To determine which geometric series converge, let’s analyze each series in detail. A geometric series converges if the absolute value of the common ratio [tex]\( r \)[/tex] is less than 1.
1. First Series:
[tex]\[
\frac{1}{81} + \frac{1}{27} + \frac{1}{9} + \frac{1}{3} + \ldots
\][/tex]
- First term [tex]\(a\)[/tex]: [tex]\(\frac{1}{81}\)[/tex]
- Common ratio [tex]\(r\)[/tex]: [tex]\(\frac{\frac{1}{27}}{\frac{1}{81}} = \frac{1}{27} \times \frac{81}{1} = 3\)[/tex]
The common ratio [tex]\( r \)[/tex] for this series is 3. Since [tex]\(|r| = 3 > 1\)[/tex], this series does not converge.
2. Second Series:
[tex]\[
1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots
\][/tex]
- First term [tex]\(a\)[/tex]: [tex]\(1\)[/tex]
- Common ratio [tex]\(r\)[/tex]: [tex]\(\frac{\frac{1}{2}}{1} = \frac{1}{2}\)[/tex]
The common ratio [tex]\( r \)[/tex] for this series is [tex]\(\frac{1}{2}\)[/tex]. Since [tex]\(|r| = \frac{1}{2} < 1\)[/tex], this series converges.
3. Third Series:
[tex]\[
\sum_{n=1}^{\infty} 7(-4)^{n-1}
\][/tex]
- First term [tex]\(a\)[/tex]: For [tex]\( n=1 \)[/tex], the term is [tex]\( 7(-4)^{0} = 7 \)[/tex]
- Common ratio [tex]\(r\)[/tex]: [tex]\(-4\)[/tex]
The common ratio [tex]\( r \)[/tex] for this series is [tex]\(-4\)[/tex]. Since [tex]\(|r| = 4 > 1\)[/tex], this series does not converge.
4. Fourth Series:
[tex]\[
\sum_{n=1}^{\infty} \frac{1}{5}(2)^{n-1}
\][/tex]
- First term [tex]\(a\)[/tex]: For [tex]\( n=1 \)[/tex], the term is [tex]\(\frac{1}{5}(2)^{0} = \frac{1}{5}\)[/tex]
- Common ratio [tex]\(r\)[/tex]: [tex]\( 2\)[/tex]
The common ratio [tex]\( r \)[/tex] for this series is [tex]\(2\)[/tex]. Since [tex]\(|r| = 2 > 1\)[/tex], this series does not converge.
Based on the analysis, the geometric series that converge are:
- [tex]\( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \)[/tex]
Thus, the second series is the only converging series among the given options.
