Answer
a)
The sequence is divergent.
b)
The sequence is convergent.
Work Step by Step
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.
