To find the sum of the series \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} + \frac{1}{256} + \frac{1}{512} + \frac{1}{1024} \), we recognize that this is a geometric series.

A geometric series has the form:
\[ a + ar + ar^2 + ar^3 + \ldots + ar^{n-1} \]

In this series:
- The first term \( a \) is \( \frac{1}{2} \).
- The common ratio \( r \) is \( \frac{1}{2} \).

The sum \( S_n \) of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{1 - r^n}{1 - r} \]

For this series, \( a = \frac{1}{2} \), \( r = \frac{1}{2} \), and \( n = 10 \).

Plugging in these values, we get:
\[ S_{10} = \frac{1}{2} \frac{1 - \left(\frac{1}{2}\right)^{10}}{1 - \frac{1}{2}} \]

Simplify the denominator:
\[ 1 - \frac{1}{2} = \frac{1}{2} \]

So,
\[ S_{10} = \frac{1}{2} \frac{1 - \left(\frac{1}{2}\right)^{10}}{\frac{1}{2}} \]

The \( \frac{1}{2} \) terms cancel out:
\[ S_{10} = 1 - \left(\frac{1}{2}\right)^{10} \]

Calculate \( \left(\frac{1}{2}\right)^{10} \):
\[ \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \]

Therefore,
\[ S_{10} = 1 - \frac{1}{1024} \]

Convert \( 1 \) to a fraction with the same denominator:
\[ 1 = \frac{1024}{1024} \]

So,
\[ S_{10} = \frac{1024}{1024} - \frac{1}{1024} = \frac{1023}{1024} \]

Thus, the sum of the series is:
\[ \frac{1023}{1024} \]