To solve the series \( \frac{1}{-2} \cdot \frac{1}{3} + \frac{1}{-3} \cdot \frac{1}{4} + \cdots + \frac{1}{-9} \cdot \frac{1}{10} \), we can simplify each term individually and then sum them up.

Each term in the series is of the form \( \frac{1}{-n} \cdot \frac{1}{n+1} \).

Let's simplify each term:
\[ \frac{1}{-n} \cdot \frac{1}{n+1} = \frac{1}{-n(n+1)} \]

So the series becomes:
\[ \sum_{n=2}^{9} \frac{1}{-n(n+1)} \]

We can rewrite each term as:
\[ \frac{1}{-n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]

Thus, the series becomes:
\[ \sum_{n=2}^{9} \left( \frac{1}{n} - \frac{1}{n+1} \right) \]

This is a telescoping series, where most terms cancel out. Let's write out the first few terms to see the pattern:
\[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{9} - \frac{1}{10} \right) \]

Notice that all intermediate terms cancel out, leaving us with:
\[ \frac{1}{2} - \frac{1}{10} \]

So the sum of the series is:
\[ \frac{1}{2} - \frac{1}{10} = \frac{5}{10} - \frac{1}{10} = \frac{4}{10} = \frac{2}{5} \]

Therefore, the value of the series is \( \frac{2}{5} \).