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*}
provided x is small is of utmost importance 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
Proof.
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
Theorem 12.1.2. Generalized Geometric Series.
For \(k \in \mathbb{N}, \sum_{k=M}^{\infty} {x^k} = \frac{x^M}{1-x}\)Proof.
\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*}
Example 12.1.3. Integrating and Differentiating to get new Power Series.
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*}
Example 12.1.4. Playing with the base.
\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*}
Example 12.1.5. Application: Converting repeating decimals to fractional form.
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*}
Example 12.1.6. Playing around with repeating decimals.
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*}
