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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 $n \ge 1$.

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 $1$.