**Definition used:**

(1) The sequence
{an}
is increasing if
an<an+1
for all
n≥1
. That is,
a1<a2<a3<...
.

(2) The sequence
{an}
is decreasing and
an>an+1
for all
n≥1
. That is,
a1>a2>a3>...
.

(3) If the sequence is either increasing or decreasing, then the sequence is called monotonic; otherwise it is not monotonic.

(4) If
{an}
is the sequence with
m≤an≤M
for
m,M∈ℕ
, then the sequence is bounded.

**Given:**

The sequence is
an=n3−3n+3
. (1)

**Calculation:**

Check whether the sequence is increasing, decreasing or not monotonic.

The graph of the sequence
an=n3−3n+3
is shown below in Figure 1.

Here, it is observed that the sequence is increasing on
(1,∞)
.

That is,
a1<a2<a3<...
.

Since the sequence increases and by definition (3), the sequence is monotonic.

Check whether the sequence is bounded or not.

Obtain the first term of the sequence by substituting 1 for *n* in equation (1).

a1=13−3⋅1+3=1−3+3=1

Thus, the first term of the sequence is
a1=1
and
a1<a2<a3<...
.

Therefore, the sequence is bounded below by 1.

That is,
1<an for all n≥1
.

Obtain the limit of the sequence (the value of the term
an
as *n* tends to infinity).

limn→∞an=limn→∞(n3−3n+3)=limn→∞(n3)−limn→∞(3n)+limn→∞(3)=∞−∞+3=∞

The limit of the sequence does not exist, the sequence is divergent. So, the sequence is not bounded above.

Thus, the sequence is not bounded.

Therefore, it can be concluded that the sequence is monotonic but not bounded.