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.