Math 101 – SOLUTIONS TO WORKSHEET 22
SEQUENCES
1. Skill 1: expression for sequences
(1) For each of the following sequences, write a formula for the general term
(a) {1, 2, 3, 4, 5, 6, · · · }
Solution: an = n. 1
1
1
, 25
, 36
,···
(b) 1, 14 , 19 , 16
Solution: an = n12
(c) {3, 7, 11, 15, 19, · · · }
Solution: an = 4n − 1. 7
7
7
7
7
(d) 79 , 27
, 81
, 243
, 729
, 3187
,···
7
Solution:
a
=
.
n+1
n
3
5
3
7
1
9
5
4
5
6
7
8
9
10
(e) 21 , 12 , 38 , 14 , 32
, 32
, 128
, 32
, 512
, 512
· · · = 12 , 24 , 38 , 16
, 32
, 64
, 128
, 256
, 512
, 1024
,···
Solution: an = 2nn .
(f) {1, −1, 1, −1, 1, −1, 1, −1, 1, −1, 1, −1, · · · }
Solution: an = (−1)n−1 .
2
5
4
7
6
9
8
11
(g) 0, 38 , 27
, 64
, 125
, 216
, 343
, 512
, 729
, 1000
,···
n
Solution: an = n+(−1)
n3
2. Skill 2: limits of sequences
(2) Determine
∞ if the sequences is convergent of divergent. If convergent, evaluate the limit.
(a) n1 n=1
Solution: limn→∞ n1 = limx→∞ x1 = 0 .
n
o∞
n
(b) n+1
n=1
Solution: limn→∞
(c)
n
n+1
= limx→∞
∞
{sin(n)}n=5
1
1
1+ x
=
1
1+0
= 1.
Solution:
∞ The function oscillates and the sequence is divergent.
sin( n1 ) n=1
Solution: Since the sine function is continouous we have limn→∞ sin
sin 0 = 0 .
(3) Further problems
n
(a) Does limn→∞ √n+1000
exist?
(d)
√ √
√
n n
n
= √n+1000
= √ n1000 −−−−→ ∞.
n+1000
1+ n
n→∞
1
√ n
q
q 1
=
=
.
We
can
see that
n
1
1000
n+1000
+ 1000
n + n2
n2
n2
Solution: No. We have
Solution:
No.
1
n
= sin limn→∞
1
n
=
√
the denominator tends to
0 so the sequence diverges to ∞.
(b) limn→∞ 2nn =
Solution: We have limn→∞ 2nn = limx→∞ 2xx = limx→∞ (log12)2x = 0 by l’Hopital.
∞
(c) (Math 103 final, 2014) Consider the sequence {an }n=1 = 1, 0, 21 , 0, 0, 31 , 0, 0, 0, 14 , 0, 0, 0, 0, 51 , · · · .
Decide whether limn→∞ an = 0.
Solution: Yes, the limit is zero.
Date: 2/3/2016, Worksheet by Lior Silberman. This instructional material is excluded from the terms of UBC Policy 81.
1
3. Tool: Squeeze Theorem
(4) Determine if the sequences
is convergent
of divergent. If convergent, evaluate the limit.
∞
1
n
(a) (Final 2013) (−1) sin n n=1 .
Solution: For n ≥ 1, sin n1 ≥ 0 so
1
1
1
n
− sin
≤ (−1) sin
≤ sin
.
n
n
n
We have seen in 1(d) that limn→∞ sin n1 = 0 and it follows that limn→∞ − sin
− limn→∞ sin n1 = 0 as well. By the squeeze theorem we conclude that
1
n
= 0.
lim (−1) sin
n→∞
n
o∞
n
sin(n)
(why do we have n ≥ 2 here?)
(b) (Final 2011) log(n)
n=2
Solution: Since limn→∞ log(n) = limx→∞ log(x) = ∞, we have limn→∞
every n we have −1 ≤ sin n ≤ 1 so that
1
sin n
1
−
≤
≤
.
log n
log n
log n
Since limn→∞ − log1 n = − limn→∞
1
log n
1
log n
1
n
=
= 0. Also, for
= 0 also, we have by the squeeze theorem that
sin n
= 0.
n→∞ log n
lim
(c) (Math 105 Final 2012) an = 1 +
n! sin(n3 )
(n+1)! .
Solution:
We have (n + 1)! = n!(n + 1) so an = 1 +
−1 ≤ sin(n3 ) ≤ 1 so that
sin(n3 )
n+1 ,
and for every n we have
sin(n3 )
1
1
≤1+
≤1+
.
1−
n+1
n+1
n+1
1
= 1 ± limx→∞ x1 = 1 and it follows from the squeeze theorem that
Now limn→∞ 1 ± n+1
lim 1 +
n→∞
n! sin(n3 )
= 1.
(n + 1)!
2