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The Squeeze Theorem is a helpful mathematical tool to determine the convergence of sequences. It asserts that if a sequence is "squeezed" between two other sequences that both converge to the same limit, then the sequence in question also converges to that limit. In simpler terms:

• If you have three sequences, say \( a_n, b_n, \) and \( c_n \), and for all large enough \( n \), you know that \( a_n \leq b_n \leq c_n \).
• If both the sequences \( a_n \) and \( c_n \) converge to the same number \( L \), then \( b_n \) will also converge to \( L \).

In the original exercise, the Squeeze Theorem was applied to the sequence \( 2^{-n} \sin n \). Since \( \sin n \) is bound between -1 and 1, the expression \( 2^{-n} \sin n \) is naturally squeezed between \(-2^{-n}\) and \(2^{-n}\). Both these bounding sequences converge to zero as \( n \to \infty \), forcing \( 2^{-n} \sin n \) to also converge to zero. The Squeeze Theorem elegantly concludes that this sequence must tend to zero for large values of \( n \).

A geometric sequence is a type of sequence where each term is obtained by multiplying the previous term by a constant number, known as the common ratio. If you start with an initial term \( a_1 \) and multiply repeatedly by the ratio \( r \), the sequence can be described as:

• \( a_1, a_1r, a_1r^2, a_1r^3, \ldots \)

Geometric sequences exhibit different behaviors depending on the value of the common ratio \( r \):

• If \(|r| < 1\), the sequence converges to 0 as \( n \to \infty \).
• If \(|r| = 1\), the terms stay constant or oscillate.
• If \(|r| > 1\), the sequence diverges.

In this case, \( 2^{-n} \) is a geometric sequence with a common ratio \( 1/2 \), since \( 2^{-n} = \left(\frac{1}{2}\right)^n \). As \( n \) increases, \( 2^{-n} \) approaches 0 because \( \left|\frac{1}{2}\right| < 1 \). This is why in our exercise, the \( 2^{-n} \) term goes to zero, playing a key role in the convergence of the sequence.

A bounded sequence is one where the terms of the sequence are contained within fixed upper and lower limits for all \( n \geq 1 \). Specifically:

• A sequence \( \{a_n\} \) is bounded from above if there exists some number \( M \) such that \( a_n \leq M \) for all \( n \).
• It is bounded from below if there exists some number \( m \) such that \( a_n \geq m \) for all \( n \).

If a sequence is both bounded above and below, it's simply called bounded. In the sequence \( \{2^{-n} \sin n\} \), the term \( \sin n \) is inherently bounded since it always satisfies \(-1 \leq \sin n \leq 1\).
Consequently, the product \( 2^{-n} \sin n \) is bounded by \(-2^{-n}\) and \(2^{-n}\). As noted in the solution, both bounds approach zero as \( n \to \infty \). The combination of being bounded and converging forms a fundamental characteristic, allowing us to utilize the Squeeze Theorem effectively to conclude that the sequence converges to zero.