The given image shows the infinite series:

$\sum_{n=1}^{\infty} \frac{\sin^2(2^n)}{n^4}$

This series involves a combination of trigonometric functions and powers of $n$. To analyze this, let's break it down:

1. Behavior of the Numerator:

The term $\sin^2(2^n)$ is bounded between 0 and 1 because for any real number $x$, $\sin^2(x)$ takes values in the range $[0, 1]$. Hence, the numerator does not grow significantly as $n$ increases.

2. Denominator Growth:

The term $n^4$ grows very rapidly as $n$ increases. The factor $n^4$ in the denominator suggests that the terms of the series decay quickly, meaning the series has a good chance of converging.

3. Comparison Test for Convergence:

We can use a comparison test with a simpler series. Consider the series:

$\sum_{n=1}^{\infty} \frac{1}{n^4}$

This is a p-series with $p = 4$, and it is known to converge since $p > 1$. Because $\sin^2(2^n)$ is bounded between 0 and 1, we have:

$0 \leq \frac{\sin^2(2^n)}{n^4} \leq \frac{1}{n^4}$

Thus, by the comparison test, since $\sum_{n=1}^{\infty} \frac{1}{n^4}$ converges, the original series also converges.

Conclusion:

The series $\sum_{n=1}^{\infty} \frac{\sin^2(2^n)}{n^4}$ converges.

Would you like more details on how to compute the exact sum, or any specific aspects of this problem?

Here are some questions you might explore further:

1. How does the rapid growth of $n^4$ ensure convergence in series?
2. Can we approximate the sum of the series numerically?
3. What if the power in the denominator was $n^2$ instead of $n^4$?
4. How do different types of tests help in determining convergence of series?
5. What are other examples of convergent series involving trigonometric functions?

Tip: When testing for convergence, comparison tests are very useful when you can bound the terms of your series with a simpler, known series.