Lesson 15 - Geometric Series and Ratio Test

Geometric Series

A geometric series is any series that we can write in the form of

a + a r + a r^{2} + a r^{3} + \dots = \sum_{n = 1}^{\infty} a r^{n - 1}

Notice here that the initial term $a$ is a ratio $r$ of the term before it. For example, the series

\sum_{n = 1}^{\infty} {(\frac{1}{2})}^{n - 1} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots

is a geometric series with initial term $a = 1$ and ratio $r = \frac{1}{2}$ .

In general:

If $| r | < 1$ , the series converges, and

\sum_{n = 1}^{\infty} a r^{n - 1}, | r | < 1

If $| r | \geq 1$ , the series diverges.

Example: Determining Convergence or Divergence of a Geometric Series

Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.

a. $\sum_{n = 1}^{\infty} \frac{(- 3)^{n + 1}}{4^{n - 1}}$
b. $\sum_{n = 1}^{\infty} e^{2 n}$

a. Writing out the first several terms in the series, we have

\sum_{n = 1}^{\infty} \frac{(- 3)^{n + 1}}{4^{n - 1}} = \frac{(- 3)^{2}}{4^{0}} + \frac{(- 3)^{3}}{4} + \frac{(- 3)^{4}}{4^{2}} + \dots

= (- 3)^{2} + (- 3)^{2} \times (- \frac{3}{4}) + (- 3)^{2} \times {(- \frac{3}{4})}^{2} + \dots

= 9 + 9 \times (- \frac{3}{4}) + 9 \times {(- \frac{3}{4})}^{2} + \dots

Recall the sum of an infinite series in the geometric form is

S = \frac{a}{1 - r}

The initial term $a = 9$ and the ratio $r = - \frac{3}{4}$ . Since $| r | = \frac{3}{4} < 1$ , the series converges to

\frac{9}{1 - (- \frac{3}{4})} = \frac{9}{\frac{7}{4}} = \frac{36}{7}

b. Try for yourself.

Ratio Test

Given the series

\sum_{n = 1}^{\infty} a_{n}

we know that $lim_{n \to \infty} = 0$ is not a sufficient condition for the series to converge. Not only do we need $a_{n} \to 0$ , but we need $a_{n} \to 0$ quickly enough.

Theorem

Let $\sum_{n = 1}^{\infty} a_{n}$ be a series with nonzero terms. Let

ρ = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |

i. If $0 \leq ρ \leq 1$ , then $\sum_{n = 1}^{\infty} a_{n}$ converges absolutely.
ii. If $ρ > 1$ or $ρ = \infty$ , then $\sum_{n = 1}^{\infty} a_{n}$ diverges.
iii. If $ρ = 1$ , the test does not provide any information.

A proof is here if you are interested.

Example: Using the Ratio Test

For each of the following series, use the ratio test to determine whether the series converges or diverges.
a. $\sum_{n = 1}^{\infty} \frac{2^{n}}{n!}$
b. $\sum_{n = 1}^{\infty} \frac{n^{n}}{n!}$
c. $\sum_{n = 1}^{\infty} \frac{(- 1)^{n} (n!)^{2}}{(2 n)!}$

a. From the ratio test, we can see that

ρ = lim_{n \to \infty} \frac{\frac{2^{n + 1}}{(n + 1)!}}{\frac{2^{n}}{n!}} = lim_{n \to \infty} \frac{2^{n + 1}}{(n + 1)!} \times \frac{n!}{2^{n}}

Since $(n + 1) = (n + 1) \times n!$ ,

ρ = lim_{n \to \infty} \frac{2}{n + 1} = 0

Since $ρ < 1$ , the series converges.

b. We can see that

ρ = lim_{n \to \infty} \frac{\frac{(n + 1)^{n + 1}}{(n + 1)!}}{\frac{n^{n}}{n!}}

= lim_{n \to \infty} \frac{(n + 1)^{n + 1}}{(n + 1)!} \times \frac{n!}{n^{n}}

= lim_{n \to \infty} {(n + \frac{1}{n})}^{n} = lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e .

Since $ρ > 1$ , the series diverges.

C. Try for yourself.

Next Lesson: Lesson 16 - Alternating Series and Alternating Series Test

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Note: This lesson adapted from OpenStax. Credit goes to OpenStax creators.