To find the value of the series \( A = \frac{2}{1 \cdot 3} - \frac{4}{3 \cdot 5} + \frac{6}{5 \cdot 7} - \cdots - \frac{20}{19 \cdot 21} \), let's first look at the general term of the series.

The general term of the series can be written as:
\[ \frac{2n}{(2n-1)(2n+1)} \]

We can simplify this term using partial fractions. Let's decompose:
\[ \frac{2n}{(2n-1)(2n+1)} = \frac{A}{2n-1} + \frac{B}{2n+1} \]

To find \( A \) and \( B \), we solve the equation:
\[ 2n = A(2n+1) + B(2n-1) \]

Expanding and equating coefficients, we get:
\[ 2n = (2A + 2B)n + (A - B) \]

This gives us two equations:
1. \( 2A + 2B = 2 \)
2. \( A - B = 0 \)

From the second equation, we get \( A = B \). Substituting \( A = B \) into the first equation, we get:
\[ 2A + 2A = 2 \]
\[ 4A = 2 \]
\[ A = \frac{1}{2} \]

So, \( A = \frac{1}{2} \) and \( B = \frac{1}{2} \). Therefore, the partial fraction decomposition is:
\[ \frac{2n}{(2n-1)(2n+1)} = \frac{1}{2(2n-1)} + \frac{1}{2(2n+1)} \]

Thus, the general term can be written as:
\[ \frac{2n}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1}{2n-1} + \frac{1}{2n+1} \right) \]

Now, let's write the series using this decomposition:
\[ A = \sum_{n=1}^{10} (-1)^{n+1} \frac{2n}{(2n-1)(2n+1)} \]
\[ = \sum_{n=1}^{10} (-1)^{n+1} \frac{1}{2} \left( \frac{1}{2n-1} + \frac{1}{2n+1} \right) \]

Let's separate the series into two parts:
\[ A = \frac{1}{2} \sum_{n=1}^{10} (-1)^{n+1} \left( \frac{1}{2n-1} + \frac{1}{2n+1} \right) \]

This can be written as:
\[ A = \frac{1}{2} \left( \sum_{n=1}^{10} (-1)^{n+1} \frac{1}{2n-1} + \sum_{n=1}^{10} (-1)^{n+1} \frac{1}{2n+1} \right) \]

Notice that the terms in each sum will cancel out in a telescoping manner. Let's look at the first sum:
\[ \sum_{n=1}^{10} (-1)^{n+1} \frac{1}{2n-1} \]

This series alternates and cancels out most terms. Similarly, the second sum:
\[ \sum_{n=1}^{10} (-1)^{n+1} \frac{1}{2n+1} \]

also alternates and cancels out most terms. The remaining terms after cancellation will be the first and last terms of each series.

Combining these results, we get:
\[ A = \frac{1}{2} \left( 1 - \frac{1}{21} \right) \]
\[ A = \frac{1}{2} \left( \frac{21-1}{21} \right) \]
\[ A = \frac{1}{2} \left( \frac{20}{21} \right) \]
\[ A = \frac{10}{21} \]

So, the value of the series \( A \) is:
\[ A = \frac{10}{21} \]