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\large 
\underline
{MA2160 \hspace{5mm}Calculus II with Technology \hspace{50mm} Ch. 9-2}

We can show from algebra that $.\bar{9}=1$.  But what is $.\bar{9}$? It is a geometric series, and its value is determined by the limit as the number of terms approaches infinity.


\Large
Goal
\large
\begin{enumerate}
\item Define geometric series and partial sums
\item Identify the means and conditions to find the finite value of a geometric series.

\end{enumerate}
What is a geometric series?  It is the sum of an infinite set of numbers such that the ratio of any two consecutive numbers is constant.  Using summation notation, where $a$ is the first term, and $n$ approaches $\infty$:

$$   \Sigma_{_{i=1}}^{^n}\,ar^{i-1}$$

We define a partial sum as the sum of a consecutive finite set of the infinite set.

How do we find the value of the partial sum?  Letting $x$ equal the partial sum and performing some algebraic manipulation, we find the value:

$$x= \frac{a(1-r^n)}{1-r}$$

What if we want to find the sum of the entire series?  We need to find the limit of the partial sum as $n \rightarrow\infty$:

$$ \lim_{n \rightarrow\infty}\,\Sigma_{_{i=1}}^{^n}\,ar^{i-1}$$

\newpage


It is quite obvious that if $|r| \geq 1$, there is no limit.  However, if $0<r < 1$, then the limit exists and the value of the sum is:

$$   \Sigma_{_{i=1}}^{^n}\,ar^{i-1}=\frac{a}{1-r}$$

You would think that if $r<0$, we may have to modify the formula, but it doesn't matter.  As long as $|r|<1$, the formula works.

Utilizing this method, we can further establish the fact that $.\bar{9}=1$, as $a=\frac{9}{10}$ and $r=\frac{1}{10}$


Find the value of the geometric series given the following conditions:

\begin{enumerate}
\item $a=1,\,r=\frac{1}{2}$
\item $a=\frac{2}{5},\,r=\frac{1}{4}$
\item $a=15,\,r=\frac{1}{3}$
\item $a=-16,\,r=\frac{2}{7}$
\item $a=1,\,r=-\frac{3}{5}$
\end{enumerate}


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