To calculate the limit of the sequence $a_n = \left( 2^{2n+1} + 3^n \right)^{\frac{1}{n}}$ as $n \to \infty$, we need to analyze the behavior of the sequence for large $n$.

Step 1: Break down the expression

The given sequence is:

$a_n = \left( 2^{2n+1} + 3^n \right)^{\frac{1}{n}}.$

We can simplify this by rewriting the expression as:

$a_n = \left( 2^{2n} \cdot 2 + 3^n \right)^{\frac{1}{n}}.$

This shows that the two terms inside the parentheses are $2^{2n+1}$ (which grows very rapidly) and $3^n$ (which grows less rapidly compared to the first term as $n$ increases).

Step 2: Asymptotic dominance of terms

As $n \to \infty$, the term $2^{2n+1} = 2 \cdot 4^n$ grows much faster than $3^n$, so for large $n$, $2^{2n+1}$ will dominate. Therefore, we can approximate the sequence as:

$a_n \approx \left( 2^{2n+1} \right)^{\frac{1}{n}}.$

Now, simplify the expression:

$a_n \approx \left( 2 \cdot 4^n \right)^{\frac{1}{n}} = 2^{\frac{1}{n}} \cdot 4.$

Step 3: Take the limit

As $n \to \infty$, $2^{\frac{1}{n}}$ approaches 1 because $\frac{1}{n}$ tends to 0. Thus, the limit of $a_n$ becomes:

$\lim_{n \to \infty} a_n = 1 \cdot 4 = 4.$

Final Answer:

The limit of the sequence $a_n = \left( 2^{2n+1} + 3^n \right)^{\frac{1}{n}}$ as $n \to \infty$ is:

$\boxed{4}.$

---

Would you like more details on any step? Here are some related questions you could explore:

1. How do we formally apply the dominance of terms in sequences and limits?
2. What is the behavior of the sequence when both terms inside the parentheses grow at similar rates?
3. How do you find the limit of a sequence involving mixed powers like $2^{2n+1}$ and $3^n$?
4. What is the general method for handling limits of sequences of the form $\left( a_n + b_n \right)^{\frac{1}{n}}$?
5. How can you use logarithms to solve sequences involving exponential terms?

Tip: When dealing with limits of sequences that involve exponential functions, look for the term with the highest growth rate. It will usually dominate the behavior of the sequence as $n \to \infty$.