The given series is a geometric series where the first term \( a = \frac{1}{2} \) and the common ratio \( r = \frac{1}{2} \). The series is:

\[ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{128} \]

To find the sum of this finite geometric series, we use the formula for the sum of the first \( n \) terms of a geometric series:

\[ S_n = a \frac{1-r^n}{1-r} \]

Here, \( a = \frac{1}{2} \), \( r = \frac{1}{2} \), and we need to determine \( n \). The last term given is \( \frac{1}{128} \), which can be written as \( \left(\frac{1}{2}\right)^7 \). Therefore, \( n = 7 \).

Now, we can plug these values into the formula:

\[ S_7 = \frac{1}{2} \frac{1 - \left(\frac{1}{2}\right)^7}{1 - \frac{1}{2}} \]

Simplify the denominator:

\[ 1 - \frac{1}{2} = \frac{1}{2} \]

So the formula becomes:

\[ S_7 = \frac{1}{2} \frac{1 - \frac{1}{128}}{\frac{1}{2}} \]

The \(\frac{1}{2}\) terms cancel out:

\[ S_7 = 1 - \frac{1}{128} \]

Convert \(\frac{1}{128}\) to a decimal:

\[ \frac{1}{128} = 0.0078125 \]

So:

\[ S_7 = 1 - 0.0078125 = 0.9921875 \]

Therefore, the sum of the series \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots + \frac{1}{128} \) is \( 0.9921875 \).