the pieces shrink fast enough that the whole strip has a finite total length, namely $\dfrac{a}{1 - r}$. The Zeno room is this exact picture with $a = r = \tfrac{1}{2}$: the half, the quarter, the eighth, all laid down, summing to the room’s full width $1$.

If the pieces do not shrink ($|r| \ge 1$), each new piece is at least as long as the last, the strip grows without end, and there is no finite total. That is the whole convergence rule in one image: shrinking pieces add to a finite length, non-shrinking pieces do not.

Connections

Built From

• Infinite series and partial sums: the geometric sum is the limit of its partial sums, and the subtraction trick gives the closed form for $s_n$ that makes the limit computable.
• Limits of sequences: the fact that $r^n \to 0$ for $|r| < 1$ is what produces the formula.

Leads To

• The divergence test: a geometric series with $|r| \ge 1$ has terms that do not go to $0$, the simplest case the divergence test rules out.
• Power series: a power series $\sum c_n x^n$ is a geometric series when the coefficients are constant, and $\sum x^n = \dfrac{1}{1 - x}$ is the seed example for representing functions as series later in the chapter.

Why It Matters

The geometric series is the most widely applied series in all of mathematics. It is the present value of a perpetual stream of payments in finance, the steady-state concentration of a repeated drug dose in pharmacology, the total distance of a damped bouncing motion in physics, and the reason every repeating decimal is a fraction. In computing and signal processing, the geometric series sum is the closed form behind the response of a simple feedback loop. Because it has an exact formula, it is also the yardstick the comparison tests use to judge other series.

Audience Notes

For students who find math intimidating: there is exactly one formula here, $\dfrac{a}{1 - r}$, and one condition, $|r| < 1$. Most problems are just reading off $a$ and $r$ and substituting.

For students who want depth: the subtraction derivation of $s_n$ is worth learning, not memorizing the result blindly. It is the same algebraic trick that produces the formula for a finite geometric sum, and it reappears when power series are manipulated.

For students aimed at a career in computing or engineering: the geometric series is the discrete analog of exponential decay and the closed-form sum behind a first-order feedback system’s total response. Recognizing it lets you replace an infinite loop of contributions with a single division.

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