To solve the series \( \frac{1}{99} - \frac{1}{99.98} - \frac{1}{98.97} - \frac{1}{97.96} - \ldots - \frac{1}{3.2} - \frac{1}{2.1} \), we can look for a pattern or a telescoping effect.

First, let's rewrite the terms in a more convenient form:
\[ \frac{1}{99} - \frac{1}{99.98} - \frac{1}{98.97} - \frac{1}{97.96} - \ldots - \frac{1}{3.2} - \frac{1}{2.1} \]

Notice that each term in the series can be written as:
\[ \frac{1}{n(n+1)} \]

So the series becomes:
\[ \sum_{n=2}^{99} \frac{1}{n(n+1)} \]

We can decompose each fraction using partial fractions:
\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]

Thus, the series can be rewritten as:
\[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \ldots + \left( \frac{1}{99} - \frac{1}{100} \right) \]

Observe that this is a telescoping series, where most terms cancel out:
\[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \ldots + \left( \frac{1}{99} - \frac{1}{100} \right) \]

After cancellation, we are left with:
\[ \frac{1}{2} - \frac{1}{100} \]

So, the sum of the series is:
\[ \frac{1}{2} - \frac{1}{100} = \frac{50}{100} - \frac{1}{100} = \frac{49}{100} = 0.49 \]

Therefore, the value of the series is:
\[ \boxed{0.49} \]