Geometric Series

Part 17, Chapter 17: Mathematical Induction and Summation

Learning objectives

State the formula for a finite geometric series
Compute finite geometric sums
Determine when an infinite geometric series converges
Find the sum of a convergent infinite geometric series

The single most useful infinite sum in all of mathematics is geometric. Compound interest, mortgages, bouncing balls, decimal expansions of fractions, probabilistic waiting times, and power series in calculus all reduce to one identity: if $|r| < 1$ , then $1 + r + r^2 + r^3 + \cdots = \frac{1}{1 - r}$ . This section earns that formula from scratch.

What makes a series geometric

A geometric series is one where each term is a fixed multiple $r$ of the previous: $a, ar, ar^2, ar^3, \ldots$ . The number $a$ is the first term and $r$ is the common ratio. Knowing $a$ and $r$ , the whole sequence is determined. Examples: $1, 2, 4, 8, \ldots$ has $a = 1, r = 2$ . The sequence $9, 3, 1, \tfrac{1}{3}, \ldots$ has $a = 9, r = \tfrac{1}{3}$ .

Finite geometric series, the clever trick

Let $S = 1 + r + r^2 + \cdots + r^n$ . Multiply both sides by $r$ : $rS = r + r^2 + \cdots + r^{n+1}$ . Subtract:

$S - rS = 1 - r^{n+1} \quad\Longrightarrow\quad S(1 - r) = 1 - r^{n+1}.$

So as long as $r \neq 1$ ,

$\sum_{k=0}^{n} r^k = \frac{1 - r^{n+1}}{1 - r}.$ k=0nrk=frac1−rn+11−r.

If the first term is $a$ instead of $1$ , factor it out and multiply through: $\sum_{k=0}^{n} a r^k = a \cdot \frac{1 - r^{n+1}}{1 - r}$ k=0nark=acdotfrac1−rn+11−r. The same telescoping idea that broke $\tfrac{1}{k(k+1)}$ in the previous section is at work here: subtract a shifted copy and the middle disappears.

The infinite case

What happens as $n \to \infty$ ? If $|r| < 1$ , the term $r^{n+1}$ shrinks to zero, and the finite sum slides into a closed-form limit:

$\sum_{k=0}^{\infty} r^k = \frac{1}{1 - r}, \qquad \text{provided } |r| < 1.$ k=0inftyrk=frac11−r,qquadtextprovided∣r∣<1.

If $|r| \geq 1$ the terms do not shrink, they stay the same size or grow, and the sum has no finite value. We say the series diverges.

Where this shows up

Finance: The present value of a perpetuity paying $C$ per year at discount rate $r$ is $\sum_{n=1}^\infty C(1+r)^{-n} = C/r$ ; the entire bond-pricing industry uses this and its finite-sum variants.
Physics: Zeno's paradox of Achilles and the tortoise dissolves once you accept that $1/2 + 1/4 + 1/8 + \ldots = 1$ is a finite geometric sum; the same convergence argument underpins time-of-flight calculations.
Algorithms: Dynamic arrays (vector::push_back, Python lists) grow geometrically, doubling capacity each resize, so the total cost of $n$ appends is a geometric series summing to $O(n)$ , amortised.

(The widget starts in Geometric mode. Slide $r$ between $-1$ and $1$ : watch the partial sums hug the dashed line at $\tfrac{a}{1 - r}$ . Push $r$ past $1$ : the dashed line disappears, the partial sums explode upward. That is divergence on screen.)

A favourite consequence: $0.999\ldots = 1$

The repeating decimal $0.999\ldots$ is the geometric series $\tfrac{9}{10} + \tfrac{9}{100} + \tfrac{9}{1000} + \cdots$ with $a = \tfrac{9}{10}$ and $r = \tfrac{1}{10}$ . The closed form gives $\tfrac{9/10}{1 - 1/10} = \tfrac{9/10}{9/10} = 1$ . The two notations $0.999\ldots$ and $1$ name the same real number. Geometric series are the bridge that makes the equality precise.

Try it

Predict first: $\sum_{k=0}^{\infty} (1/2)^k$ k=0infty(1/2)k converges to what? Set $a = 1, r = \tfrac{1}{2}$ in the widget and confirm the limit reads $2$ with partial sums approaching from below.
Predict first: with $a = 1, r = -\tfrac{1}{2}$ , what does the series sum to, and how should the partial sums approach the limit? Set $r = -\tfrac{1}{2}$ and verify they oscillate above and below $\tfrac{2}{3}$ .
Predict first: for $|r| \geq 1$ , does a geometric series converge or diverge? Set $r = 1.1$ and verify the partial sums blow up, with the readout showing $S_\infty = \text{diverges}$ infty=textdiverges.
Compute by hand: $\sum_{k=0}^{6} 3^k$ k=063k. Compare with the formula $\tfrac{3^7 - 1}{3 - 1}$ .

Pause: the formula $\sum r^k = \tfrac{1}{1-r}$ fails at $r = 1$ . Why is that exactly the point where it has to fail? Look at what each side of the equality becomes.

Try it in code

A trap to watch for

The convergence condition is $|r| < 1$ , not $r < 1$ . Beginners write the formula and forget to check the absolute value. Plug $r = -2$ into $\tfrac{1}{1 - r}$ and you get $\tfrac{1}{3}$ , a perfectly innocent-looking number. But the series $1 - 2 + 4 - 8 + 16 - \cdots$ does not add up to $\tfrac{1}{3}$ ; the terms grow without bound and the sum has no value at all. The formula is meaningless for $|r| \geq 1$ . Always check absolute value before reaching for $\tfrac{1}{1-r}$ .

A second trap: when the series starts at $k = 1$ instead of $k = 0$ , the answer changes. $\sum_{k=1}^{\infty} r^k = \tfrac{r}{1 - r}$ k=1inftyrk=tfracr1−r, not $\tfrac{1}{1-r}$ . Always note where the index starts.

What you now know

You can compute any finite geometric sum in closed form, decide whether an infinite geometric series converges by looking at $|r|$ , evaluate the limit when it does, and recognise hidden geometric structure (repeating decimals, half-life, compound interest). The next chapter turns away from numbers and toward objects you can pile into rectangles: matrices. Linear algebra is what you get when you let geometric and arithmetic ideas live together in two dimensions and beyond.

Quick check

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References

Lang, S. (1971). Basic Mathematics. Springer. Chapter 16, §3, the derivation of finite and infinite geometric sums.
Rudin, W. (1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill. §3.26, geometric series as the canonical example of a convergent series and the model for the ratio test.
Spivak, M. (1994). Calculus (3rd ed.). Publish or Perish. Chapter 22: geometric series as the gateway to power series.
Knuth, D. E.; Graham, R. L.; Patashnik, O. (1994). Concrete Mathematics (2nd ed.). Addison-Wesley. §2.3, closed forms for finite sums, including geometric.