#### To determine

Whether the geometric series is convergent or divergent and obtain the sum if the series is convergent.

#### Answer

The series is divergent.

#### Explanation

**Given:**

The series is
∑n=0∞6⋅22n−13n
.

**Result used:**

The geometric series
∑n=1∞arn−1 (or) a+ar+ar2+⋯
is divergent if
|r|≥1
, where *a* is the first term and *r* is the common ratio of the series.

**Calculation:**

Consider the given series
∑n=0∞6⋅22n−13n
.

∑n=1∞6⋅22n−13n=6⋅22⋅1−131+6⋅22⋅2−132+6⋅22⋅3−133+6⋅22⋅4−134+⋯=2⋅22−1+2⋅24−13+2⋅26−132+2⋅28−133+⋯=22+243+2632+2833+⋯=4+163+649+25627+⋯

Here, the first term of the series is *a* = 4 and the common ratio of the series is,

r=Second termFirst term=1634=163⋅14=43

The absolute value of *r* is,

|r|=|43|=43≈1.3333>1

Since
|r|>1
and by using Result (2) stated above, the series is divergent.

Therefore, the series is divergent.