Answer:
OPTION B
Step-by-step explanation:
Geometric series : [tex]$ \displaystyle \sum_{n = 0}^{ \infty} ar^n $[/tex]
where [tex]a[/tex] is the first term of the series and
[tex]r[/tex] is common difference.
A geometric series is convergent if |r| < 1.
It is divergent otherwise.
Since the first term of the series is a and the second term is ar, the ration of second term and first term, [tex]\frac{ar}{r}[/tex] = r.
OPTION A:
[tex]$ \frac{1}{81} + \frac{1}{27} + \frac{1}{9} + \frac{1}{3} + \hdots $[/tex].
Here, [tex]$ a = \frac{1}{81} $[/tex] and [tex]$ ar = \frac{1}{27} $[/tex]
[tex]$ \implies r= \frac{27}{81} = 3 $[/tex]
r > 1. So, this series is divergent.
OPTION B:
[tex]$ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \hdots $[/tex]
a = 1; ar = [tex]$ \frac{1}{2} $[/tex].
[tex]$ \implies r = \frac{1}{2} $[/tex].
Since, r < 1, we can say that the series is convergent.
OPTION C:
We can easily see that |r| =4. So, it is not convergent.
OPTION D:
Again |r| = 2. So, the series should be divergent.
