**Given:**

The sequence is
an=1−n2+n
. (1)

**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.

**Calculation:**

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

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

an+1=1−(n+1)2+(n+1)=1−n−12+n+1=−n3+n<1−n2+n

Therefore,
an>an+1 for all n≥1
.

Since
an>an+1 for all n≥1
and by definition (1), the sequence is decreasing.

Check whether the sequence is bounded or not.

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

a1=1−12+1=03=0

Thus, the first term of the sequence is
a1=0
and
a1>a2>a3>...
.

Therefore,
an≤0 for all n≥1
. (2)

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

Compute,
limn→∞an=limn→∞1−n2+n

Divide the numerator and denominator by *n.*

limn→∞1−n2+n=limn→∞1−nn2+nn=limn→∞1n−12n+1=limn→∞(1n)−1limn→∞(2n)+1

Apply infinity property
limn→∞(kna)=0
and simplify the terms as shown below.

limn→∞1−n2+n=1∞−11∞+1=0−10+1=−11=−1

Thus, the limit of the sequence is −1.

Therefore,
−1<an for all n≥1
. (3)

Combine the equations (2) and (3),
−1<an≤0 for all n≥1
.

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

Hence, it can be concluded that the sequence is decreasing and bounded.