The series given is:

$\sum_{n=0}^{\infty} \frac{7^n}{8^n \cdot n!}$

This expression can be simplified by rewriting the ratio $\frac{7^n}{8^n}$ as $\left(\frac{7}{8}\right)^n$, resulting in:

$\sum_{n=0}^{\infty} \frac{\left(\frac{7}{8}\right)^n}{n!}$

This series resembles the Taylor series expansion for $e^x$, which is:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

In our case, $x = \frac{7}{8}$. Therefore, we can rewrite the sum as:

$\sum_{n=0}^{\infty} \frac{\left(\frac{7}{8}\right)^n}{n!} = e^{\frac{7}{8}}$

Thus, the sum of the series is:

$e^{\frac{7}{8}}$

Would you like more details on any step, or do you have questions?

Here are some additional questions to extend your understanding:

1. What is the Taylor series expansion of $e^x$ and why is it useful in solving infinite series?
2. How does the ratio $\frac{7}{8}$ affect the convergence of the series?
3. Can we apply this method to other exponential series, and what would change?
4. How is this series related to the probability distribution functions?
5. What other types of functions have similar series representations?

Tip: Recognizing series forms like $e^x$ or $\sin(x)$ in problems can simplify calculations significantly.