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Section 12.1 Geometric Series

Knowledge of the use of power series is very important when dealing with both probability functions.

\begin{gather*} S = \sum_{k=0}^{\infty} {x^k} = \frac{1}{1-x} \end{gather*}

as is its extension know as the negative binomial series \(( n \in \mathbb{N} )\text{.}\)

\begin{gather*} NB = \sum_{k=0}^{\infty} (-1)^k \binom{-n + k - 1}{k} {x^k b^{-n-k}} = \frac{1}{(x+b)^n} \end{gather*}

In this section, we review this series, develop its properties, and explore some of its extensions.

Subsection 12.1.1 Geometric Series

Consider the partial sum

\begin{gather*} S_n = \sum_{k=0}^{n} {x^k} = 1 + x + x^2 + ... + x^n \\ (1-x)S_n = S_n - x S_n = 1 + x + x^2 + ... + x^n - (x + x^2 + ... + x^n + x^{n+1}) = 1 - x^{n+1} \\ \Rightarrow S_n = \frac{1-x^{n+1}}{1-x} \end{gather*}

and so as \(n \rightarrow \infty \text{,}\)

\begin{gather*} S_n \rightarrow S = \frac{1}{1-x} \end{gather*}

The interactive activity below shows how well the partial sums approximate \(\frac{1}{1-x}\) as the number of terms increases.

Subsection 12.1.2 Alternate Forms for the Geometric Series

\begin{align*} \sum_{k=M}^{\infty} {x^k} & = x^M \sum_{k=0}^{\infty} {x^k}\\ & = x^M \frac{1}{1-x}\\ & = \frac{x^M}{1-x} \end{align*}

The geometric power series is a nice function which is relatively easily differentiated and integrated. In doing so, one can obtain new power series which might also be very useful in their own right. Here we develop a few which are of special interest.

Let \(f(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x}\text{.}\) Then,

\begin{gather*} f'(x) = \sum_{k=1}^{\infty} {kx^{k-1}} = \frac{1}{(1-x)^2}\\ f''(x) = \sum_{k=2}^{\infty} {k(k-1)x^{k-1}} = \frac{2}{(1-x)^3}\\ f^{(n)}(x) = \sum_{k=n}^{\infty} {k(k-1)...(k-n+1)x^{k-n}} = \frac{n!}{(1-x)^{n+1}}\\ \int f(x) dx = \sum_{k=0}^{\infty} {\frac{x^{k+1}}{k+1}} = -ln(1-x) \end{gather*}
\begin{align*} \sum_{k=0}^{\infty} {a^k x^k} & = \sum_{k=0}^{\infty} {(ax)^k}\\ & = \frac{1}{1-ax}, |x| \lt \frac{1}{a} \end{align*}

or perhaps

\begin{gather*} \sum_{k=0}^{\infty} {(x-b)^k} = \frac{1}{1-(x-b)}, |x-b| \lt 1 \end{gather*}

Consider this example:

\begin{align*} 2.48484848... & = 2 + 0.48 + 0.0048 + 0.000048 + ...\\ & = 2 + 0.48(1 + 0.01 + 0.0001 + ... ) = 2 + 0.48 \sum_{k=0}^\infty (0.01)^k \end{align*}

Therefore, applying the Geometric Series

\begin{align*} 2.48484848... & = 2 + 0.48 \frac{1}{1-0.01} \\ & = 2 + 0.48 \frac{100}{99} = 2 + \frac{48}{99} \end{align*}

Certainly most students would agree that \(0.333333... = \frac{1}{3} \text{.}\) So, what about \(0.999999...\text{?}\) Simply follow the pattern above

\begin{align*} 0.999999... & = 0.9 + 0.09 + 0.009 + 0.0009 + ... = 0.9(1 + 0.1 + 0.1^2 + 0.1^3 + ...\\ & = 0.9 \frac{1}{1-0.1} = 0.9 \frac{1}{0.9} = 1 \end{align*}