#### To determine

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

#### Answer

The series is convergent to the sum
17
.

#### Explanation

**Given:**

The series is
∑n=1∞(−3)n−14n
.

**Result used:**

The geometric series
∑n=1∞arn−1 (or) a+ar+ar2+⋯
is convergent if
|r|<1
and its sum is
a1−r
, where *a* is the first term and *r* is the common ratio of the series.

**Calculation:**

Consider the given series
∑n=1∞(−3)n−14n
.

∑n=1∞(−3)n−14n=(−3)1−141+(−3)2−142+(−3)3−143+(−3)4−144+⋯=(−3)04+(−3)142+(−3)243+(−3)344+⋯=14−316+964−27256+⋯

Then, the first term of the series is
a=14
and the common ratio of the series is,

r=Second termFirst term=−31614=−316⋅41=−34

The absolute value of *r* is,

|r|=|−34|=34<1

Since
|r|<1
and by using the Result stated above, the series is convergent.

Obtain the sum of the series.

Since
a=14
and
r=−34
,

∑n=1∞(−3)n−14n=141−(−34)=141+34=1444+34

Perform the arithmetic operations and simplify the terms,

∑n=1∞(−3)n−14n=144+34=1474=14⋅47=17

Therefore, the series is convergent to the sum
17
.