To find the sum of the series \( \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \cdots + \frac{1}{18 \cdot 19 \cdot 20} \), we can start by expressing the general term of the series.

The general term of the series is:
\[ \frac{1}{n(n+1)(n+2)} \]

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

To find the constants \( A \), \( B \), and \( C \), we multiply both sides by \( n(n+1)(n+2) \):
\[ 1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1) \]

Expanding and combining like terms, we get:
\[ 1 = A(n^2 + 3n + 2) + B(n^2 + 2n) + C(n^2 + n) \]
\[ 1 = (A + B + C)n^2 + (3A + 2B + C)n + 2A \]

For this equation to hold for all \( n \), the coefficients of \( n^2 \), \( n \), and the constant term must match on both sides. Therefore, we have the system of equations:
\[ A + B + C = 0 \]
\[ 3A + 2B + C = 0 \]
\[ 2A = 1 \]

Solving these equations, we first find \( A \):
\[ 2A = 1 \]
\[ A = \frac{1}{2} \]

Substitute \( A = \frac{1}{2} \) into the first two equations:
\[ \frac{1}{2} + B + C = 0 \]
\[ 3 \cdot \frac{1}{2} + 2B + C = 0 \]

Simplify these equations:
\[ \frac{1}{2} + B + C = 0 \]
\[ \frac{3}{2} + 2B + C = 0 \]

Subtract the first equation from the second:
\[ \left( \frac{3}{2} + 2B + C \right) - \left( \frac{1}{2} + B + C \right) = 0 \]
\[ \frac{3}{2} + 2B + C - \frac{1}{2} - B - C = 0 \]
\[ \frac{2}{2} + B = 0 \]
\[ B = -1 \]

Now substitute \( B = -1 \) back into the first equation:
\[ \frac{1}{2} - 1 + C = 0 \]
\[ -\frac{1}{2} + C = 0 \]
\[ C = \frac{1}{2} \]

So, the partial fractions decomposition is:
\[ \frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} \]

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

Now, we sum the series from \( n = 1 \) to \( n = 18 \):
\[ \sum_{n=1}^{18} \left( \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)} \right) \]

This series will telescope. Let's write out the first few terms to see the pattern:
\[ \left( \frac{1}{2 \cdot 1} - \frac{1}{2} + \frac{1}{2 \cdot 3} \right) + \left( \frac{1}{2 \cdot 2} - \frac{1}{3} + \frac{1}{2 \cdot 4} \right) + \left( \frac{1}{2 \cdot 3} - \frac{1}{4} + \frac{1}{2 \cdot 5} \right) + \cdots + \left( \frac{1}{2 \cdot 18} - \frac{1}{19} + \frac{1}{2 \cdot 20} \right) \]

Notice that many terms will cancel out. Specifically, the terms involving \( \frac{1}{n+1} \) and \( \frac{1}{2(n+2)} \) will cancel with terms from previous and subsequent fractions.

After cancellation, we are left with:
\[ \frac{1}{2 \cdot 1} + \frac{1}{2 \cdot 2} - \frac{1}{19} + \frac{1}{2 \cdot 20} \]

Simplify these remaining terms:
\[ \frac{1}{2} + \frac{1}{4} - \frac{1}{19} + \frac{1}{40} \]

Combine these fractions:
\[ \frac{20}{40} + \frac{10}{40} - \frac{2.105}{40} + \frac{1}{40} \]
\[ \frac{20 + 10 - 2.105 + 1}{40} \]
\[ \frac{28.895}{40} \]

Thus, the sum of the series is:
\[ \boxed{\frac{28.895}{40}} \]