The problem gives the general term of a sequence $a_n = \frac{1}{n(n+1)}$, and asks us to find the sum of the first 8 terms, $S_8$.

1. Problem Description

The problem gives the general term of a sequence $a_n = \frac{1}{n(n+1)}$, and asks us to find the sum of the first 8 terms, $S_8$.

2. Solution Steps

We are given the general term $a_n = \frac{1}{n(n+1)}$. We can decompose this fraction into partial fractions:

$$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$$

Multiplying both sides by $n(n+1)$, we get:

$$1 = A(n+1) + Bn$$

If $n = 0$, then $1 = A(0+1) + B(0) \Rightarrow A = 1$.

If $n = -1$, then $1 = A(-1+1) + B(-1) \Rightarrow 1 = -B \Rightarrow B = -1$.

So,

$$a_n = \frac{1}{n} - \frac{1}{n+1}$$

Now, we need to find the sum of the first 8 terms, $S_8$:

$$S_8 = \sum_{n=1}^{8} a_n = \sum_{n=1}^{8} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$

$$S_8 = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{8} - \frac{1}{9}\right)$$

This is a telescoping sum, so all the intermediate terms cancel out:

$$S_8 = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}$$

3. Final Answer

The sum of the first 8 terms is $\frac{8}{9}$.

So the answer is D.