Answer:
Option (1)
Step-by-step explanation:
In a geometric series,
[tex]\sum_{n=0}^{\infty}ar^n=a+ar^1+ar^2+....[/tex]
Here r = [tex]\frac{a_n}{a_{n-1}}[/tex]
For |r| < 1, series will converge.
For |r| > 1, series will diverge.
Option (1)
Given geometric series is,
[tex]\frac{1}{2}+\frac{1}{4}+ \frac{1}{8}+ \frac{1}{16}.......[/tex]
Common ratio = [tex]\frac{\frac{1}{4}}{\frac{1}{2} }[/tex]
= [tex]\frac{1}{2}[/tex]
Since, [tex]\frac{1}{2}<1[/tex]
Series will converge.
Option (2)
[tex]\frac{1}{2}+1+2+4+......[/tex]
r = [tex]\frac{1}{\frac{1}{2}}=2[/tex]
Since, 2 > 1,
Series will diverge.
Option (3)
[tex]\frac{1}{2}+ \frac{3}{2}+ \frac{9}{2}+ \frac{27}{2}+......[/tex]
Common ratio 'r' = [tex]\frac{\frac{3}{2} }{\frac{1}{2} }=3[/tex]
Since, 3 > 1,
Series will diverge.
Option (4)
[tex]\frac{1}{2}+3+18+108+.....[/tex]
Common ratio 'r' = [tex]\frac{3}{\frac{1}{2}}=6[/tex]
Since, 6 > 1
Series will diverge.
Therefore, Option (1) will be the correct option.
