To solve the problem (assuming it's the classic telescoping series sum), let's proceed step by step:

Problem Context

We need to compute the sum of the series:
$$S = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{n \times (n+1)}$$

Key Insight: Partial Fraction Decomposition

Each term in the series can be split into two parts:
$$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$

Telescoping Sum

Substitute this into the series:
$$S = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$

All intermediate terms cancel out (telescope):
$$S = 1 - \frac{1}{n+1} = \frac{n}{n+1}$$

Example Case

If the series ends at $\frac{1}{99 \times 100}$ (n=99), then:
$$S = \frac{99}{100} = 0.99$$

Answer: $\boxed{\dfrac{99}{100}}$ (or $\boxed{0.99}$ if decimal is preferred, but fraction is more precise).

Assuming the problem asked for the sum up to 99 terms, the answer is $\boxed{99/100}$.

$\boxed{99/100}$

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