#### 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∞3n+1(−2)n
.

**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∞3n+1(−2)n
.

∑n=0∞3n+1(−2)n=30+1(−2)0+31+1(−2)1+32+1(−2)2+33+1(−2)3+⋯=311+32−2+334+34−8+⋯=3−92+274−818+⋯

Here, the first term of the series is *a* = 3 and the common ratio of the series is,

r=Second termFirst term=−923=−92⋅13=−32

The absolute value of *r* is,

|r|=|−32|=32=1.5>1

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

Therefore, the series is divergent.