The problem presented is to evaluate the infinite series:

$\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2 + 1}.$

Solution:

To analyze the convergence and possibly the value of this series, we'll use the comparison test or limit comparison test.

1. Convergence Check:

   • Note that for all $n$, $\sin^2 n \leq 1$. So, we can compare this series with a simpler series:

   $\frac{\sin^2 n}{n^2 + 1} \leq \frac{1}{n^2 + 1}.$

   • The series $\sum \frac{1}{n^2 + 1}$ converges because $\frac{1}{n^2 + 1} \leq \frac{1}{n^2}$, and $\sum \frac{1}{n^2}$ (the p-series with $p = 2 > 1$) converges.

2. Applying the Comparison Test:

   • Since $\frac{\sin^2 n}{n^2 + 1} \leq \frac{1}{n^2 + 1}$ and $\sum \frac{1}{n^2 + 1}$ converges, by the comparison test, the series $\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2 + 1}$ also converges.

Conclusion:

The series $\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2 + 1}$ converges, but it is not simple to find a closed form for the sum. However, convergence is assured by comparison with a convergent p-series.

Would you like a more detailed explanation on any step or further analysis of the series?

Here are 5 related questions you might find interesting:

1. What is the p-series test and how does it determine the convergence of series?
2. How can the limit comparison test be applied to similar trigonometric series?
3. Why does the $\sin^2 n$ function ensure that the series remains bounded?
4. What are some other techniques to evaluate or estimate the sum of infinite series?
5. Can this series be approximated numerically, and what methods are used for such approximations?

Tip: When comparing series, always check if a simpler known series (like p-series) can bound the original series to establish convergence.