MATH10242 Sequences and Series:
Solutions 4
Question 1: In each part of the question, we first try to manipulate the expression (say
by dividing top and bottom by some function) in order to get it into a sum of terms for
which the limit is obvious. Then we use the Theorem 4.2.1 to give the answer.
(a) So, here we first expand the top and bottom and then divide both by n3 .
4
+ n42 + n13
(2n + 1)2
4n2 + 4n + 1
n
an =
=
=
.
n(n2 − n + 1)
n3 − n2 + n
1 − n1 + n12
Now we use the fact that n1r → 0 as n → ∞ for r > 0 (see Example 4.2.3). Thus by
The Algebra of Limits Theorem 4.2.1 parts (ii, iii, v) we get
4
n
+ n42 + n13
0
0+0+0
=
= 0
→
1
1
1−0+0
1
1 − n + n2
as n → ∞.
Remark: In these sorts of questions you do need to be careful about the power of n
by which you divide—a higher power than the third would end up with 00 which is useless,
while dividing top and bottom by (say) n2 would leave a term of n in the denominator.
Since that does not tend to a limit, you cannot (directly) use 4.2.1.
(2n+1)2 (2n+1)2
(b) Here, if the sequence is (bn ) = cos(n) n(n2 −n+1) then notice that |bn | ≤ n(n2 −n+1) ,
which is the sequence (an ) from part (a). Thus, by the Sandwich Theorem (the most convenient form is 4.1.4) limn→∞ bn = 0 as well.
(c) Again, we first expand the top and bottom and then divide both by n3 .
8 + 12 n1 + 6 n12 + n13
(2n + 1)3
8n3 + 12n2 + 6n + 1
an =
=
=
.
n(n2 − n + 1)
n3 − n2 + n
1 − n1 + n12
Applying the same logic as in part (a) we see that
lim an =
n→∞
8+0+0+0
= 8.
1−0+0
(d) Here we divide through top and bottom by 9n to get
an =
3n + 5n
7n + 9n
=
( 31 )n + ( 59 )n
.
( 79 )n + 1
Now by the notes (5.1.2) all but the final entry tends to zero as n → ∞. Thus, by
4.2.1 again
0+0
0
lim an =
= = 0.
n→∞
0+1
1
1
(e) Using 4.2.1 and 5.1.1,
n3 + ( 13 )n
n3 + 1
=
1 + ( n13 )( 31 )n
1 + n13
−→
1+0·0
= 1.
1+0
(f) We use the same trick for this as on the second sheet:
√
√
√
√
2+n−
2−n
2+n+
2−n
n
n
n
n
√
√
√
an = ( n2 + n − n2 − n ) =
√
n2 + n + n2 − n
=
(n2 + n) − (n2 − n) )
2n
2
p
√
√
√
= √
= p
2
2
2
2
( n + n + n − n)
( n + n + n − n)
( 1 + 1/n + 1 − 1/n )
(In the final step, I did the usual thing of dividing top and bottom by n.)
Now, (1 ± 1/n) → 1 as√n → ∞ (use Theorem 4.2.1 again). But we are also told that
if bn → 1 as n → ∞ then bn → 1. So,
2
2
p
p
→
= 1 as n → ∞ .
1+1
( 1 + 1/n + 1 − 1/n )
(g) This does not seem to follow directly from our rules, so let’s begin by simplifying
a bit, by using (n + 1)! = (n!)(n + 1) and dividing through by n!:
1 + n!1
n! + 1
n! + 1
an =
=
=
.
(†)
(n + 1)!
(n + 1)(n!)
(n + 1)
Now, for example since 0 ≤ n!1 ≤ n1 , the Sandwich Rule says lim n!1 = 0. So now writing
n→∞
1 an = 1 + n!1 1+n
and using 4.2.1(iv) gives lim an = (1 + 0) · 0 = 0.
n→∞
Remark Note here that in order to use Theorem 4.2.1(v) on the displayed equation (†)
lim an
an
=
since (bn ) = (n + 1) does not converge.
I cannot directly use the fact that lim
bn
lim bn
Question 2:
Theorem 4.2.2. Let (an ) be a null sequence and let (bn ) be a bounded sequence (not
necessarily convergent). Then (an · bn )n∈N is a null sequence.
Proof. We are given that there exists B > 0 such that |bn | ≤ B for all n. Also, for any
η > 0 there exists N ∈ N such that |an | < η for all n ≥ N .
So, if > 0 is given, take η = B . Then with N as above we get
|an bn | = |an | · |bn | < ηB = ,
as required.
(b) True or false: Let (an ) be a convergent sequence and let (bn ) be a bounded sequence
(not necessarily convergent). Then (an · bn )n∈N is a convergent sequence.
This is false. For example take an = 1 for all n (which is certainly convergent) and
bn = (−1)n (which is certainly bounded). Then (an bn ) = ((−1)n ) which is our favourite
non-convergent sequence.
2
Question 3*: Probably I should have given the following hint on the question sheet: b
and c are the roots of the equation x2 − x − 1 = 0.
Define (an )n∈N inductively by a1 = 1, a2 = 3 and, for n ≥ 1, an+2 = an+1 + an .
√
√
1+ 5
1− 5
(a) Let b =
and c =
. Prove that ∀n ∈ N, P(n) : an = bn + cn .
2
2
Solution. Routinely a1 = b + c and a2 = 3 = b2 + c2 . So for some k ≥ 2 assume that
P(k) and P(k − 1) are true. Then
ak+1 = ak + ak−1 = bk + ck + bk−1 + ck−1 = bk−1 (b + 1) + ck−1 (c + 1).
But b and c are the roots of the equation x2 − x − 1 = 0 and so b + 1 = b2 and c + 1 = c2 .
Thus
ak+1 = bk−1 (b + 1) + ck−1 (c + 1) = bk+1 + ck+1 .
Thus, P(k + 1) holds and, by induction P(n) holds for all n.
an+1
.
(b) Find lim
n→∞ an
√
5
Solution. Here we first compute that cb = · · · = 6−2
and hence that | cb | ≤ 12 . This
−4
c n
means that, by the Sandwich Rule, lim ( b ) → 0 as n → ∞.
n→∞
This gives us the clue about how to manipulate:
an+1
n→∞ an
Hence lim
b + ( cb )n · c
an+1
bn+1 + cn+1
=
=
.
an
bn + c n
1 + ( cb )n
√
b+0·c
1+ 5
=
=b=
.
1+0
2
3