Answer:-[tex]1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}[/tex] is a converging geometric series.
Explanation:-
The geometric series is given by
[tex]\sum_{n=0}^{\infty }ar^n=a+ar+ar^2+...[/tex]
If |r| < 1 then the geometric series converges to.
If |r| ≥1 then the geometric series diverges.
1)[tex]\frac{1}{81}+\frac{1}{27}+\frac{1}{9}+\frac{1}{3}[/tex]
here[tex]r=\frac{ar}{a}=\frac{\frac{1}{27}}{\frac{1}{81}}=3>1[/tex]
⇒The geometric series diverges.
2)[tex]1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}[/tex]
here [tex]r=\frac{ar}{a}=\frac{\frac{1}{2}}{1}=\frac{1}{2}<1[/tex]
⇒The geometric series converges.
