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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 write
it as
lim
n→∞
xn = s.
Condition Variable Example
Nil lim
n→∞
1
n
= 0 lim
n→∞
1
3 + 2n
= 0
0 ≤ a < 1 lim
n→∞
an
= 0 lim
n→∞
1
3
n
= 0
b > 1 lim
n→∞
bn
= ∞ lim
n→∞
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.

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−x and 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,

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 decimal
places.
(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 briefly
how 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
2
or
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. Using
GC, the sequence converges to 2.3028.

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
= 0.92
un−2 + 0.9(90) + 90
= 0.92
[0.9un−3 + 90] + 0.9(90) + 90
= 0.93
un−3 + 0.92
(90) + 0.9(90) + 90
=
...
= 0.9n
u0 + 0.9n−1
(90) + 0.9n−2
(90) + · · · + 90
= 0.9n
u0 +
90(1 − 0.9n)
1 − 0.9
(GP: a = 90, r = 0.9, no. terms = n)
= 0.9n
u0 + 900(1 − 0.9n
)
= 0.9n
(u0 − 900) + 900
= 100(0.9n
) + 900 (Since u0 = 1000)

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.

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