The problem provided requires the use of the Direct Comparison Test to determine the convergence or divergence of the given series:

$\sum_{n=0}^{\infty} \frac{8^n}{9^n + 7}.$

Step-by-step solution:

1. Expression Simplification: The general term of the series is:

   $a_n = \frac{8^n}{9^n + 7}.$

   For large $n$, the $9^n$ term dominates over the constant $7$ in the denominator. So, for large $n$, we can approximate the general term as:

   $a_n \approx \frac{8^n}{9^n} = \left(\frac{8}{9}\right)^n.$

2. Comparison Series: The term $\left(\frac{8}{9}\right)^n$ is a geometric series with common ratio $r = \frac{8}{9}$. The sum of a geometric series $\sum_{n=0}^{\infty} r^n$ converges if $|r| < 1$. In this case, since $\frac{8}{9} < 1$, the geometric series:

   $\sum_{n=0}^{\infty} \left(\frac{8}{9}\right)^n$

   converges.

3. Direct Comparison: Now we use the Direct Comparison Test. For large $n$, we know:

   $\frac{8^n}{9^n + 7} < \frac{8^n}{9^n} = \left(\frac{8}{9}\right)^n.$

   Since $\sum \left(\frac{8}{9}\right)^n$ converges and $\frac{8^n}{9^n + 7} \leq \left(\frac{8}{9}\right)^n$ for all $n \geq 0$, by the Direct Comparison Test, the given series also converges.

Final Answer:

The series $\sum_{n=0}^{\infty} \frac{8^n}{9^n + 7}$ converges.

Would you like more details on any of the steps? Here are some related questions:

1. How do you apply the Direct Comparison Test in other types of series?
2. What is the difference between the Direct Comparison Test and the Limit Comparison Test?
3. Why does the geometric series with $r = \frac{8}{9}$ converge?
4. Can you explain how to identify dominant terms in a series?
5. What are some other tests for determining the convergence of series?

Tip: When using the Direct Comparison Test, always choose a comparison series that behaves similarly to the given series for large $n$.