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Subtract $a_n$ from both sides of the recurrence relation. Distribute the $\frac{1}{3}$ and combine like terms to simplify the expression to $-\frac{2}{3}a_n + \frac{1}{3}$.

Conditions: The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.; The goal is to analyze the sign of the difference to determine monotonicity.

Subtract $a_n$ from both sides of the recurrence relation. Distribute the $\frac{1}{3}$ and combine like terms to simplify the expression to $-\frac{2}{3}a_n + \frac{1}{3}$.

Conditions: The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.; The goal is to analyze the sign of the difference to determine monotonicity.

To compute the terms, substitute the initial value into the recurrence relation iteratively. First, plug in $a_1=1$ to find $a_2$.

Conditions: The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.; The initial value is $a_1=1$.; The index $n$ starts at $1$.

To compute the terms, substitute the initial value into the recurrence relation iteratively. First, plug in $a_1=1$ to find $a_2$.

Conditions: The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.; The initial value is $a_1=1$.; The index $n$ starts at $1$.

The proof uses mathematical induction. First, the base case is verified by checking that $a_1=1 \ge \frac{1}{2}$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ for $n \ge 1$.; The initial value is $a_1=1$.; The goal is to prove $a_n \ge \frac{1}{2}$ for all $n \ge 1$.

The proof uses mathematical induction. First, the base case is verified by checking that $a_1=1 \ge \frac{1}{2}$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ for $n \ge 1$.; The initial value is $a_1=1$.; The goal is to prove $a_n \ge \frac{1}{2}$ for all $n \ge 1$.

Once convergence to a limit $L$ is established, pass to the limit in the recurrence relation $a_{n+1}=\frac{1}{3}(a_n+1)$. Since both $a_n$ and $a_{n+1}$ approach $L$, substitute $L$ for both terms to get $L=\frac{1}{3}(L+1)$, and then solve this linear equation for $L$.

Conditions: The sequence is known to converge to some real number $L$.; The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.

Once convergence to a limit $L$ is established, pass to the limit in the recurrence relation $a_{n+1}=\frac{1}{3}(a_n+1)$. Since both $a_n$ and $a_{n+1}$ approach $L$, substitute $L$ for both terms to get $L=\frac{1}{3}(L+1)$, and then solve this linear equation for $L$.

Conditions: The sequence is known to converge to some real number $L$.; The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.

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 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 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 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.