#### To determine

Whether the 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=1∞2n+4nen
.

**Result used:**

(1) The geometric series
∑n=1∞arn−1 (or) a+ar+ar2+⋯
is convergent if
|r|<1
and its sum is
a1−r
.

(2) 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:**

The given series can be written as follows,

∑n=1∞2n+4nen=∑n=1∞[2nen+4nen]=∑n=1∞[(2e)n+(4e)n]

∑n=1∞2n+4nen=∑n=1∞(2e)n+∑n=1∞(4e)n
(1)

Here, both the series
∑n=1∞(2e)n and ∑n=1∞(4e)n
are geometric series.

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

Here, the first term of the series is,
a1=2e
and the common ratio of the series is,

r1=(2e)22e=2e

The absolute value of
r1
is,

|r1|=|2e|=2e=0.74<1

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

Therefore, the series
∑n=1∞(2e)n is convergent
.

Consider the series,
∑n=1∞(4e)n=(4e)+(4e)2+(4e)3+⋯
.

Here, the first term of the series is,
a2=4e
and the common ratio of the series is,

r2=(4e)24e=4e

The absolute value of
r2
is,

|r2|=|4e|=4e=1.47>1

Since
|r2|>1
and by the Result (2) stated above, the series is divergent.

Therefore, the series
∑n=1∞(4e)n is divergent
.

Note that, the sum of convergent and divergent series is divergent.

Since the series
∑n=1∞(2e)n
is convergent and
∑n=1∞(4e)n
is divergent, then their sum
∑n=1∞(2e)n+∑n=1∞(4e)n
is divergent.

From equation (1), the series
∑n=1∞2n+4nen is divergent
.

Therefore, the series is divergent.