**Given:**

The sequence is
a1=1, an+1=3−1an
(1)

**Theorem used: Monotonic Sequence Theorem**

“If the sequence is bounded and monotonic, then the sequence is convergent.”

**Proof: Induction Method**

Claim: Let
Pn
be the statement that
an+1≥an
and
an≤3
(2)

*Base case:*
n=1

Substitute 1 for *n* in equation (2).

a1+1≥a1 and a1≤3a2≥a1 and a1≤3

Obtain the value of
a2
.

Substitute 1 for *n* in equation (1),

a1+1=3−1a1a2=3−11 [∵a1=1]=3−1=2

Thus, the value of
a2=2
.

Since the value of
a1=1 and a2=2
,
a2≥a1 and a1≤3
.

Therefore, the claim is true for
n=1
.

*Induction hypothesis:*
n=k

Assume that the claim is true when
n=k
.

That is, assume that
Pk
is true;
ak+1≥ak
and
ak≤3
.

*Inductive step:*
n=k+1

To prove that the claim is true when
n=k+1
.

That is, if
Pk+1
is the statement, then it is enough to prove that
a(k+1)+1≥ak+1
and
ak+1≤3
is true.

Substitute
k+1
for *n* in equation (1),

a(k+1)+1=3−1ak+1
(3)

From induction hypothesis,

ak+1≥ak1ak+1<1ak

−1ak+1≥−1ak
(4)

Substitute the inequality (4) in the equation (3) and substitute
ak+1
for
3−1ak
,

a(k+1)+1≥3−1ak

a(k+1)+1≥ak+1
(5)

Consider
ak+1
, apply induction hypothesis
ak≤3
and equation (1).

ak+1=3−1ak≤3−13=83<3

Therefore,
ak+1≤3
. (6)

From (5) and (6),
a(k+1)+1≥ak+1 and ak+1≤3
.

Hence, the claim is true for all *n* and
0<an≤3 for n≥1
.

Since the sequence
{an}
is bounded and monotonic and by Monotonic Sequence Theorem, the sequence is convergent.

That is,
limn→∞an
exists.

Obtainthe limit of the sequence.

limn→∞an+1=limn→∞(3−1an)=limn→∞(3)−limn→∞(1an)=3−1limn→∞an

Since the sequence
{an}
is convergent,
limn→∞an=L
.

Thus,
limn→∞an+1=L
.

Substitute *L* for
limn→∞an and limn→∞an+1
,

L=3−1LL=3L−1LL2=3L−1

Rearrange the terms and form the quadratic equation such as
L2−3L+1=0
.

Solve the equation for *L* by using the quadratic formula.

L=−(−3)±(−3)2−4⋅1⋅12⋅1=3±9−42=3±52

Since
1≤an≤3 for all n≥1
, the value of
L=3+52
.

Therefore, the limit of the sequence is
3+52
.