Infinite Geometric Series

A geometric series multiplies by the same ratio $r$ each time, like $6 + 3 + 1.5 + 0.75 + \dots$ You already know how to add the first $n$ terms from finite geometric series. Now let $n$ go to infinity. Geometric series are the one big family of series where you can find the exact sum easily, and they’re the model that the ratio test and power series are built on.

Key ideas

The form

An infinite geometric series has a first term $a$ and a common ratio $r$:

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

To find $r$, divide any term by the one before it.

When it converges, and the sum

The $n$th partial sum (from algebra) is

$$S_n = \frac{a(1 - r^n)}{1 - r}, \qquad r \ne 1 .$$

Everything depends on what $r^n$ does as $n \to \infty$:

• If $\lvert r \rvert \lt 1$, then $r^n \to 0$, so $S_n \to \dfrac{a}{1 - r}$. The series converges.
• If $\lvert r \rvert \ge 1$ (and $a \ne 0$), the terms $ar^n$ don’t approach $0$, so the series diverges by the nth term test.

$$\sum_{n=0}^{\infty} a r^n = \frac{a}{1 - r} \quad \text{if } \lvert r \rvert \lt 1 .$$

A handy way to remember it: sum = first term ÷ (1 − ratio). That version works no matter where the index starts, as long as “first term” means the actual first term of the series.

With $a = 6$: for $r = \tfrac{1}{2}$ the partial sums climb to $\frac{6}{1 - 1/2} = 12$; for $r = -\tfrac{1}{2}$ they zig-zag toward $\frac{6}{1 + 1/2} = 4$.

Spotting a geometric series

A series is geometric when the variable $n$ appears only in an exponent, like $\dfrac{3^n}{5^{n+1}}$ or $4\left(-\tfrac{2}{3}\right)^{n-1}$. If $n$ appears anywhere else (like $\dfrac{1}{n^2}$ or $\dfrac{2^n}{n}$), it isn’t geometric and the formula doesn’t apply.

Repeating decimals

A repeating decimal is an infinite geometric series in disguise. For example,

$$0.272727\ldots = \frac{27}{100} + \frac{27}{100^2} + \frac{27}{100^3} + \dots$$

has $a = \dfrac{27}{100}$ and $r = \dfrac{1}{100}$.

Worked examples

Example 1: Converge or diverge?

Decide whether each series converges. If it does, find its sum.

• (a) $\displaystyle\sum_{n=0}^{\infty} 5\left(\frac{2}{3}\right)^n$
• (b) $8 - 4 + 2 - 1 + \dots$
• (c) $\displaystyle\sum_{n=1}^{\infty} \left(\frac{5}{4}\right)^n$

Solution.

(a) $a = 5$ (the $n = 0$ term) and $r = \tfrac{2}{3}$. Since $\lvert r \rvert \lt 1$, it converges:

$$\frac{5}{1 - \frac{2}{3}} = \frac{5}{\frac{1}{3}} = 15$$

(b) $a = 8$ and $r = \dfrac{-4}{8} = -\dfrac{1}{2}$. Since $\lvert r \rvert \lt 1$, it converges:

$$\frac{8}{1 - \left(-\frac{1}{2}\right)} = \frac{8}{\frac{3}{2}} = \frac{16}{3}$$

(c) $r = \tfrac{5}{4}$ and $\lvert r \rvert \ge 1$, so the series diverges. (The terms grow, so they certainly don’t approach $0$.)

Example 2: A shifted index

Find the sum of $\displaystyle\sum_{n=2}^{\infty} \frac{3^n}{5^{n+1}}$.

Solution. Don’t try to read $a$ off the formula. Instead, write out the first couple of terms:

$$n = 2: \ \frac{9}{125}, \qquad n = 3: \ \frac{27}{625}$$

The ratio is $\dfrac{27/625}{9/125} = \dfrac{3}{5}$, so $r = \tfrac{3}{5}$ and the first term is $\tfrac{9}{125}$:

$$\text{sum} = \frac{\frac{9}{125}}{1 - \frac{3}{5}} = \frac{9}{125} \cdot \frac{5}{2} = \frac{9}{50}$$

Example 3: Repeating decimals as fractions

Write (a) $0.\overline{27}$ and (b) $1.2\overline{3} = 1.2333\ldots$ as fractions.

Solution.

(a) $a = \dfrac{27}{100}$, $r = \dfrac{1}{100}$:

$$0.\overline{27} = \frac{\frac{27}{100}}{1 - \frac{1}{100}} = \frac{27}{100} \cdot \frac{100}{99} = \frac{27}{99} = \frac{3}{11}$$

Check: $3 \div 11 = 0.2727\ldots$ ✓

(b) Only the $3$s repeat, so split off the $1.2$:

$$1.2\overline{3} = 1.2 + \left(0.03 + 0.003 + 0.0003 + \dots\right) = \frac{6}{5} + \frac{\frac{3}{100}}{1 - \frac{1}{10}} = \frac{6}{5} + \frac{1}{30} = \frac{37}{30}$$

Example 4: A series with a variable

For which values of $x$ does $\displaystyle\sum_{n=0}^{\infty} \frac{(x - 1)^n}{2^n}$ converge? Find its sum in terms of $x$.

Solution. This is geometric with $a = 1$ and $r = \dfrac{x - 1}{2}$. It converges when

$$\left\lvert \frac{x - 1}{2} \right\rvert \lt 1 \quad\Rightarrow\quad \lvert x - 1 \rvert \lt 2 \quad\Rightarrow\quad -1 \lt x \lt 3 .$$

For those $x$, the sum is

$$\frac{1}{1 - \frac{x - 1}{2}} = \frac{2}{2 - (x - 1)} = \frac{2}{3 - x}$$

So on $-1 \lt x \lt 3$, the function $\dfrac{2}{3 - x}$ is equal to an infinite “polynomial.” That’s the first glimpse of power series.

Common mistakes

Using the coefficient as the first term. In $\sum_{n=1}^{\infty} 4\left(\tfrac{1}{2}\right)^n$, the first term is $4 \cdot \tfrac{1}{2} = 2$, not $4$. The sum is $\dfrac{2}{1 - 1/2} = 4$. Always plug in the starting value of $n$.

Using the formula when the ratio is too big. Plugging $r = 2$ into $\dfrac{a}{1 - r}$ would say $1 + 2 + 4 + 8 + \dots = -1$, which is nonsense. Check $\lvert r \rvert \lt 1$ first, and state it in your answer.

Losing the sign of r. For $8 - 4 + 2 - \dots$, $r = -\tfrac{1}{2}$, so $1 - r = \tfrac{3}{2}$, not $\tfrac{1}{2}$.

Calling a series geometric when it isn’t. $\sum \frac{1}{n^2}$ and $\sum \frac{2^n}{n}$ are not geometric, because the ratio between terms isn’t constant. Geometric means $n$ appears only in exponents.

Getting the repeating decimal’s ratio wrong. A block of $k$ repeating digits has $r = \dfrac{1}{10^k}$. For $0.\overline{27}$, two digits repeat, so $r = \tfrac{1}{100}$, not $\tfrac{1}{10}$.

Practice

1. (Warm-up) Find the sum of $\displaystyle\sum_{n=0}^{\infty} 6\left(\frac{1}{4}\right)^n$.

Solution

$a = 6$, $r = \tfrac{1}{4}$, and $\lvert r \rvert \lt 1$:

$$\frac{6}{1 - \frac{1}{4}} = \frac{6}{\frac{3}{4}} = 8$$

2. (Warm-up) Does each series converge? If so, find its sum.

• (a) $\displaystyle\sum_{n=0}^{\infty} 3(-1.1)^n$
• (b) $\displaystyle\sum_{n=0}^{\infty} (-0.9)^n$

Solution

(a) $\lvert r \rvert = 1.1 \ge 1$, so it diverges.

(b) $a = 1$, $r = -0.9$, $\lvert r \rvert \lt 1$, so it converges:

$$\frac{1}{1 - (-0.9)} = \frac{1}{1.9} = \frac{10}{19}$$

3. (Warm-up) Write $0.\overline{6}$ as a fraction using a geometric series.

Solution

$0.666\ldots = 0.6 + 0.06 + 0.006 + \dots$ with $a = \tfrac{6}{10}$, $r = \tfrac{1}{10}$:

$$\frac{\frac{6}{10}}{1 - \frac{1}{10}} = \frac{6}{10} \cdot \frac{10}{9} = \frac{6}{9} = \frac{2}{3}$$

4. (Core) Find the sum of $\displaystyle\sum_{n=1}^{\infty} \frac{2^{n+1}}{3^n}$.

Solution

The first term ($n = 1$) is $\tfrac{4}{3}$, and the next is $\tfrac{8}{9}$, so $r = \tfrac{8/9}{4/3} = \tfrac{2}{3}$:

$$\frac{\frac{4}{3}}{1 - \frac{2}{3}} = \frac{\frac{4}{3}}{\frac{1}{3}} = 4$$

5. (Core) Write $2.\overline{45}$ as a fraction.

Solution

$2.\overline{45} = 2 + 0.4545\ldots$, and $0.4545\ldots$ has $a = \tfrac{45}{100}$, $r = \tfrac{1}{100}$:

$$2 + \frac{\frac{45}{100}}{1 - \frac{1}{100}} = 2 + \frac{45}{99} = 2 + \frac{5}{11} = \frac{27}{11}$$

6. (Core) A ball is dropped from a height of $2$ m. Each time it hits the floor, it bounces back up to $60\%$ of the height it fell from. Assuming it bounces forever, find the total vertical distance it travels.

Solution

The first drop is $2$ m. After that, each bounce goes up and comes back down the same distance. The bounce heights are $1.2, 0.72, 0.432, \dots$ m, a geometric series with $a = 1.2$ and $r = 0.6$.

$$\text{total} = 2 + 2\left(\frac{1.2}{1 - 0.6}\right) = 2 + 2(3) = 8 \text{ m}$$

7. (Core) Find the sum of $\displaystyle\sum_{n=0}^{\infty} \frac{2^n + 3^n}{6^n}$.

Solution

Split it into two geometric series, both with $\lvert r \rvert \lt 1$:

$$\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n + \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{1 - \frac{1}{3}} + \frac{1}{1 - \frac{1}{2}} = \frac{3}{2} + 2 = \frac{7}{2}$$

Splitting is allowed because both pieces converge.

8. (Challenge) Consider $\displaystyle\sum_{n=0}^{\infty} (2x - 1)^n$.

• (a) For which values of $x$ does it converge?
• (b) For which $x$ is the sum equal to $4$?

Solution

(a) It’s geometric with $a = 1$, $r = 2x - 1$. It converges when $\lvert 2x - 1 \rvert \lt 1$, that is, $-1 \lt 2x - 1 \lt 1$, so $0 \lt x \lt 1$.

(b) The sum is $\dfrac{1}{1 - (2x - 1)} = \dfrac{1}{2 - 2x}$. Set it equal to $4$:

$$\frac{1}{2 - 2x} = 4 \quad\Rightarrow\quad 2 - 2x = \frac{1}{4} \quad\Rightarrow\quad x = \frac{7}{8}$$

Since $0 \lt \tfrac{7}{8} \lt 1$, this value is allowed. Check: $r = \tfrac{3}{4}$, and $\dfrac{1}{1 - 3/4} = 4$. ✓

9. (Challenge) An infinite geometric series has sum $12$. The series formed by squaring each of its terms has sum $48$. Find the first term $a$ and the ratio $r$.

Solution

Squaring each term of $a + ar + ar^2 + \dots$ gives $a^2 + a^2r^2 + a^2r^4 + \dots$, a geometric series with ratio $r^2$. So

$$\frac{a}{1 - r} = 12 \qquad\text{and}\qquad \frac{a^2}{1 - r^2} = 48 .$$

Since $1 - r^2 = (1 - r)(1 + r)$, divide the second equation by the first:

$$\frac{a}{1 + r} = 4$$

So $a = 12(1 - r)$ and $a = 4(1 + r)$. Setting these equal: $12 - 12r = 4 + 4r$, so $r = \tfrac{1}{2}$ and $a = 6$.

Check: $\dfrac{6}{1/2} = 12$ and $\dfrac{36}{3/4} = 48$. ✓