Answer :
To determine which of the given geometric series diverges, we need to check the common ratio ([tex]\(r\)[/tex]) of each series. A geometric series converges if the absolute value of the common ratio is less than 1 ([tex]\(|r| < 1\)[/tex]). Conversely, it diverges if the absolute value of the common ratio is greater than or equal to 1 ([tex]\(|r| \geq 1\)[/tex]).
Let's analyze each series one by one.
### Series 1:
[tex]\[ \frac{3}{5}+\frac{3}{10}+\frac{3}{20}+\frac{3}{40}+\ldots \][/tex]
To find the common ratio for the first series, we divide the second term by the first term:
[tex]\[ r = \frac{\frac{3}{10}}{\frac{3}{5}} = \frac{3}{10} \times \frac{5}{3} = \frac{5}{10} = 0.5 \][/tex]
The absolute value of the common ratio is:
[tex]\[ |r| = |0.5| = 0.5 \][/tex]
Since [tex]\(0.5 < 1\)[/tex], this series converges.
### Series 2:
[tex]\[ -10+4-\frac{8}{5}+\frac{16}{25}+\ldots \][/tex]
To find the common ratio for this series, we divide the second term by the first term:
[tex]\[ r = \frac{4}{-10} = -0.4 \][/tex]
The absolute value of the common ratio is:
[tex]\[ |r| = |-0.4| = 0.4 \][/tex]
Since [tex]\(0.4 < 1\)[/tex], this series converges.
### Series 3:
[tex]\[ \sum_{n=1}^{\infty} \frac{2}{3}(-4)^{n-1} \][/tex]
The series given in summation notation has a common ratio specified directly:
[tex]\[ r = -4 \][/tex]
The absolute value of the common ratio is:
[tex]\[ |r| = |-4| = 4 \][/tex]
Since [tex]\(4 \geq 1\)[/tex], this series diverges.
### Series 4:
[tex]\[ \sum_{n=1}^{\infty}(-12)\left(\frac{1}{5}\right)^{n-1} \][/tex]
The series given in summation notation has a common ratio specified directly:
[tex]\[ r = \frac{1}{5} \][/tex]
The absolute value of the common ratio is:
[tex]\[ |r| = \left|\frac{1}{5}\right| = 0.2 \][/tex]
Since [tex]\(0.2 < 1\)[/tex], this series converges.
### Conclusion:
The series that diverges is the one with the common ratio [tex]\(-4\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \frac{2}{3}(-4)^{n-1} \][/tex]
Let's analyze each series one by one.
### Series 1:
[tex]\[ \frac{3}{5}+\frac{3}{10}+\frac{3}{20}+\frac{3}{40}+\ldots \][/tex]
To find the common ratio for the first series, we divide the second term by the first term:
[tex]\[ r = \frac{\frac{3}{10}}{\frac{3}{5}} = \frac{3}{10} \times \frac{5}{3} = \frac{5}{10} = 0.5 \][/tex]
The absolute value of the common ratio is:
[tex]\[ |r| = |0.5| = 0.5 \][/tex]
Since [tex]\(0.5 < 1\)[/tex], this series converges.
### Series 2:
[tex]\[ -10+4-\frac{8}{5}+\frac{16}{25}+\ldots \][/tex]
To find the common ratio for this series, we divide the second term by the first term:
[tex]\[ r = \frac{4}{-10} = -0.4 \][/tex]
The absolute value of the common ratio is:
[tex]\[ |r| = |-0.4| = 0.4 \][/tex]
Since [tex]\(0.4 < 1\)[/tex], this series converges.
### Series 3:
[tex]\[ \sum_{n=1}^{\infty} \frac{2}{3}(-4)^{n-1} \][/tex]
The series given in summation notation has a common ratio specified directly:
[tex]\[ r = -4 \][/tex]
The absolute value of the common ratio is:
[tex]\[ |r| = |-4| = 4 \][/tex]
Since [tex]\(4 \geq 1\)[/tex], this series diverges.
### Series 4:
[tex]\[ \sum_{n=1}^{\infty}(-12)\left(\frac{1}{5}\right)^{n-1} \][/tex]
The series given in summation notation has a common ratio specified directly:
[tex]\[ r = \frac{1}{5} \][/tex]
The absolute value of the common ratio is:
[tex]\[ |r| = \left|\frac{1}{5}\right| = 0.2 \][/tex]
Since [tex]\(0.2 < 1\)[/tex], this series converges.
### Conclusion:
The series that diverges is the one with the common ratio [tex]\(-4\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \frac{2}{3}(-4)^{n-1} \][/tex]
