Let's analyze the given infinite series and determine if its sum approaches 2:
[tex]\[
1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \cdots
\][/tex]
This is a geometric series where each term after the first is obtained by multiplying the preceding term by a common ratio. In this case, the first term [tex]\( a \)[/tex] is 1, and the common ratio [tex]\( r \)[/tex] is [tex]\(\frac{1}{2}\)[/tex].
To find the sum of an infinite geometric series, we use the formula:
[tex]\[
S = \frac{a}{1 - r}
\][/tex]
where
- [tex]\( S \)[/tex] is the sum of the series,
- [tex]\( a \)[/tex] is the first term,
- [tex]\( r \)[/tex] is the common ratio, and the series will converge if [tex]\( |r| < 1 \)[/tex].
Here, [tex]\( a = 1 \)[/tex] and [tex]\( r = \frac{1}{2} \)[/tex].
Let's plug these values into the formula:
[tex]\[
S = \frac{1}{1 - \frac{1}{2}}
\][/tex]
Simplify the denominator:
[tex]\[
1 - \frac{1}{2} = \frac{1}{2}
\][/tex]
Thus, the sum of the series is:
[tex]\[
S = \frac{1}{\frac{1}{2}} = 2
\][/tex]
So, the sum of the infinite series [tex]\( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \)[/tex] indeed approaches 2.
