#### 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
4009
.

#### Explanation

**Given:**

The series is
∑n=1∞12(0.73)n−1
.

**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∞12(0.73)n−1
.

∑n=1∞12(0.73)n−1=12(0.73)1−1+12(0.73)2−1+12(0.73)3−1+⋯=12(0.73)0+12(0.73)1+12(0.73)2+⋯=12(1)+12(0.73)+12(0.5329)+⋯=12+8.76+6.3948+⋯

Here, the first term of the series is
a=12
.

The common ratio of the series is,

r=Second termFirst term=8.7612=0.73

The absolute value of *r* is,

|r|=|0.73|=0.73<1

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

Obtain the sum of the series.

Since *a* = 12 and *r* = 0.73,

∑n=1∞12(0.73)n−1=121−0.73=120.27×100100=120027=4009

Therefore, the series is convergent to the sum
4009
.