#### To determine

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

#### Answer

The series is convergent to the sum
73
.

#### Explanation

**Given:**

The series is
∑n=1∞[(−0.2)n+(0.6)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:**

The given series can be expressed as follows,

∑n=1∞[(−0.2)n+(0.6)n−1]=∑n=1∞(−0.2)n+∑n=1∞(0.6)n−1
(1)

Here, both the series
∑n=1∞(−0.2)n and ∑n=1∞(0.6)n−1
are geometric series.

Obtain the sum of the geometric series
∑n=1∞(−0.2)n
if the series is convergent.

Consider the series,
∑n=1∞(−0.2)n=(−0.2)+(−0.2)2+(−0.2)3+⋯
.

Here, the first term of the series is,
a1=−0.2
and the common ratio of the series is,

r1=(−0.2)2(−0.2)=−0.2

The absolute value of
r1
is,

|r1|=|−0.2|=0.2<1

Since
|r1|<1
and by the Result stated above, the series is convergent.

Obtain the sum of the series.

Compute the value of
∑n=1∞(−0.2)n=a11−r1
.

Since
a1=−0.2
and
r1=−0.2
,

a11−r1=−0.21−(−0.2)=−0.21+0.2=−0.21.2=−16

Thus, the sum of the series is
∑n=1∞(−0.2)n=−16
. (2)

Obtain the sum of the geometric series
∑n=1∞(0.6)n−1
if the series is convergent.

Consider the series,
∑n=1∞(0.6)n−1=1+(0.6)+(0.6)2+(0.6)3+⋯
.

Here, the first term of the series is,
a2=1
and the common ratio of the series is,

r2=0.61=0.6

The absolute value of
r2
is,

|r2|=|0.6|=0.6<1

Since
|r2|<1
and by the Result stated above, the series is convergent.

Obtain the sum of the series.

Compute the value of
∑n=1∞(0.6)n−1=a21−r2
.

Since
a2=1
and
r2=0.6
,

a21−r2=11−0.6=10.4=104=52

Thus, the sum of the series is
∑n=1∞(0.6)n−1=52
. (3)

Since both the geometric series
∑n=1∞(−0.2)n and ∑n=1∞(0.6)n−1
are convergent, the sum
∑n=1∞[(−0.2)n+(0.6)n−1]
is also convergent.

Obtain the sum of the series
∑n=1∞[(−0.2)n+(0.6)n−1]
.

Substitute equation (2) and (3) in equation (1),

∑n=1∞[(−0.2)n+(0.6)n−1]=−16+52=−16+156=146=73

Thus, the sum of the series is
∑n=1∞[(−0.2)n+(0.6)n−1]=73
.

Therefore, the series is convergent to the sum
73
.