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Geometric Series

Calculus 2

Geometric Series

A geometric series is a series of the form

a+ar+ar2+ar3+⋯+arn+⋯=∑n=1∞arn−1=∑n=0∞arna + ar + ar^2 + ar^3 + \cdots + ar^n + \cdots = \sum_{n=1}^\infty ar^{n-1}=\sum_{n=0}^\infty ar^n

in which aa and rr are fixed real numbers and a≠0a \neq 0.

The ratio rr can be positive or negative, and can be determined by dividing any term in the series by the term before it. For example:

∑n=0∞2n=1+2+4+8+⋯+2n+⋯  ⇒  r=21=42=84=⋯=2∑n=1∞(−13)n−1=1−13+19−127+⋯+(−13)n−1+⋯  ⇒  r=−1/31=1/9−1/3=−1/271/9=⋯=−13\sum_{n=0}^\infty 2^n = 1+2+4+8+\cdots+2^n+\cdots \ \ \Rightarrow \ \ r=\frac{2}{1}=\frac{4}{2}=\frac{8}{4}=\cdots=2\\ \sum_{n=1}^\infty \left(-\frac{1}{3}\right)^{n-1} = 1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\cdots+\left(-\frac{1}{3}\right)^{n-1} + \cdots \ \ \Rightarrow \ \ r=\frac{-1/3}{1}=\frac{1/9}{-1/3}=\frac{-1/27}{1/9}=\cdots=-\frac{1}{3}

If r=1r=1, the infinite geometric diverges. Why?

∑n=0∞a(1)n=∑n=0∞a=a+a+a+a+⋯⇒∞\sum_{n=0}^\infty a(1)^n = \sum_{n=0}^\infty a = a+a+a+a+\cdots \Rightarrow \infty

Note: This can also be shown using the nth Term Test. Take the limit when r=1r=1 and see that the limit approaches infinity.

If r=−1r=-1, the infinite geometric diverges. Why?

∑n=0∞a(−1)n=∑n=0∞a=a−a+a−a+⋯⇒oscillates\sum_{n=0}^\infty a(-1)^n = \sum_{n=0}^\infty a = a-a+a-a+\cdots \Rightarrow \text{oscillates}

Note: This can also be shown using the nth Term Test. Take the limit when r=−1r=-1 and see that the limit does not exist.

If ∣r∣>1|r|>1, the infinite geometric diverges. Why?

Note: This can also be shown using the nth Term Test. Take the limit when r=1r=1 and see that the limit approaches infinity.

If ∣r∣<1|r|<1, the infinite geometric converges. Why and to what?

Geometric Series Test

The logic we laid out above leads us to the Geometric Series Test.

The Geometric Series Test (GST) says that:

  • If ∣r∣<1|r|<1, the infinite geometric series ∑n=0∞arn\sum_{n=0}^\infty ar^n converges to a1−r\frac{a}{1-r} (where aa is the first term of the series)

  • If ∣r∣≥1|r| \geq 1, the infinite geometric series ∑n=0∞arn\sum_{n=0}^\infty ar^n diverges.

Example 1

Determine the convergence/divergence of the series 1+2+4+8+…1+2+4+8+\dots using the Geometric Series Test. If it converges, find its sum.

View Answer

Diverges

View Solution

1+2+4+8+…1+2+4+8+\dots diverges according to the Geometric Series Test as ∣r∣=2≥1|r|=2 \geq 1.

Example 2

Determine the convergence/divergence of the series ∑k=1∞35k\sum_{k=1}^\infty \frac{3}{5^k} using the Geometric Series Test. If it converges, find its sum.

View Answer

Converges to 34\frac{3}{4}

View Solution

∑k=1∞35k\sum_{k=1}^\infty \frac{3}{5^k} converges to 34\frac{3}{4} according to the Geometric Series Test as ∣r∣=15<1|r|=\frac{1}{5} < 1.

Example 3

Determine the convergence/divergence of the series ∑n=0∞(−1.1)n\sum_{n=0}^\infty (-1.1)^n using the Geometric Series Test. If it converges, find its sum.

View Answer

Diverges

View Solution

∑n=0∞(−1.1)n\sum_{n=0}^\infty (-1.1)^n diverges according to the Geometric Series Test as ∣r∣=1.1≥1|r|=1.1 \geq 1.

Example 4

Determine the convergence/divergence of the series ∑n=1∞13n15n\sum_{n=1}^\infty \frac{13^n}{15^n} using the Geometric Series Test. If it converges, find its sum.

View Answer

Converges to 132\frac{13}{2}

View Solution

∑n=1∞13n15n\sum_{n=1}^\infty \frac{13^n}{15^n} converges to 132\frac{13}{2} according to the Geometric Series Test as ∣r∣=1315<1|r|=\frac{13}{15} < 1.

Example 5

Determine the convergence/divergence of the series ∑n=1∞1000(0.75)n\sum_{n=1}^\infty 1000(0.75)^n using the Geometric Series Test. If it converges, find its sum.

View Answer

Converges to 30003000

View Solution

∑n=1∞1000(0.75)n\sum_{n=1}^\infty 1000(0.75)^n converges to 30003000 according to the Geometric Series Test as ∣r∣=0.75<1|r|=0.75 < 1.

Example 6

Determine the convergence/divergence of the series ∑i=1∞2i−13i\sum_{i=1}^\infty \frac{2^i-1}{3^i} using the Geometric Series Test. If it converges, find its sum.

View Answer

Converges to 32\frac{3}{2}

View Solution

∑i=1∞2i−13i=∑i=1∞(23)i−∑i=1∞(13)i\sum_{i=1}^\infty \frac{2^i-1}{3^i} = \sum_{i=1}^\infty \left( \frac{2}{3} \right)^i - \sum_{i=1}^\infty \left( \frac{1}{3} \right)^i converges to 32\frac{3}{2} according to the Geometric Series Test (as it is the difference of two convergent Geometric Series with ∣r∣=23<1|r|=\frac{2}{3} < 1 and ∣r∣=13<1|r|=\frac{1}{3} < 1).

Example 7

Determine the convergence/divergence of the series ∑n=1∞7n+3(−2)n−1\sum_{n=1}^\infty \frac{7^{n+3}}{(-2)^{n-1}} using the Geometric Series Test. If it converges, find its sum.

View Answer

Diverges

View Solution

∑n=1∞7n+3(−2)n−1\sum_{n=1}^\infty \frac{7^{n+3}}{(-2)^{n-1}} diverges according to the Geometric Series Test as ∣r∣=72≥1|r|=\frac{7}{2} \geq 1.

Example 8

Find the values of xx for which the series ∑n=1∞(x+1)n\sum_{n=1}^\infty (x+1)^n converges. Then, find the sum of the series for those values of xx.

View Answer

Converges to −x+1x-\frac{x+1}{x} for −2<x<0-2 < x < 0

Decimals as the Ratio of Two Integers

An interesting application of geometric series is to express decimals as the ratio of two integers. For example:

5.232323 . . .=5+23100+231002++231003+⋯=5+23100(1+1100+11002+⋯ )=5+23100∑n=0∞(1100)n=5+23100(11−1100)=5+23100⋅10099=51899\begin{aligned} 5.232323 \ . \ . \ . &= 5 + \frac{23}{100} + \frac{23}{100^2} + + \frac{23}{100^3} + \cdots\\ &= 5 + \frac{23}{100}\left( 1 + \frac{1}{100} + \frac{1}{100^2} + \cdots \right)\\ &= 5 + \frac{23}{100}\sum_{n=0}^\infty \left(\frac{1}{100}\right)^n\\ &= 5 + \frac{23}{100}\left( \frac{1}{1-\frac{1}{100}} \right)\\ &= 5 + \frac{23}{100}\cdot\frac{100}{99}\\ &= \frac{518}{99} \end{aligned}

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