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Math Analysis 1

The geometric series: when it converges and what it adds up to

Closed form of the partial sum, convergence for |r| less than 1, divergence and oscillation, examples such as 0.999... = 1, common mistakes and where it returns (Z transform, feedback).

The geometric series is the first series you meet in calculus, and one of the very few whose sum can be computed exactly. That is why it serves as a yardstick for many others: the ratio test and the root test, for example, do nothing more than compare a series with a geometric one.

Statement

Let a \neq 0 and r be real numbers. The series

    \[ \sum_{n=0}^{\infty} a r^n = a + ar + ar^2 + ar^3 + \dots \]

  • converges if |r| < 1, and its sum is \dfrac{a}{1-r};
  • diverges (to +\infty or -\infty, depending on the sign of a) if r \ge 1;
  • oscillates, that is it has no limit, if r \le -1.

The number r is called the common ratio: it is the ratio between a term and the previous one.

The trick: the partial sum has a closed form

Studying a series means studying the limit of its partial sums. For the geometric series the sum of the first N+1 terms can be computed exactly. Write

    \[ S_N = a + ar + ar^2 + \dots + ar^N \]

and multiply by r:

    \[ r S_N = ar + ar^2 + \dots + ar^N + ar^{N+1}. \]

Subtracting the second line from the first, all the middle terms cancel in pairs and only this is left:

    \[ (1 - r)\,S_N = a - ar^{N+1} \quad\Longrightarrow\quad S_N = a\,\frac{1 - r^{N+1}}{1 - r} \qquad (r \neq 1). \]

The case r = 1 has to be treated separately, but it is trivial: every term equals a and S_N = (N+1)\,a, which diverges.

Taking the limit

Everything depends on what r^{N+1} does as N \to \infty:

  • if |r| < 1, r^{N+1} \to 0 and therefore S_N \to \dfrac{a}{1-r}: the series converges;
  • if r > 1, r^{N+1} \to +\infty and S_N diverges;
  • if r = -1, r^{N+1} jumps between +1 and -1: the partial sums alternate between a and 0, so there is no limit;
  • if r < -1, r^{N+1} oscillates with growing amplitude and again there is no limit.

Why does r^{n} \to 0 when |r| < 1? A quick way: write \frac{1}{|r|} = 1 + h with h > 0; by Bernoulli’s inequality (1+h)^n \ge 1 + nh, so |r|^n \le \frac{1}{1 + nh}, which tends to zero. ∎

Three examples

1. \frac12 + \frac14 + \frac18 + \dots Here a = \frac12 and r = \frac12: the sum is \frac{1/2}{1 - 1/2} = 1. This is Zeno’s paradox of Achilles and the tortoise: infinitely many shorter and shorter stretches can add up to a finite length.

2. Why 0.999… = 1. The repeating decimal is the series \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \dots, with a = \frac{9}{10} and r = \frac{1}{10}. The sum is \frac{9/10}{9/10} = 1. The same method gives the rule for turning any repeating decimal into a fraction.

3. A series starting at a different index. \sum_{n=2}^{\infty} \left(\frac{2}{3}\right)^n: the first term is a = \frac49, the ratio \frac23, so the sum is \frac{4/9}{1/3} = \frac43. It is always better to identify the first term than to memorise a different formula for every starting index.

Common mistakes

  • Using \frac{a}{1-r} even when |r| \ge 1: with r = 2 it would give 1 + 2 + 4 + \dots = -1, an obviously meaningless result.
  • Confusing a with the term of index 0 when the series starts at n = 1 or later.
  • Forgetting that r can be negative: \sum (-\frac12)^n = \frac{1}{1 + 1/2} = \frac23.

Where you will meet it again

In later courses the geometric series keeps coming back. In the Z transform the transform of an exponential a^n u[n] is precisely a geometric series in a z^{-1}, and the condition |a z^{-1}| < 1 defines the region of convergence. In feedback circuits the gain \frac{A}{1 - A\beta} can be read as the sum of the contributions of the signal going round the loop once, twice, three times. And in an echo, or a line with multiple reflections, every pass attenuates the signal by the same factor: the total energy is again a geometric sum.

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