## Calculus 8th Edition

a) $a_{1}=1$, $a_{n+1}=4-a_{n}$ for $n\geq 1$ $a_{2}=4-a_{1}$ $=4-1$ $=3$ $a_{3}=4-a_{2}$ $=4-3$ $=1$ $a_{4}=4-a_{3}$ $=4-1$ $=3$ $a_{5}=4-a_{4}$ $=4-3$ $=1$ The sequence is divergent because the given sequence oscillates between 1 and 3 forever. b) $a_{2}=4-a_{1}$ $=4-2$ $=2$ $a_{3}=4-a_{2}$ $=4-2$ $=2$ $a_{4}=4-a_{3}$ $=4-2$ $=2$ $a_{5}=4-a_{4}$ $=4-2$ $=2$ The sequence is convergent because the given sequence is bounded and has finite unit limit points.