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Sure, let's go through the questions step-by-step:
### Fill in the Blanks:
- A sequence is a list of numbers.
- A series is a sum of numbers. A series converges if its partial sum has a limit.
### Geometric Series Convergence:
1. A geometric series is a series of the form [tex]\(\sum_{n=0}^{\infty} a r^n\)[/tex] for constants [tex]\(a\)[/tex] and [tex]\(r\)[/tex]. When does such a series converge? If the series does converge, what does it converge to?
A geometric series converges when the absolute value of the common ratio [tex]\(r\)[/tex] is less than 1, i.e., [tex]\(|r| < 1\)[/tex]. If the series converges, it converges to the sum:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
### Convergence of the Given Series:
2. Decide whether or not [tex]\(\sum_{i=1}^{\infty} \left(8 \cdot 3^{-2n}\right)\)[/tex] converges or diverges. If it converges, find the sum.
To analyze the convergence of the given series, let's rewrite it more clearly. The series can be written as:
[tex]\[ \sum_{n=1}^{\infty} \left(8 \cdot 3^{-2n}\right) = 8 \sum_{n=1}^{\infty} \left(3^{-2n}\right) \][/tex]
Notice that [tex]\(3^{-2n}\)[/tex] can be written as [tex]\((3^{-2})^n\)[/tex]. Let's set [tex]\(r = 3^{-2}\)[/tex], which simplifies to [tex]\(r = \frac{1}{9}\)[/tex].
The series now looks like:
[tex]\[ 8 \sum_{n=1}^{\infty} \left(\frac{1}{9}\right)^n \][/tex]
We see that this is a geometric series with [tex]\(a = 8 \cdot \left(\frac{1}{9}\right)^1 = \frac{8}{9}\)[/tex] and [tex]\(r = \frac{1}{9}\)[/tex]. A geometric series converges if [tex]\(|r| < 1\)[/tex], which is true in this case since [tex]\(\frac{1}{9}\)[/tex] is less than 1.
The sum of an infinite geometric series starting from [tex]\(n = 0\)[/tex] is given by:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
But our series starts from [tex]\(n = 1\)[/tex]. To adjust for this, we can consider summing an infinite geometric series from [tex]\(n = 0\)[/tex] and subtracting the [tex]\(n = 0\)[/tex] term. The sum from [tex]\(n = 0\)[/tex] to [tex]\(\infty\)[/tex] of [tex]\(8 \left(\frac{1}{9}\right)^n\)[/tex] is:
[tex]\[ S_{\text{total}} = \frac{8}{1 - \frac{1}{9}} = \frac{8}{\frac{8}{9}} = \frac{8 \cdot 9}{8} = 9 \][/tex]
Since we want the sum from [tex]\(n=1\)[/tex] to [tex]\(\infty\)[/tex], we subtract the term at [tex]\(n = 0\)[/tex] which is [tex]\(8 \left(\frac{1}{9}\right)^0 = 8\)[/tex]:
[tex]\[ S = 9 - 8 = 1 \][/tex]
Hence, the series [tex]\(\sum_{i=1}^{\infty} \left(8 \cdot 3^{-2n}\right)\)[/tex] converges and its sum is:
[tex]\[ S = 1 \][/tex]
### Fill in the Blanks:
- A sequence is a list of numbers.
- A series is a sum of numbers. A series converges if its partial sum has a limit.
### Geometric Series Convergence:
1. A geometric series is a series of the form [tex]\(\sum_{n=0}^{\infty} a r^n\)[/tex] for constants [tex]\(a\)[/tex] and [tex]\(r\)[/tex]. When does such a series converge? If the series does converge, what does it converge to?
A geometric series converges when the absolute value of the common ratio [tex]\(r\)[/tex] is less than 1, i.e., [tex]\(|r| < 1\)[/tex]. If the series converges, it converges to the sum:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
### Convergence of the Given Series:
2. Decide whether or not [tex]\(\sum_{i=1}^{\infty} \left(8 \cdot 3^{-2n}\right)\)[/tex] converges or diverges. If it converges, find the sum.
To analyze the convergence of the given series, let's rewrite it more clearly. The series can be written as:
[tex]\[ \sum_{n=1}^{\infty} \left(8 \cdot 3^{-2n}\right) = 8 \sum_{n=1}^{\infty} \left(3^{-2n}\right) \][/tex]
Notice that [tex]\(3^{-2n}\)[/tex] can be written as [tex]\((3^{-2})^n\)[/tex]. Let's set [tex]\(r = 3^{-2}\)[/tex], which simplifies to [tex]\(r = \frac{1}{9}\)[/tex].
The series now looks like:
[tex]\[ 8 \sum_{n=1}^{\infty} \left(\frac{1}{9}\right)^n \][/tex]
We see that this is a geometric series with [tex]\(a = 8 \cdot \left(\frac{1}{9}\right)^1 = \frac{8}{9}\)[/tex] and [tex]\(r = \frac{1}{9}\)[/tex]. A geometric series converges if [tex]\(|r| < 1\)[/tex], which is true in this case since [tex]\(\frac{1}{9}\)[/tex] is less than 1.
The sum of an infinite geometric series starting from [tex]\(n = 0\)[/tex] is given by:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
But our series starts from [tex]\(n = 1\)[/tex]. To adjust for this, we can consider summing an infinite geometric series from [tex]\(n = 0\)[/tex] and subtracting the [tex]\(n = 0\)[/tex] term. The sum from [tex]\(n = 0\)[/tex] to [tex]\(\infty\)[/tex] of [tex]\(8 \left(\frac{1}{9}\right)^n\)[/tex] is:
[tex]\[ S_{\text{total}} = \frac{8}{1 - \frac{1}{9}} = \frac{8}{\frac{8}{9}} = \frac{8 \cdot 9}{8} = 9 \][/tex]
Since we want the sum from [tex]\(n=1\)[/tex] to [tex]\(\infty\)[/tex], we subtract the term at [tex]\(n = 0\)[/tex] which is [tex]\(8 \left(\frac{1}{9}\right)^0 = 8\)[/tex]:
[tex]\[ S = 9 - 8 = 1 \][/tex]
Hence, the series [tex]\(\sum_{i=1}^{\infty} \left(8 \cdot 3^{-2n}\right)\)[/tex] converges and its sum is:
[tex]\[ S = 1 \][/tex]
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