Last time we studied "geometric series" of a special form. In general, a geometric series is a series where the ratio between consecutive terms is a fixed constant. In this lecture we will study such more general cases.
Tianran Chen
Department of Mathematics
Auburn University at Montgomery
Last time we learned that the geometric series $a + ar + ar^2 + ar^3 + \cdots$ (with $a \ne 0, |r| \ne 0,1$) converges to \[ \frac{a}{1-r} \] if $|r| < 1$. Otherwise, it diverges.
For example, \[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = \sum_{k=0}^\infty \left( \frac{1}{2} \right)^k = \frac{1}{ 1 - \frac{1}{2} } = \frac{1}{ \frac{1}{2} } = 2. \]
For example, \[ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = \sum_{\color{red}{k=1}}^\infty \left( \frac{1}{2} \right)^k \]
We can simply factor out $\frac{1}{2}$ from each term and get \[ \frac{1}{2} \cdot 1 + \frac{1}{2} \frac{1}{2} + \frac{1}{2} \frac{1}{4} + \frac{1}{2} \frac{1}{8} + \cdots = \sum_{\color{red}{k=0}}^\infty \frac{1}{2} \left( \frac{1}{2} \right)^k = \frac{\frac{1}{2}}{ 1 - \frac{1}{2} } = 1. \]
For real numbers $a,r$ with $a \ne 0$, $|r| \ne 0,1$, and integer $k_0$, the series \[ ar^{k_0} + ar^{k_0+1} + ar^{k_0+2} + ar^{k_0+3} + \cdots \] converges to \[ \frac{a \, r^{k_0} }{ 1-r } \] if $|r| < 1$. Otherwise, this series diverges.
For example, \[ \sum_{k=3}^\infty 4 \left( \frac{2}{3} \right)^k = \frac{ 4 \cdot \left( \frac{2}{3} \right)^3 }{ 1 - \frac{2}{3} } = \frac{ 4 \cdot \left( \frac{8}{27} \right) }{ \frac{1}{3} } = \frac{32}{9} \]
What can we say about the series \[ \sum_{k=0}^\infty \frac{ 2^{k+2} }{ 3^{k-1} } ? \]
Even though it does not look a geometric series, with some algebraic manipulation, we can turn it into one. \[ \sum_{k=0}^\infty \frac{ 2^{k+2} }{ 3^{k-1} } = \sum_{k=0}^\infty \frac{ 2^2 \cdot 2^{k} }{ 3^{-1} \cdot 3^{k} } = \sum_{k=0}^\infty 2^2 \cdot 3 \cdot \frac{ 2^{k} }{ 3^{k} } = \sum_{k=0}^\infty 12 \left( \frac{ 2 }{ 3 } \right)^k. \] We can see it is a geometric series of the form $\sum_{k=0}^\infty a r^k$ with $a=12$ and $r = 2/3$. Since $|r| < 1$, we can conclude that this series converges.
You have already used geometric series in elementary school: Every time you write down a repeating decimal, you are actually using a series. \[ 0.9999\cdots = 0.9 + 0.09 + 0.009 + 0.0009 + \cdots = \sum_{k=1}^\infty 9 (0.1)^{k} \]
Using the formula given above, we can compute that \[ 0.9999\cdots = \sum_{k=1}^\infty 9 (0.1)^{k} = \frac{ 9 \cdot (0.1)^1 }{ 1 - 0.1 } = \frac{ 0.9 }{ 0.9 } = 1. \] This provides a rigorous answers to the question we probably all had: $0.9999\cdot$ is indeed exactly 1 (not just an approximation).
Problem. Which fraction is equivalent to $0.222\cdots$?
Solution. Note that \[ 0.222\cdots = 0.2 + 0.02 + 0.002 + \cdots = \sum_{k=1}^\infty 2 (0.1)^{k}. \] Using the formula given above, we can easily calculate that \[ 0.222\cdots = \sum_{k=1}^\infty 2 (0.1)^{k} = \frac{2 \cdot 0.1}{1 - 0.1} = \frac{0.2}{0.9} = \frac{2}{9}. \]
Problem. Which fraction is equivalent to $0.123123123\cdots$?
Solution. Note that \[ 0.\overline{123} = 0.123 + 0.000123 \cdots = \sum_{k=1}^\infty 123 (0.001)^{k} . \] Using the formula given above, we can easily calculate that \[ 0.\overline{123} = \sum_{k=1}^\infty 123 (0.001)^{k} = \frac{123 \cdot 0.001}{1 - 0.001} = \frac{0.123}{0.999} = \frac{123}{999}. \]
Here, we can also see geometric series, or infinite series, in general, are quite natural and necessary: we have to use them if we want to express some fractions in decimal notation.