Answer:
absolutely convergent
Step-by-step explanation:
given data
sin(n)/3^n
solution
we have given term [tex]\frac{sin(n)}{3^n}[/tex]
when n = 1
and we know that
value of sin(n) ≤ 1
so that we can say that
[tex]\frac{sin(n)}{3^n}[/tex] ≤ [tex]\frac{1}{3^n}[/tex] or [tex](\frac{1}{3})^n[/tex]
here [tex]\frac{1}{3^n}[/tex] is converges this is because common ratio in geometric series
here r is [tex]\frac{1}{3}[/tex] and here it satisfy that -1 < r < 1
so it is converges
and
[tex]\frac{sin(n)}{3^n}[/tex] is also similar
so it is converges
and here no [tex](-1)^n[/tex] term is
so we can say series is absolutely convergent
