To find the value of the series \( A = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{99 \cdot 100} \), we can use a telescoping series approach.

First, let's rewrite the general term \(\frac{1}{n(n+1)}\) in a form that makes it easier to sum. We can use partial fraction decomposition:

\[
\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}
\]

Multiplying both sides by \(n(n+1)\) to clear the denominators:

\[
1 = A(n+1) + Bn
\]

Expanding and combining like terms:

\[
1 = An + A + Bn
\]
\[
1 = (A + B)n + A
\]

For this equation to hold for all \(n\), the coefficients of \(n\) and the constant term must match on both sides. Therefore, we have:

\[
A + B = 0
\]
\[
A = 1
\]

Solving these equations, we get \(A = 1\) and \(B = -1\). Thus,

\[
\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
\]

Now, we can rewrite the series \(A\) using this decomposition:

\[
A = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{99} - \frac{1}{100} \right)
\]

Notice that this is a telescoping series, where most terms cancel out:

\[
A = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{99} - \frac{1}{100} \right)
\]

After cancellation, we are left with:

\[
A = 1 - \frac{1}{100}
\]

So,

\[
A = 1 - 0.01 = 0.99
\]

Therefore, the value of the series \( A \) is \( \boxed{0.99} \).