Question

Show that if \(a_{n} \geq 0\) and \(\sum_{n=1}^{\infty} a^{2}{ }_{n}\) converges, then \(\sum_{n=1}^{\infty} \sin ^{2}\left(a_{n}\right)\) converges.

Short answer

The series \(\sum_{n=1}^{\infty} \sin^2(a_n)\) converges by comparison test.

Step-by-step solution

Analyze Given Series

We are provided with a series \(\sum_{n=1}^{\infty} a_n^2\) which converges. This implies that the sequence \(a_n^2\) tends to zero as \(n\) tends to infinity and the terms are summable.

Consider the Behavior of \(\sin^2(a_n)\)

Recall that \(\sin^2(a_n)\) can be bounded: \(\sin^2(a_n) \leq a_n^2\) for small positive \(a_n\). This is because \(\sin(x) \approx x\) when \(x\) is near zero.

Use Comparison Test

Since \(\sin^2(a_n) \leq a_n^2\) for all \(n\) and \(\sum_{n=1}^{\infty} a_n^2\) converges, we can use the comparison test. The comparison test states that if \(0 \leq b_n \leq a_n\) and \(\sum a_n\) converges, then \(\sum b_n\) also converges.

Apply Final Conclusion

Using the comparison test, we conclude that since \(0 \leq \sin^2(a_n) \leq a_n^2\) and the series \(\sum_{n=1}^{\infty} a_n^2\) converges, it follows that \(\sum_{n=1}^{\infty} \sin^2(a_n)\) also converges.

Key concepts

Comparison Test

The Comparison Test is a powerful tool used in determining the convergence of a series. It helps us compare a complex series with another series that we already know converges. Imagine you have two series, \( \sum a_n \) and \( \sum b_n \). According to the Comparison Test, if:

• every term of the second series \( b_n \) is less than or equal to a corresponding term in the first series \( a_n \), and
• the series \( \sum a_n \) is known to converge,

you can confidently say that \( \sum b_n \) also converges.
In our exercise, we are given that the series \( \sum_{n=1}^{\infty} a_n^2 \) converges. By bounding \( \sin^2(a_n) \) such that \( \sin^2(a_n) \leq a_n^2 \), and knowing that \( a_n^2 \) converges, we use the Comparison Test to conclude the convergence of \( \sum_{n=1}^{\infty} \sin^2(a_n) \). It’s like having a friend guide you based on their experience with a similar situation.

Sequence Behavior

Understanding the behavior of a sequence can be crucial when determining series convergence. When we say a sequence "behaves well", we often mean that the sequence approaches zero or a finite number as \( n \) increases indefinitely.
In this problem, we observe how \( a_n^2 \) tends to zero, since the series \( \sum_{n=1}^{\infty} a_n^2 \) converges. This behavior provides us significant insights because if the terms of \( a_n^2 \) become negligible as \( n \) grows, any function of smaller terms—like \( \sin^2(a_n) \)—will also share this diminishing trait.
The trick is to leverage this behavior for the function \( \sin^2(x) \) which approximates to \( x^2 \) when \( x \) is very small. Therefore, recognizing this behavior helps predict how \( \sin^2(a_n) \) aligns closely to \( a_n^2 \) when \( a_n \) is near zero, simplifying the comparison of their convergence.

Bounding Functions

Bounding functions is a technique used to compare a function with another simpler function that bounds or restricts it, particularly in terms of size and growth. When dealing with convergence, bounding is useful to approximate and make fair assessments of how one function compares to another.
In the context of our exercise, we use the bounding relation \( \sin^2(a_n) \leq a_n^2 \). We chose this bounding technique because when \( a_n \) is small, \( \sin(a_n) \) approximates to \( a_n \) itself. The sine function doesn’t exceed its input when the value is near zero—thus, \( \sin^2(a_n) \) is naturally bounded by \( a_n^2 \).
By ensuring \( \sin^2(a_n) \) is always smaller or equal to \( a_n^2 \), and knowing that \( \sum_{n=1}^{\infty} a_n^2 \) converges, we find confidence to claim the convergence of the series \( \sum_{n=1}^{\infty} \sin^2(a_n) \) through using these bounds effectively.