Let's explore the sum of the given series step-by-step.
The series provided is:
[tex]\[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \ldots \][/tex]
This series is an example of an infinite geometric series. In a geometric series, each term after the first is found by multiplying the previous term by a constant called the common ratio (r).
For this series:
- The first term (a) is 1.
- The common ratio (r) is [tex]\(\frac{1}{2}\)[/tex].
The sum [tex]\(S\)[/tex] of an infinite geometric series can be found using the formula:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Substituting the values from the given series:
[tex]\[ a = 1 \][/tex]
[tex]\[ r = \frac{1}{2} \][/tex]
Now, let's plug these values into the formula:
[tex]\[ S = \frac{1}{1 - \frac{1}{2}} \][/tex]
[tex]\[ S = \frac{1}{\frac{1}{2}} \][/tex]
[tex]\[ S = 2 \][/tex]
Therefore, the sum of the infinitely continuing series:
[tex]\[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \ldots \][/tex]
will approach [tex]\(2\)[/tex] as the number of terms goes to infinity.
So, the sum of this series reaches 2 when it is extended to infinity.
