To solve this series quickly, let's first understand the pattern of the series:

\[ S = \frac{1}{5} + \frac{1}{10} + \frac{1}{20} + \frac{1}{40} + \ldots + \frac{1}{1280} \]

You might notice that each denominator is doubling. It can be expressed as:

\[ S = \frac{1}{5} + \frac{1}{5 \times 2} + \frac{1}{5 \times 2^2} + \frac{1}{5 \times 2^3} + \ldots + \frac{1}{5 \times 2^n} \]

where \( 5 \times 2^n = 1280 \).

First, solve for \( n \) in \( 5 \times 2^n = 1280 \):
\[ 2^n = \frac{1280}{5} = 256 \]

Since \( 256 = 2^8 \), we have \( n = 8 \).

Thus, we are summing a geometric series with the first term \( a = \frac{1}{5} \) and common ratio \( r = \frac{1}{2} \). The number of terms is \( n + 1 = 9 \).

The sum of a geometric series is given by the formula:
\[ S = a \frac{1 - r^{n+1}}{1 - r} \]

Plugging in the values, we get:
\[ S = \frac{1}{5} \cdot \frac{1 - \left(\frac{1}{2}\right)^9}{1 - \frac{1}{2}} \]

\[ S = \frac{1}{5} \cdot \frac{1 - \frac{1}{512}}{\frac{1}{2}} \]

\[ S = \frac{1}{5} \cdot 2 \left(1 - \frac{1}{512}\right) \]

\[ S = \frac{2}{5} \left(1 - \frac{1}{512}\right) \]

\[ S = \frac{2}{5} \cdot \frac{511}{512} \]

\[ S = \frac{1022}{2560} \]

Divide the numerator and the denominator by 2 to simplify:

\[ S = \frac{511}{1280} \]

Therefore, the sum of the series is \(\frac{511}{1280}\).