Which of the sequences converge, and which diverge? Give...

Which of the sequences converge, and which diverge? Give reasons for your answers. The first term of a sequence is $x_{1}=\cos (1) .$ The next terms are $x_{2}=x_{1}$ or $\cos (2),$ whichever is larger; and $x_{3}=x_{2}$ or $\cos (3)$ whichever is larger (farther to the right). In general, $$ x_{n+1}=\max \left\{x_{n}, \cos (n+1)\right\} $$

Which of the sequences converge, and which diverge? Give reasons for your answers.
The first term of a sequence is $x_{1}=\cos (1) .$ The next terms are
$x_{2}=x_{1}$ or $\cos (2),$ whichever is larger; and $x_{3}=x_{2}$ or $\cos (3)$
whichever is larger (farther to the right). In general,
$$
x_{n+1}=\max \left\{x_{n}, \cos (n+1)\right\}
$$

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Key Concepts

Monotonicity

The concept of a monotone sequence is used here because the sequence is constructed so that each term is at least as large as the previous one, making it non-decreasing. This property is crucial in determining convergence, as monotone sequences that are bounded will converge by the Monotone Convergence Theorem.

Bounded Sequences

A sequence is bounded if there exists a real number that acts as an upper bound for all terms in the sequence. In this context, since the cosine function yields values between -1 and 1 and the sequence takes the maximum of previous terms and cosine values, the sequence is bounded above by 1. Boundedness ensures that even if the sequence is non-decreasing, it will not diverge to infinity.

Density of Cosine Function Values

The cosine function, when evaluated at integer values, can be shown to be dense in the interval [-1, 1]. This means that for any number in the interval, there exists an integer such that the cosine of that integer is arbitrarily close. In the sequence given, this density implies that the sequence will eventually encounter cosine values that come arbitrarily close to 1, which helps establish that its limit is 1.

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