RECURRENCE RELATION

1 Limits and Convergence

For a sequence of real numbers x1, x2, x3, . . . ,

when n −→ ∞, if xn −→ s,

then the sequence is said to be convergent (to s) and s is called the limit of the sequence. We writeit as

limn→∞

xn = s.

Condition Variable Example

Nil limn→∞

1

n= 0 lim

n→∞

1

3 + 2n= 0

0 ≤ a < 1 limn→∞

an = 0 limn→∞

(

1

3

)n

= 0

b > 1 limn→∞

bn = ∞ limn→∞

3n = ∞

Strictly increasing sequence

A strictly increasing sequence is one where

xn < xn+1, for all n ∈ N.

To show that a sequence is increasing, instead of showing xn < xn+1, it suffices to show that

xn − xn+1 < 0.

Remark: When asked to describe the behaviour of a sequence we can (possibly) mention 2 things:

1. It is an increasing or decreasing sequence.

2. Is it convergent? If yes, what value does it converge to? If no, is it oscillating about 2 points?

These can be checked using a GC.

www.MathAcademy.sg 1 c© 2014 Math Academy

Example 1 (2008/TJC/MYE(J1)/Q6).The numbers xn satisfy the relation xn+1 =

12

7−xn

for all positive integers n.It is given that as n −→ ∞, xn −→ s.

(i) Find the exact value(s) of s.

(ii) Show that if 3 < xn < 4, then xn+1 < xn.

Solution:

(i)

xn+1 =12

7− xn

As n −→ ∞, xn, xn+1 −→ s, therefore,

s =12

7− s

s(7− s) = 12

−s2 + 7s− 12 = 0

(s− 3)(s − 4) = 0

s = 3 or 4

(ii) Method 1: sketch y = xn+1 − xn

We want to sketch y = xn+1 − xn = 12

7−xn

− xn. Replacing xn by x, we get y = 12

7−x− x.

3 4

y = 12

7−x− x

x

y

From the graph, if 3 < xn < 4, then y < 0.

y < 0

=⇒ 12

7− xn− xn < 0

=⇒ xn+1 − xn < 0

=⇒ xn+1 < xn.

www.MathAcademy.sg 2 c© 2014 Math Academy

Method 2: Sketch y = xn+1 and y = xn

We want to sketch y = xn+1 =12

7−xn

and y = xn. Replacing xn by x, we get y = 12

7−xand y = x.

y = x

y = 12

7−x

3 4 7

x

y

From the graph, if 3 < xn < 4, then,

12

7− xn< xn

=⇒ xn+1 < xn.

Method 3: Show algebraically

xn+1 − xn =12

7− xn− xn

=12− xn(7− xn)

7− xn

Since 3 < xn < 4,

www.MathAcademy.sg 3 c© 2014 Math Academy

Example 2 (2011/PJC/II/2 modified).A sequence of positive real numbers x1, x2, x3, . . . satisfies the recurrence relation xn+1 =

√3 + xn for

n ≥ 1.

(a) If the first term x1 is 1.5, write down the 6th term, that is, x6, leaving your answer to 4 decimalplaces.

(b) If the sequence converges to α, determine the exact value of α. [2]

(c) By using a graphical method, prove that xn+1 > xn if 0 < x < α. [2]

(d) Use a calculator to determine the behaviour of the sequence {xn} when x1 = 1. Hence state brieflyhow the results in (c) relate to the behaviour of the sequence. [2]

Solution:

(a) Using GC, x6 = 2.3009.

(b) As n −→ ∞, xn, xn+1 −→ α, therefore,

α =√3 + α

α2 = 3 + α

α2 − α− 3 = 0

α =1±

√

(−1)2 − 4(1)(−3)

2

=1±

√13

2

α =1 +

√13

2or

1−√13

2(rej, since the sequence is positive)

(c) We want to sketch y = xn+1 − xn =√3 + xn − xn. Replacing xn by x, we get y =

√3 + x− x.

1+√13

2

y =√3 + x− x

x

y

From the graph, if 0 < xn < α, then y > 0.

y > 0

=⇒√3 + xn − xn > 0

=⇒ xn+1 − xn > 0

=⇒ xn+1 > xn.

(d) Since 0 < x1 < α, by part (c), xn+1 > xn. Therefore the sequence is strictly increasing. UsingGC, the sequence converges to 2.3028.

www.MathAcademy.sg 4 c© 2014 Math Academy

Example 3.

A sequence of real numbers u1, u2, u3, . . . satisfies the recurrence relation un+1 = 0.9un + 90, a ∈ R

for all positive integers n and u0 = 1000. Express un in terms n.

Solution:

un = 0.9un−1 + 90

= 0.9[0.9un−2 + 90] + 90

= 0.92un−2 + 0.9(90) + 90

= 0.92[0.9un−3 + 90] + 0.9(90) + 90

= 0.93un−3 + 0.92(90) + 0.9(90) + 90

=...

= 0.9nu0 + 0.9n−1(90) + 0.9n−2(90) + · · ·+ 90

= 0.9nu0 +90(1 − 0.9n)

1− 0.9(GP: a = 90, r = 0.9, no. terms = n)

= 0.9nu0 + 900(1 − 0.9n)

= 0.9n(u0 − 900) + 900

= 100(0.9n) + 900 (Since u0 = 1000)

www.MathAcademy.sg 5 c© 2014 Math Academy

2 Using GC

A sequence of positive real numbers x1, x2, x3, . . . satisfies the recurrence relation xn+1 =√3 + xn for

n ≥ 1.If the first term x1 is 1.5, write down the 6th term, that is, x6, leaving your answer to 4 decimal places.

REMEMBER: We must first convert the equation to un =√3 + un−1.

Step What to do Screen

1 Press ‘MODE’ button.Select ‘SEQ’ in the 4th row.

2 Press ‘Y=’ button.Key in nMin = 1. (this means ur sequence starts from x1)Key in u(n) =

√

3 + u(n− 1).-Press ‘2nd’, ‘7’ for u.-Press ‘X,T, θ, n’ for n.-un−1 is keyed as u(n− 1).Key in u(nMin) = 1.5 (the value of u1)

3 Press ‘2nd’, ‘GRAPH’ to get the table.Scroll down to view more terms.

www.MathAcademy.sg 6 c© 2014 Math Academy