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]
