To find the value of \( A = \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + \frac{1}{99} \), we can start by expressing each term in the series in a more manageable form.

Notice that each denominator can be written as a product of two consecutive integers:
- \(\frac{1}{3} = \frac{1}{1 \cdot 3}\)
- \(\frac{1}{15} = \frac{1}{3 \cdot 5}\)
- \(\frac{1}{35} = \frac{1}{5 \cdot 7}\)
- \(\frac{1}{63} = \frac{1}{7 \cdot 9}\)
- \(\frac{1}{99} = \frac{1}{9 \cdot 11}\)

We can use partial fraction decomposition to simplify each term. For a general term \(\frac{1}{n(n+2)}\), we can write:
\[
\frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}
\]
Multiplying both sides by \(n(n+2)\) gives:
\[
1 = A(n+2) + Bn
\]
\[
1 = An + 2A + Bn
\]
\[
1 = (A + B)n + 2A
\]

By comparing coefficients, we get the system of equations:
\[
A + B = 0
\]
\[
2A = 1
\]

Solving for \(A\) and \(B\):
\[
A = \frac{1}{2}
\]
\[
B = -\frac{1}{2}
\]

Thus,
\[
\frac{1}{n(n+2)} = \frac{1/2}{n} - \frac{1/2}{n+2}
\]

Applying this to each term in the series:
\[
\frac{1}{3} = \frac{1}{1 \cdot 3} = \frac{1/2}{1} - \frac{1/2}{3}
\]
\[
\frac{1}{15} = \frac{1}{3 \cdot 5} = \frac{1/2}{3} - \frac{1/2}{5}
\]
\[
\frac{1}{35} = \frac{1}{5 \cdot 7} = \frac{1/2}{5} - \frac{1/2}{7}
\]
\[
\frac{1}{63} = \frac{1}{7 \cdot 9} = \frac{1/2}{7} - \frac{1/2}{9}
\]
\[
\frac{1}{99} = \frac{1}{9 \cdot 11} = \frac{1/2}{9} - \frac{1/2}{11}
\]

Adding these together:
\[
A = \left( \frac{1/2}{1} - \frac{1/2}{3} \right) + \left( \frac{1/2}{3} - \frac{1/2}{5} \right) + \left( \frac{1/2}{5} - \frac{1/2}{7} \right) + \left( \frac{1/2}{7} - \frac{1/2}{9} \right) + \left( \frac{1/2}{9} - \frac{1/2}{11} \right)
\]

Notice that this forms a telescoping series, where most terms cancel out:
\[
A = \frac{1/2}{1} - \frac{1/2}{11}
\]

Simplifying:
\[
A = \frac{1}{2} \left( 1 - \frac{1}{11} \right)
\]
\[
A = \frac{1}{2} \left( \frac{11 - 1}{11} \right)
\]
\[
A = \frac{1}{2} \cdot \frac{10}{11}
\]
\[
A = \frac{10}{22}
\]
\[
A = \frac{5}{11}
\]

Thus, the value of \( A \) is:
\[
A = \frac{5}{11}
\]