Rút gọn A

Let's simplify the expression \( A \) given:

\[
A = \frac{1}{1 \cdot 2} + \frac{2}{1 \cdot 2 \cdot 3} + \frac{3}{1 \cdot 2 \cdot 3 \cdot 4} + \ldots + \frac{19}{1 \cdot 2 \cdot 3 \cdot \ldots \cdot 20}
\]

Each term in the series can be rewritten as:

\[
\frac{n}{1 \cdot 2 \cdot 3 \cdots n \cdots (n+1)} = \frac{1}{(n-1)!} \cdot \frac{n}{n!} = \frac{1}{(n-1)!} \cdot \frac{1}{(n-1)!} = \frac{1}{(n-1)! \cdot n!}
\]

The sum becomes:

\[
A = \sum_{n=1}^{19} \frac{1}{(n-1)! \cdot n!}
\]

Using the identity \( \frac{1}{(n-1)! \cdot n!} = \frac{1}{(n)!} \cdot \frac{1}{n} \), we can identify that:

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

Therefore, we can write:

\[
A = \sum_{n=1}^{19} \left( \frac{1}{(n-1)!} - \frac{1}{n!} \right)
\]

This series telescopes, leading to:

\[
A = 1 - \frac{1}{20!}
\]

Thus, the simplified result is:

\[
A = 1 - \frac{1}{20!}
\]