Understanding Engineering Mathematics

(iii) In this caseun= 1 +(n− 1 )3 is pretty obvious.

(iv) In this case it is perhaps not quite so easy to spot a pattern for an

explicit expression forun. You might spot more easily that there is a

relation betweenunandun− 1 ,un− 2. Explicitly we find the recurrence

relationun= 2 un− 1 +un− 2 ;u 1 = 2 ,u 2 =4.

In case you are wondering what the explicit expression forunwould be in (iv) that is
equivalent to the recurrence relation, it is actually

un=

1

√

2

( 1 +

√

2 )n−

1

√

2

( 1 −

√

2 )n

Not something that readily springs to mind, as do (i) – (iii)! This illustrates why recurrence
relations are sometimes to be preferred. As a useful exercise in surds you might like to
check that this explicit result does generate the given sequence. Also, you may like to
check, using the binomial expansion that an equivalent form to the above result, not
involving surds, is

un=

∑n

r= 1

Cr 2 (r+^1 )/^2

where the sum ranges only over odd values ofr. Again, not a form that springs imme-
diately to mind!
An obvious question is, as we continue along an infinite sequence, does thenth term
tend to a definite limit? That is, what is the behaviour ofunasn→∞? Symbolically,
what is limn→∞un? Clearly this depends on the form of thenth term.

Problem 14.9

Examine the limits of the following sequences as n increases indefinitely:

(i) 1,

1

2

,

1

3

,

1

4

,...,

1

n

,... (ii)

1

2

,

3

4

,

7

8

,...,1−

(

1

2

)n

,....

(iii) 0, 3, 8,...,n^2 −1,...

Essentially we have to examine the limit asntends to infinity.

(i) In this case we have limn→∞un=limn→∞

1

n

= 0

(ii) In this case we have limn→∞un=limn→∞

(

1 −

( 1

2

)n)

=lim

n→∞

1 −lim

n→∞

( 1

2

)n

= 1 − 0 = 1

where we have used the fact that an ‘infinite power’ of any number

between 0 and 1 is zero.

(iii) Fairly straightforward in this case

lim

n→∞

un=lim

n→∞

(

n^2 − 1

)

=∞

So this sequence ‘diverges’ – the terms get larger and larger indefinitely.