To show that $a_n < \frac{1}{n} \quad \forall \quad n = 1, 2, 3, \ldots$, we will proceed with the following steps:

1. Boundedness and Initial Inequality:
Given that $\{a_n\}_{n \geq 1}$ is a bounded sequence, there exists a constant $D$ such that $a_n < D$ for all $n$. The given inequality is:
$$a_n < \sum_{k=n}^{2n+2006} \frac{a_k}{k+1} + \frac{1}{2n+2007}.$$

2. Applying the Boundedness:
Using the boundedness of $a_n$, we can write:
$$a_n < \frac{1}{2n+2007} + \sum_{k=n}^{2n+2006} \frac{D}{k+1}.$$

3. Estimating the Sum:
The sum $\sum_{k=n}^{2n+2006} \frac{1}{k+1}$ can be approximated using the integral test, which gives us:
$$\sum_{k=n}^{2n+2006} \frac{1}{k+1} \approx \ln\left(\frac{2n+2007}{n}\right).$$
Since $\ln\left(\frac{2n+2007}{n}\right) = \ln(2 + \frac{2007}{n})$, for large $n$, this can be approximated by $\ln 2.5$.

4. Combining the Estimates:
Therefore, we have:
$$a_n < \frac{1}{2n+2007} + D \cdot \ln 2.5.$$

5. Iterative Process:
We now consider the iterative process. Suppose $a_n < \frac{C}{n}$ for some constant $C$. Then:
$$a_n < \frac{1}{2n+2007} + \sum_{k=n}^{2n+2006} \frac{C}{k+1}.$$
Using the same approximation for the sum, we get:
$$a_n < \frac{1}{2n+2007} + C \cdot \ln 2.5.$$

6. **Refining the Constant $C$:**
We need to show that $C$ can be reduced to 1. Assume $C = 1 + \epsilon$ for some small $\epsilon > 0$. Then:
$$a_n < \frac{1}{2n+2007} + (1 + \epsilon) \cdot \ln 2.5.$$
For large $n$, $\frac{1}{2n+2007}$ becomes negligible, and we can iterate this process to show that $C$ approaches 1.

7. Final Step:
By iterating the inequality and reducing $C$ step by step, we can show that for any $\epsilon > 0$, there exists an $N$ such that for all $n > N$, $a_n < \frac{1 + \epsilon}{n}$. Taking the limit as $\epsilon \to 0$, we get:
$$a_n < \frac{1}{n}.$$

Thus, we have shown that $a_n < \frac{1}{n}$ for all $n$.

$\blacksquare$