**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=3−2ne−n
. (1)

**Calculation:**

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

The graph of the sequence
an=3−2ne−n
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=3−2(1)e−1=3−2e≈3−0.74=2.26

Thus, the first term of the sequence is
a1≈2.26
and
a1<a2<a3<...
.

Therefore, the sequence is bounded below by 2.26.

That is,
2.26<an for all n≥1
. (2)

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

limn→∞an=limn→∞(3−2ne−n)=limn→∞(3)−limn→∞(2ne−n)

=3−2⋅limn→∞(ne−n)
(3)

Obtain the limit of
limn→∞(ne−n)
.

limn→∞(ne−n)=limn→∞(nen)

Since
∞∞
is in indeterminate form, apply L’Hospital’s Rule.

limn→∞(ne−n)=limn→∞(ddn(n)ddn(en))=limn→∞(1en)=1∞=0

Substitute 0 for
limn→∞(ne−n)
in equation (2),

limn→∞3−2ne−n=3−2⋅0=3−0=3

Thus, the limit of the sequence is 3.

Therefore, the sequence is bounded above by 3.

That is,
an<3 for all n≥1
. (4)

Combine the equations (2) and (4),
2.26<an<3 for all n≥1
.

Thus, by definition (4), the sequence is bounded.

Therefore, it can be concluded that the sequence is increasing, monotonic and bounded.