Answer :
To determine which geometric series converges, we need to check the absolute value of the common ratio \( r \) for each series. A geometric series \( \sum_{n=0}^\infty ar^n \) converges if and only if \( |r| < 1 \).
Let's analyze each given series step by step:
1. The series \( \frac{1}{81}+\frac{1}{27}+\frac{1}{9}+\frac{1}{3}+\ldots \)
- Common ratio \( r_1 \) can be found by dividing any term by the previous term.
- \( r_1 = \frac{\frac{1}{27}}{\frac{1}{81}} = 3 \)
- Since \( |3| > 1 \), this series diverges.
2. The series \( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \)
- Common ratio \( r_2 \) can be found by dividing any term by the previous term.
- \( r_2 = \frac{\frac{1}{2}}{1} = \frac{1}{2} \)
- Since \( |\frac{1}{2}| < 1 \), this series converges.
3. The series \( \sum_{n=1}^{\infty} 7(-4)^{n-1} \)
- Common ratio \( r_3 \) comes from the expression inside the sum.
- \( r_3 = -4 \)
- Since \( |-4| > 1 \), this series diverges.
4. The series \( \sum_{n=1}^{\infty} \frac{1}{5}(2)^{n-1} \)
- Common ratio \( r_4 \) comes from the expression inside the sum.
- \( r_4 = 2 \)
- Since \( |2| > 1 \), this series diverges.
### Conclusion
Only the second geometric series [tex]\( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \)[/tex] converges because its common ratio [tex]\( r = \frac{1}{2} \)[/tex] lies within the interval [tex]\(-1 < r < 1\)[/tex].
Let's analyze each given series step by step:
1. The series \( \frac{1}{81}+\frac{1}{27}+\frac{1}{9}+\frac{1}{3}+\ldots \)
- Common ratio \( r_1 \) can be found by dividing any term by the previous term.
- \( r_1 = \frac{\frac{1}{27}}{\frac{1}{81}} = 3 \)
- Since \( |3| > 1 \), this series diverges.
2. The series \( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \)
- Common ratio \( r_2 \) can be found by dividing any term by the previous term.
- \( r_2 = \frac{\frac{1}{2}}{1} = \frac{1}{2} \)
- Since \( |\frac{1}{2}| < 1 \), this series converges.
3. The series \( \sum_{n=1}^{\infty} 7(-4)^{n-1} \)
- Common ratio \( r_3 \) comes from the expression inside the sum.
- \( r_3 = -4 \)
- Since \( |-4| > 1 \), this series diverges.
4. The series \( \sum_{n=1}^{\infty} \frac{1}{5}(2)^{n-1} \)
- Common ratio \( r_4 \) comes from the expression inside the sum.
- \( r_4 = 2 \)
- Since \( |2| > 1 \), this series diverges.
### Conclusion
Only the second geometric series [tex]\( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \)[/tex] converges because its common ratio [tex]\( r = \frac{1}{2} \)[/tex] lies within the interval [tex]\(-1 < r < 1\)[/tex].
