C. Infinite Geometric Series
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C. Infinite Geometric Series
A ball is dropped from a height of \(1\) m. When it bounces back up, it reaches a height of \(0.5\) m. The next bounce reaches \(0.25\) m, and so on. The real question is, will it ever stop bouncing? Practically speaking, you will probably say yes because experience has taught you that it will. However, in theory, the ball will never stop bouncing. It will always return to half of its previous height. But, as the number of bounces increases, the subsequent heights reached by the ball get closer and closer to zero.
In other words, as the number of bounces goes to infinity, the heights (the sequence) approach, or have a limit of, zero. In order for an infinite sequence to have a limit, it must be convergent, such as the bouncing ball example. A convergent series (or sequence) has terms that approach a particular value. On the other hand, an infinite sequence will not have a limit if it is divergent. A divergent series (or sequence) has terms that do not approach a particular value. A sum cannot be calculated for infinite divergent geometric series. For infinite geometric series that are convergent, a finite sum can be determined (and there is a formula).
Convergent geometric series have the unique property that the common ratio, \(r\), is between \(–1\) and \(1\). As the number of terms, \(n \), increase indefinitely, \(r^n \) approaches zero when \(−1 < r < 1 \). The symbol, \(\infty\), is a mathematical symbol representing infinity.
To test that \(r^n \) approaches \(0\) when \(r \) is between \(-1\) and \(1\), calculate \(0.5^{99} \), \(0.5^{999} \), \(0.5^{9999} \), ...
What do you notice about the values as the value of the exponent increases towards infinity?
In other words, as the number of bounces goes to infinity, the heights (the sequence) approach, or have a limit of, zero. In order for an infinite sequence to have a limit, it must be convergent, such as the bouncing ball example. A convergent series (or sequence) has terms that approach a particular value. On the other hand, an infinite sequence will not have a limit if it is divergent. A divergent series (or sequence) has terms that do not approach a particular value. A sum cannot be calculated for infinite divergent geometric series. For infinite geometric series that are convergent, a finite sum can be determined (and there is a formula).
Convergent geometric series have the unique property that the common ratio, \(r\), is between \(–1\) and \(1\). As the number of terms, \(n \), increase indefinitely, \(r^n \) approaches zero when \(−1 < r < 1 \). The symbol, \(\infty\), is a mathematical symbol representing infinity.
\(\begin{align}
S_n &= \frac{{t_1 r^n - t_1 }}{{r - 1}} \\
S_\infty &= \frac{{0 - t_1 }}{{r - 1}} \\
S_\infty &= \frac{{ - t_1 }}{{ - \left( {1 - r} \right)}} \\
S_\infty &= \frac{{t_1 }}{{1 - r}} \\
\end{align}\)
S_n &= \frac{{t_1 r^n - t_1 }}{{r - 1}} \\
S_\infty &= \frac{{0 - t_1 }}{{r - 1}} \\
S_\infty &= \frac{{ - t_1 }}{{ - \left( {1 - r} \right)}} \\
S_\infty &= \frac{{t_1 }}{{1 - r}} \\
\end{align}\)
To test that \(r^n \) approaches \(0\) when \(r \) is between \(-1\) and \(1\), calculate \(0.5^{99} \), \(0.5^{999} \), \(0.5^{9999} \), ...
What do you notice about the values as the value of the exponent increases towards infinity?
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Key Lesson Marker
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Infinite Geometric Series
\[S_\infty = \frac{{t_1 }}{{1 - r}} \]
\(t_1 \) is the first term in the series
\(r \) is the common ratio \(\left(−1 < r < 1 \right)\)