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A positive-term series is a mathematical summation where all terms are positive numbers. When dealing with series, convergence is a common question. For a series with terms denoted as \( \sum a_n \), convergence means that as you add more and more terms from the series, you approach a certain value, rather than continue indefinitely. In simpler terms, think of convergence like pouring water into a cup. As you pour, there comes a point when you can't add more without the water spilling over -- this is where the 'limit' is. However, in a convergent series, you're able to keep pouring, but despite an infinite amount of terms, the sum finds its "limit" and does not overflow.
It's important to highlight that for a positive-term series to converge, the sequence of its terms, \((a_n)\), must approach zero as \( n \), which represents the number of terms, becomes very large. This means that eventually, the additional water doesn't significantly add to the total, so to speak. Positive-term series are often straightforward to analyze because their non-negative nature makes certain tests and properties more manageable, as in the exercise at hand.

The sine function, denoted as \( \sin(x) \), is a well-known trigonometric function that is periodic and oscillates between -1 and 1. In geometry, it relates to the ratio of the length of the opposite side to the hypotenuse of a right-angled triangle. However, in calculus and real analysis, it's often studied in terms of its behavior for small values. For small angles -- those approaching zero -- the sine of an angle is approximately equal to the angle itself when measured in radians. This leads to the approximation \( \sin(x) \approx x \) for small values of \( x \).
This property is especially helpful when evaluating the convergence of series where the sine function appears. In the given exercise, since \( a_n \to 0 \), applying the notion \( \sin(a_n) \approx a_n \) assists in evaluating the behavior of the series \( \sum \sin(a_n) \).
When \( a_n \) is small, \( \sin(a_n) \) doesn't deviate much from \( a_n \), allowing us to use this approximation effectively. This contributes to determining whether the series formed by the sine of terms of a convergent series also converges.

The comparison test is a vital tool in calculus to determine the convergence of a series. It provides a method to compare a series whose convergence we want to establish with another series whose convergence behavior is already known. Here's how it works:

• If \( \sum b_n \) is a convergent series and \( 0 \leq a_n \leq b_n \) for all \( n \), then the series \( \sum a_n \) also converges.
• Conversely, if \( \sum b_n \) diverges and \( a_n \geq b_n \geq 0 \) for all \( n \), then \( \sum a_n \) also diverges.

In the context of our exercise, since all terms \( a_n \) are positive and \( \sum a_n \) converges, we also know that \( \sin(a_n) \leq a_n \). This means that each term of the series \( \sum \sin(a_n) \) is smaller than the corresponding term in the convergent series \( \sum a_n \).
Utilizing the comparison test, it follows that \( \sum \sin(a_n) \) must also converge. The test works effectively here because the bounded nature of \( \sin(a_n) \) ensures that the greater series' (\( \sum a_n \)) convergence properties envelop the sine series, proving its convergence as well.