(a)

#### To determine

Whether the sequence is convergent or divergent.

#### Answer

The sequence
{an}
is convergent.

#### Explanation

**Given:**

The sequence is
an=2n3n+1
.

**Definition used:**

If
an
is a sequence and
limn→∞an
exists, then the sequence
an
is said to be converges; otherwise it is diverges.

**Calculation:**

Obtain the limit of the sequence to investigate whether the sequence converges or diverges.

Compute the value of
limn→∞an=limn→∞2n3n+1
.

Divide the numerator and the denominator by the highest power.

limn→∞2n3n+1=limn→∞2nn3n+1n=limn→∞23nn+1n=limn→∞23+1n

Redefine the terms as follows,

limn→∞2n3n+1=limn→∞23+1n=limn→∞(2)limn→∞(3)+limn→∞(1n)=23+0=23

Since
limn→∞an
exists and by using the definition of the sequence, it can be concluded that the sequence is converges to the limit
23
.

Therefore, the sequence is convergent.

(b)

#### To determine

Whether the series is convergent or divergent.

#### Answer

The series
∑n=1∞an
is divergent.

#### Explanation

**Given:**

The series is
∑n=1∞an=∑n=1∞2n3n+1
.

Here,
an=2n3n+1
.

**Theorem used:** Series test for Divergence

If
limn→∞an
does not exist or
limn→∞an≠0
, then the series
∑n=1∞an
is divergent.

**Calculation:**

Compute the value of
limn→∞an=limn→∞2n3n+1
.

From part (a),
limn→∞2n3n+1=23
.

Since
limn→∞an≠0
and by using the series test for Divergence, the series is divergent.

Therefore, the series
∑n=1∞an
is divergent.