Answer :
To solve this problem, we need to analyze the given series [tex]$\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\ldots$[/tex] and determine which expression defines [tex]\( S_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity.
This series is a geometric series where the first term ([tex]\( a \)[/tex]) is [tex]\( \frac{1}{8} \)[/tex] and the common ratio ([tex]\( r \)[/tex]) is [tex]\( \frac{1}{2} \)[/tex]. The sum of the first [tex]\( n \)[/tex] terms ([tex]\( S_n \)[/tex]) of a geometric series can be given by:
[tex]\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \][/tex]
For a geometric series with [tex]\( |r| < 1 \)[/tex], as [tex]\( n \)[/tex] approaches infinity, the term [tex]\( r^n \)[/tex] approaches 0. Therefore, the infinite sum ([tex]\( S \)[/tex]) is:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
When we plug in the values [tex]\( a = \frac{1}{8} \)[/tex] and [tex]\( r = \frac{1}{2} \)[/tex], we get:
[tex]\[ S = \frac{\frac{1}{8}}{1 - \frac{1}{2}} = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{8} \times \frac{2}{1} = \frac{1}{4} \][/tex]
Now, we need to check which of the given options approaches [tex]\( \frac{1}{4} \)[/tex] as [tex]\( n \)[/tex] approaches infinity.
1. [tex]\(\lim_{n \to \infty} \frac{1}{2^n}\)[/tex]
As [tex]\( n \)[/tex] approaches infinity, [tex]\( 2^n \)[/tex] grows very large, causing [tex]\( \frac{1}{2^n} \)[/tex] to approach 0.
2. [tex]\(\lim_{n \to \infty} \frac{1}{4(2^n)}\)[/tex]
Similarly, [tex]\( 4(2^n) \)[/tex] grows very large as [tex]\( n \)[/tex] approaches infinity, causing [tex]\( \frac{1}{4(2^n)} \)[/tex] to also approach 0.
3. [tex]\(\lim_{n \to \infty} \frac{2^n}{4(2^n)}\)[/tex]
Here, the term [tex]\( \frac{2^n}{4(2^n)} \)[/tex] simplifies to [tex]\( \frac{2^n}{4 \cdot 2^n} = \frac{2^n}{2^{n+2}} = \frac{1}{4} \)[/tex]. As [tex]\( n \)[/tex] approaches infinity, this limit equals [tex]\( \frac{1}{4} \)[/tex].
4. [tex]\(\lim_{n \to \infty} \frac{2^n - 1}{4(2^n)}\)[/tex]
This expression simplifies to [tex]\( \frac{2^n - 1}{4 \cdot 2^n} = \frac{2^n}{4 \cdot 2^n} - \frac{1}{4 \cdot 2^n} = \frac{1}{4} - \frac{1}{4 \cdot 2^n} \)[/tex]. As [tex]\( n \)[/tex] approaches infinity, [tex]\( \frac{1}{4 \cdot 2^n} \)[/tex] approaches 0, so the whole limit approaches [tex]\( \frac{1}{4} \)[/tex].
Thus, the two expressions that define [tex]\( S_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity are:
[tex]\[ \lim_{n \to \infty} \frac{2^n}{4 \cdot 2^n} \][/tex]
[tex]\[ \lim_{n \to \infty} \frac{2^n - 1}{4 \cdot 2^n} \][/tex]
Both of these limits approach [tex]\( \frac{1}{4} \)[/tex], making an expression defining [tex]\( S_n \)[/tex]:
Answer:
[tex]\[ \lim_{n \to \infty} \frac{2^n}{4(2^n)} \quad \text{and} \quad \lim_{n \to \infty} \frac{2^n - 1}{4(2^n)} \][/tex]
Given that the second form is exactly provided in the options, the correct answer thus is:
[tex]\[ \lim_{n \to \infty} \frac{2^n - 1}{4(2^n)} \][/tex]
This series is a geometric series where the first term ([tex]\( a \)[/tex]) is [tex]\( \frac{1}{8} \)[/tex] and the common ratio ([tex]\( r \)[/tex]) is [tex]\( \frac{1}{2} \)[/tex]. The sum of the first [tex]\( n \)[/tex] terms ([tex]\( S_n \)[/tex]) of a geometric series can be given by:
[tex]\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \][/tex]
For a geometric series with [tex]\( |r| < 1 \)[/tex], as [tex]\( n \)[/tex] approaches infinity, the term [tex]\( r^n \)[/tex] approaches 0. Therefore, the infinite sum ([tex]\( S \)[/tex]) is:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
When we plug in the values [tex]\( a = \frac{1}{8} \)[/tex] and [tex]\( r = \frac{1}{2} \)[/tex], we get:
[tex]\[ S = \frac{\frac{1}{8}}{1 - \frac{1}{2}} = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{8} \times \frac{2}{1} = \frac{1}{4} \][/tex]
Now, we need to check which of the given options approaches [tex]\( \frac{1}{4} \)[/tex] as [tex]\( n \)[/tex] approaches infinity.
1. [tex]\(\lim_{n \to \infty} \frac{1}{2^n}\)[/tex]
As [tex]\( n \)[/tex] approaches infinity, [tex]\( 2^n \)[/tex] grows very large, causing [tex]\( \frac{1}{2^n} \)[/tex] to approach 0.
2. [tex]\(\lim_{n \to \infty} \frac{1}{4(2^n)}\)[/tex]
Similarly, [tex]\( 4(2^n) \)[/tex] grows very large as [tex]\( n \)[/tex] approaches infinity, causing [tex]\( \frac{1}{4(2^n)} \)[/tex] to also approach 0.
3. [tex]\(\lim_{n \to \infty} \frac{2^n}{4(2^n)}\)[/tex]
Here, the term [tex]\( \frac{2^n}{4(2^n)} \)[/tex] simplifies to [tex]\( \frac{2^n}{4 \cdot 2^n} = \frac{2^n}{2^{n+2}} = \frac{1}{4} \)[/tex]. As [tex]\( n \)[/tex] approaches infinity, this limit equals [tex]\( \frac{1}{4} \)[/tex].
4. [tex]\(\lim_{n \to \infty} \frac{2^n - 1}{4(2^n)}\)[/tex]
This expression simplifies to [tex]\( \frac{2^n - 1}{4 \cdot 2^n} = \frac{2^n}{4 \cdot 2^n} - \frac{1}{4 \cdot 2^n} = \frac{1}{4} - \frac{1}{4 \cdot 2^n} \)[/tex]. As [tex]\( n \)[/tex] approaches infinity, [tex]\( \frac{1}{4 \cdot 2^n} \)[/tex] approaches 0, so the whole limit approaches [tex]\( \frac{1}{4} \)[/tex].
Thus, the two expressions that define [tex]\( S_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity are:
[tex]\[ \lim_{n \to \infty} \frac{2^n}{4 \cdot 2^n} \][/tex]
[tex]\[ \lim_{n \to \infty} \frac{2^n - 1}{4 \cdot 2^n} \][/tex]
Both of these limits approach [tex]\( \frac{1}{4} \)[/tex], making an expression defining [tex]\( S_n \)[/tex]:
Answer:
[tex]\[ \lim_{n \to \infty} \frac{2^n}{4(2^n)} \quad \text{and} \quad \lim_{n \to \infty} \frac{2^n - 1}{4(2^n)} \][/tex]
Given that the second form is exactly provided in the options, the correct answer thus is:
[tex]\[ \lim_{n \to \infty} \frac{2^n - 1}{4(2^n)} \][/tex]
