To determine whether the given geometric series converges or diverges, we start by rewriting the series.

The series can be expressed as:

$\sum_{n=0}^{\infty} 3^n 4^{-n+1} = \sum_{n=0}^{\infty} \frac{3^n}{4^{n-1}} = \sum_{n=0}^{\infty} \frac{3^n}{4^n} \cdot 4 = 4 \sum_{n=0}^{\infty} \left(\frac{3}{4}\right)^n$

This is a geometric series where the first term $a = 4$ and the common ratio $r = \frac{3}{4}$.

Convergence

A geometric series converges if the absolute value of the common ratio $r$ is less than 1:

$|r| < 1$

In this case:

$\left|\frac{3}{4}\right| < 1$

Since this condition holds, the series converges.

Sum of the Series

The sum $S$ of a convergent geometric series can be calculated using the formula:

$S = \frac{a}{1 - r}$

Substituting the values:

$S = \frac{4}{1 - \frac{3}{4}} = \frac{4}{\frac{1}{4}} = 4 \cdot 4 = 16$

Conclusion

The geometric series converges, and its sum is:

$\boxed{16}$

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. What are the conditions for convergence of a geometric series?
2. How does the value of the common ratio affect the sum of the series?
3. Can you provide examples of divergent geometric series?
4. What other series types can be analyzed for convergence?
5. How would the series change if the common ratio was greater than 1?

Tip: Always check the common ratio to quickly assess convergence in geometric series.