Let's first understand the series given:

$\sum_{n=3}^{\infty} \frac{4^{3n+2}}{4^{3n-1}}$

This is an infinite series starting from $n = 3$. Let's break it down step by step:

1. Simplify the general term: $\frac{4^{3n+2}}{4^{3n-1}} = 4^{(3n+2) - (3n-1)} = 4^{3n+2 - 3n+1} = 4^3 = 64$ So each term of the series is 64.

2. The series is now: $\sum_{n=3}^{\infty} 64$

3. This is a constant series where every term is 64, starting from $n = 3$ to infinity.

   Since the series consists of an infinite number of terms, and each term is a constant 64, this sum diverges to infinity. In other words, it does not converge to a finite number. The sum of an infinite series with constant terms that are non-zero will tend to infinity.

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Questions to consider:

1. Why does an infinite series with constant terms diverge?
2. How can we determine if a series converges or diverges?
3. What happens if we replace the constant terms with a function that decays as $n$ increases?
4. Can this series be transformed into a convergent one by modifying the powers of 4?
5. What are some examples of convergent infinite series?

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Tip:

To check if a series converges, you can apply the n-th term test for divergence. If the terms of the series do not approach zero as $n$ increases, the series will diverge.