**Proof:** By mathematical induction.

Consider the sequence,
pn+1=bpna+pn
. (6)

(1)

*Base case:*
n=0

To show that the claim is true when
n=0
.

Substitute 1 for *n* in equation (6),

p0+1=bp0a+p0p1=bp0a+p0

Subtract
p0
on both sides and simplify the terms.

p1−p0=bp0a+p0−p0=bp0−p0(a+p0)a+p0=p0(b−(a+p0))a+p0=p0((b−a)−p0)a+p0

Since
p0<b−a
,
(b−a)−p0>0
.

p1−p0>0p1>p0

Thus, the inequality is
0<p0<p1
.

Therefore, the claim is true when
n=0
.

*Inductive hypothesis:*
n=k

Assume that the claim is true when
n=k
.

That is,
0<pk<b−a and pk<pk+1
is true

*Inductive step:*
n=k+1

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

That is, to prove that
0<pk+1<b−a and pk+1<pk+2
is true.

(b−a)−pk+1=(b−a)−bpka+pk=(b−a)(a+pk)−bpka+pk

Simplify the terms of the expression.

(b−a)−pk+1=a(b−a)+pk(b−a)−bpka+pk=a(b−a)+pk(b−a−b)a+pk=a(b−a)+pk(−a)a+pk=a(b−a−pk)a+pk

Since
pk<b−a
,
b−a−pk>0
.

(b−a)−pk+1>0b−a>pk+1

Thus, the inequality is
pk+1<b−a
.

Consider,
pk+2−pk+1
.

pk+2−pk+1=bpk+1a+pk+1−pk+1=bpk+1−pk+1(a+pk+1)a+pk+1=pk+1(b−a+pk+1)a+pk+1

Since
pk+1<b−a
,
(b−a)− pk+1>0
.

pk+2−pk+1>0pk+2>pk+1

Therefore, the inequality is
pk+1<pk+2
.

Hence, by mathematical induction, the claim is true for all *n.*

(2)

*Base case:*
n=0

To show that the claim is true when
n=0
.

Substitute 1 for *n* in equation (5),

p0+1=bp0a+p0p1=bp0a+p0

Subtract
p0
on both sides.

p1−p0=bp0a+p0−p0=bp0−p0(a+p0)a+p0=p0(b−(a+p0))a+p0=p0((b−a)−p0)a+p0

Since
p0>b−a
,
(b−a)−p0<0
.

p1−p0<0p1<p0

Thus, the inequality is
p0>p1
.

Therefore, the claim is true when
n=0
.

*Inductive hypothesis:*
n=k

Assume that the claim is true when
n=k
.

That is,
pk>b−a and pk>pk+1
is true.

*Inductive step:*
n=k+1

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

That is, to prove that
pk+1>b−a and pk+1>pk+2
is true.

(b−a)−pk+1=(b−a)−bpka+pk=(b−a)(a+pk)−bpka+pk

Simplify the term of the expression.

(b−a)−pk+1=a(b−a)+pk(b−a)−bpka+pk=a(b−a)+pk(b−a−b)a+pk=a(b−a)+pk(−a)a+pk=a(b−a−pk)a+pk

Since
pk>b−a
,
b−a−pk<0
.

(b−a)−pk+1<0b−a<pk+1

Therefore,
pk+1>b−a
.

pk+2−pk+1=bpk+1a+pk+1−pk+1=bpk+1−pk+1(a+pk+1)a+pk+1=pk+1(b−a+pk+1)a+pk+1

Since
pk+1>b−a
,
(b−a)− pk+1<0
.

pk+2−pk+1<0pk+2<pk+1

Therefore,
pk+1>pk+2
.

Hence, by mathematical induction, the claim is true for all *n.*

(3)

Both the cases (1) and (2), the sequence
{pn}
is monotonic and bounded.

Therefore, the sequence
{pn}
is convergent by the Monotonic Sequence Theorem.

From part (a),
limn→∞pn=b−a
.

Hence showed.