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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 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 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.