Thank you for the clarification. Let's solve the problem with the correct expression for $a_n$, which is given as:

$a_n = \frac{1}{2^{2n}}$

Step 1: Analyze the Sequence $a_n$

Given $a_n = \frac{1}{2^{2n}}$, we can rewrite this as:

$a_n = \frac{1}{(2^2)^n} = \frac{1}{4^n}$

Step 2: Analyze the Sum $S_n$

The sum $S_n$ is defined as the sum of the first $n$ terms of the sequence:

$S_n = \sum_{k=0}^{n-1} a_k = \sum_{k=0}^{n-1} \frac{1}{4^k}$

Step 3: Recognize the Sum as a Geometric Series

This sum $S_n$ is a geometric series with the first term $a_0 = 1$ (since $4^0 = 1$) and the common ratio $r = \frac{1}{4}$. The sum of the first $n$ terms of a geometric series is given by:

$S_n = \frac{1-r^n}{1-r}$

Substituting $r = \frac{1}{4}$:

$S_n = \frac{1 - \left(\frac{1}{4}\right)^n}{1 - \frac{1}{4}} = \frac{1 - \frac{1}{4^n}}{\frac{3}{4}} = \frac{4}{3} \left(1 - \frac{1}{4^n}\right)$

Step 4: Take the Limit as $n \to \infty$

Now, take the limit of $S_n$ as $n \to \infty$:

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{4}{3} \left(1 - \frac{1}{4^n}\right)$

As $n \to \infty$, $\frac{1}{4^n}$ approaches 0. Therefore:

$\lim_{n \to \infty} S_n = \frac{4}{3} \times 1 = \frac{4}{3}$

Final Answer

The limit of $S_n$ as $n \to \infty$ is $\boxed{\frac{4}{3}}$.

Would you like further details or have any questions? Here are some related questions that might interest you:

1. What are the conditions for the convergence of a geometric series?
2. How would the result change if the common ratio were greater than 1?
3. Can you find the sum of an infinite geometric series when $r < 1$?
4. How does the initial term affect the sum of a geometric series?
5. What would happen if $a_n$ was defined as $\frac{1}{2^{n^2}}$ instead?

Tip: For any geometric series, always verify the common ratio to determine convergence before summing the series.