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The Monotone Convergence Theorem states that if a sequence $(a_n)_{n\in\mathbb{N}}$ is monotonically decreasing (i.e., $a_{n+1} \le a_n$ for all $n$) and bounded from below, then the sequence is convergent. This is a direct application of the existence of the infimum for bounded sets.

Conditions: The sequence $(a_n)$ is monotonically decreasing.; The sequence $(a_n)$ is bounded from below.

The Monotone Convergence Theorem states that if a sequence $(a_n)_{n\in\mathbb{N}}$ is monotonically decreasing (i.e., $a_{n+1} \le a_n$ for all $n$) and bounded from below, then the sequence is convergent. This is a direct application of the existence of the infimum for bounded sets.

Conditions: The sequence $(a_n)$ is monotonically decreasing.; The sequence $(a_n)$ is bounded from below.

The Monotone Convergence Theorem is used after establishing that the sequence is both bounded below and non-increasing. It is invoked to guarantee the existence of a real limit $L$ for the sequence, which is a prerequisite for solving the limit equation.

Conditions: The sequence has been proven to be bounded from below.; The sequence has been proven to be non-increasing.

The Monotone Convergence Theorem is used after establishing that the sequence is both bounded below and non-increasing. It is invoked to guarantee the existence of a real limit $L$ for the sequence, which is a prerequisite for solving the limit equation.

Conditions: The sequence has been proven to be bounded from below.; The sequence has been proven to be non-increasing.

The Monotone Convergence Theorem states that every bounded, monotone sequence of real numbers converges. Therefore, to prove convergence using this theorem, one must establish both boundedness and monotonicity as the necessary hypotheses.

Conditions: The sequence consists of real numbers.; The goal is to prove convergence using the Monotone Convergence Theorem.

The Monotone Convergence Theorem states that every bounded, monotone sequence of real numbers converges. Therefore, to prove convergence using this theorem, one must establish both boundedness and monotonicity as the necessary hypotheses.

Conditions: The sequence consists of real numbers.; The goal is to prove convergence using the Monotone Convergence Theorem.

The bisection method constructs a sequence of upper bounds $(b_n)$ that is proven to be a Cauchy sequence. However, being Cauchy only means the terms get close to each other; it does not guarantee they converge to a specific real number.

Conditions: The sequence $(b_n)$ has been proven to be a Cauchy sequence.; The context is the field of real numbers $\mathbb{R}$.

The bisection method constructs a sequence of upper bounds $(b_n)$ that is proven to be a Cauchy sequence. However, being Cauchy only means the terms get close to each other; it does not guarantee they converge to a specific real number.

Conditions: The sequence $(b_n)$ has been proven to be a Cauchy sequence.; The context is the field of real numbers $\mathbb{R}$.

The example proves the sequence is non-increasing. The mathematical derivation shows $a_{n+1}-a_n \le 0$, which implies $a_{n+1} \le a_n$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ with $a_1=1$.; The lower bound $a_n \ge \frac{1}{2}$ has been proven.

The example proves the sequence is non-increasing. The mathematical derivation shows $a_{n+1}-a_n \le 0$, which implies $a_{n+1} \le a_n$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ with $a_1=1$.; The lower bound $a_n \ge \frac{1}{2}$ has been proven.

The inequality $-\frac{2}{3}a_n+\frac{1}{3}\le0$ is algebraically equivalent to $\frac{1}{3}\le\frac{2}{3}a_n$, which simplifies to $\frac{1}{2}\le a_n$. Since the lower bound $a_n\ge\frac{1}{2}$ was already proven by induction, the difference $a_{n+1}-a_n$ is guaranteed to be non-positive.

Conditions: The expression for the difference is $a_{n+1}-a_n = -\frac{2}{3}a_n + \frac{1}{3}$.; The lower bound $a_n \ge \frac{1}{2}$ has been established for all $n \ge 1$.

The inequality $-\frac{2}{3}a_n+\frac{1}{3}\le0$ is algebraically equivalent to $\frac{1}{3}\le\frac{2}{3}a_n$, which simplifies to $\frac{1}{2}\le a_n$. Since the lower bound $a_n\ge\frac{1}{2}$ was already proven by induction, the difference $a_{n+1}-a_n$ is guaranteed to be non-positive.

Conditions: The expression for the difference is $a_{n+1}-a_n = -\frac{2}{3}a_n + \frac{1}{3}$.; The lower bound $a_n \ge \frac{1}{2}$ has been established for all $n \ge 1$.

The distance between any two terms $b_n$ and $b_m$ (with $m > n$) is bounded by the length of the interval $[a_n, b_n]$. Because the bisection method halves the interval length at each step, this distance is bounded by $\left(\frac{1}{2}\right)^{n-1}|b_1 - a_1|$.

Conditions: $m > n \in \mathbb{N}$.; The sequences $(a_n)$ and $(b_n)$ are constructed via the bisection method.; $b_n$ is an upper bound and $a_n$ is a lower bound for the set $M$.

The distance between any two terms $b_n$ and $b_m$ (with $m > n$) is bounded by the length of the interval $[a_n, b_n]$. Because the bisection method halves the interval length at each step, this distance is bounded by $\left(\frac{1}{2}\right)^{n-1}|b_1 - a_1|$.

Conditions: $m > n \in \mathbb{N}$.; The sequences $(a_n)$ and $(b_n)$ are constructed via the bisection method.; $b_n$ is an upper bound and $a_n$ is a lower bound for the set $M$.