To determine which geometric series converge, let's analyze each series and their common ratios.
1. First Geometric Series:
[tex]\[
\frac{1}{81} + \frac{1}{27} + \frac{1}{9} + \frac{1}{3} + \ldots
\][/tex]
This series can be written as:
[tex]\[
a, ar, ar^2, ar^3, \ldots
\][/tex]
where \(a = \frac{1}{81}\) and \(r = \frac{\frac{1}{27}}{\frac{1}{81}} = 3.0\).
For a geometric series to converge, the absolute value of the common ratio \(r\) must be less than 1. Here, the common ratio is \(3.0\), which is greater than 1.
Hence, this series does not converge.
2. Second Geometric Series:
[tex]\[
1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots
\][/tex]
This series can be written as:
[tex]\[
a, ar, ar^2, ar^3, \ldots
\][/tex]
where \(a = 1\) and \(r = \frac{\frac{1}{2}}{1} = 0.5\).
For a geometric series to converge, the absolute value of the common ratio \(r\) must be less than 1. Here, the common ratio is \(0.5\), which is less than 1.
Hence, this series converges.
3. Third Geometric Series:
[tex]\[
\sum_{n=1}^{\infty} 7(-4)^{n-1}
\][/tex]
This series can be written as:
[tex]\[
a, ar, ar^2, ar^3, \ldots
\][/tex]
where \(a = 7\) and \(r = -4\).
For a geometric series to converge, the absolute value of the common ratio \(r\) must be less than 1. Here, the common ratio is \(-4\), which has an absolute value of \(4\), which is greater than 1.
Hence, this series does not converge.
4. Fourth Geometric Series:
[tex]\[
\sum_{n=1}^{\infty} \frac{1}{5}(2)^{n-1}
\][/tex]
This series can be written as:
[tex]\[
a, ar, ar^2, ar^3, \ldots
\][/tex]
where \(a = \frac{1}{5}\) and \(r = 2\).
For a geometric series to converge, the absolute value of the common ratio \(r\) must be less than 1. Here, the common ratio is \(2\), which is greater than 1.
Hence, this series does not converge.
In summary, the only geometric series among the given options that converges is:
[tex]\[
1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots
\][/tex]
