Sure! Let's analyze each geometric series to determine if it diverges.
### Series 1: [tex]\( 27 + 9 + 3 + 1 + \ldots \)[/tex]
This can be rewritten as:
[tex]\[ 27 + 27 \left(\frac{1}{3}\right) + 27 \left(\frac{1}{3}\right)^2 + 27 \left(\frac{1}{3}\right)^3 + \ldots \][/tex]
The common ratio [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{9}{27} = \frac{1}{3} \][/tex]
Since [tex]\( |r| = \frac{1}{3} < 1 \)[/tex], this series converges.
### Series 2: [tex]\( 1 + \frac{4}{3} + \frac{16}{9} + \frac{64}{27} + \ldots \)[/tex]
This can be rewritten as:
[tex]\[ 1 + \left(\frac{4}{3}\right) + \left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^3 + \ldots \][/tex]
The common ratio [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{4}{3} \][/tex]
Since [tex]\( |r| = \frac{4}{3} > 1 \)[/tex], this series diverges.
### Series 3: [tex]\( 2.2 + 0.22 + 0.022 + \ldots \)[/tex]
This can be rewritten as:
[tex]\[ 2.2 + 2.2 \left(\frac{1}{10}\right) + 2.2 \left(\frac{1}{10}\right)^2 + \ldots \][/tex]
The common ratio [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{0.22}{2.2} = \frac{1}{10} \][/tex]
Since [tex]\( |r| = \frac{1}{10} < 1 \)[/tex], this series converges.
### Series 4: [tex]\( \sum_{n=1}^{\infty}\left(\frac{1}{64}\right) \cdot(2)^{n-1} \)[/tex]
This can be rewritten as:
[tex]\[ \frac{1}{64} + \frac{1}{64} \cdot 2 + \frac{1}{64} \cdot 2^2 + \frac{1}{64} \cdot 2^3 + \ldots \][/tex]
The common ratio [tex]\( r \)[/tex] is:
[tex]\[ r = 2 \][/tex]
Since [tex]\( |r| = 2 > 1 \)[/tex], this series diverges.
### Conclusion
The geometric series that diverge are:
[tex]\[ 1 + \frac{4}{3} + \frac{16}{9} + \frac{64}{27} + \ldots \][/tex]
[tex]\[ \sum_{n=1}^{\infty}\left(\frac{1}{64}\right) \cdot(2)^{n-1} \][/tex]
Thus, we have determined that the second and fourth series diverge.
