Question
why can't the formula for an infinite geometric series be applied to 1/81 + 1/27 + 1/9 + 1/3 + ...
Answers
GPT-4o mini
To determine whether the formula for an infinite geometric series can be applied to the series \( S = \frac{1}{81} + \frac{1}{27} + \frac{1}{9} + \frac{1}{3} + \ldots \), we need to check if the series meets the criteria of an infinite geometric series.
An infinite geometric series can be represented as:
\[
S = a + ar + ar^2 + ar^3 + \ldots
\]
where:
- \( a \) is the first term,
- \( r \) is the common ratio between consecutive terms.
For the series to be classified as a geometric series, the ratio between consecutive terms must be constant and satisfy \( |r| < 1 \) for the series to converge.
Let's analyze the given series:
1. **Identifying the first term \( a \)**:
The first term \( a \) is \( \frac{1}{81} \).
2. **Finding the common ratio \( r \)**:
To find the common ratio, we can look at the ratio of successive terms:
- Second term: \( \frac{1}{27} \)
- Ratio \( r = \frac{\frac{1}{27}}{\frac{1}{81}} = \frac{1}{27} \times \frac{81}{1} = 3 \)
Similarly, calculating the ratio for other successive terms:
- For the third term: \( r = \frac{\frac{1}{9}}{\frac{1}{27}} = 3 \)
- For the fourth term: \( r = \frac{\frac{1}{3}}{\frac{1}{9}} = 3 \)
We observe that the ratio between consecutive terms is \( r = 3 \).
3. **Check the condition for convergence**:
Since \( r = 3 \), we see that:
\[
|r| = 3 \not < 1
\]
Because the common ratio is greater than 1, the series does not converge.
### Conclusion
The formula for an infinite geometric series cannot be applied in this case because the series does not meet the convergence criterion (i.e., the common ratio \( r \) is not less than 1 in absolute value). Instead, the series diverges to infinity.
An infinite geometric series can be represented as:
\[
S = a + ar + ar^2 + ar^3 + \ldots
\]
where:
- \( a \) is the first term,
- \( r \) is the common ratio between consecutive terms.
For the series to be classified as a geometric series, the ratio between consecutive terms must be constant and satisfy \( |r| < 1 \) for the series to converge.
Let's analyze the given series:
1. **Identifying the first term \( a \)**:
The first term \( a \) is \( \frac{1}{81} \).
2. **Finding the common ratio \( r \)**:
To find the common ratio, we can look at the ratio of successive terms:
- Second term: \( \frac{1}{27} \)
- Ratio \( r = \frac{\frac{1}{27}}{\frac{1}{81}} = \frac{1}{27} \times \frac{81}{1} = 3 \)
Similarly, calculating the ratio for other successive terms:
- For the third term: \( r = \frac{\frac{1}{9}}{\frac{1}{27}} = 3 \)
- For the fourth term: \( r = \frac{\frac{1}{3}}{\frac{1}{9}} = 3 \)
We observe that the ratio between consecutive terms is \( r = 3 \).
3. **Check the condition for convergence**:
Since \( r = 3 \), we see that:
\[
|r| = 3 \not < 1
\]
Because the common ratio is greater than 1, the series does not converge.
### Conclusion
The formula for an infinite geometric series cannot be applied in this case because the series does not meet the convergence criterion (i.e., the common ratio \( r \) is not less than 1 in absolute value). Instead, the series diverges to infinity.