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Worksheet 10.1.
Sequences
1. Calculate the first four terms of the sequence bn = cos πn, starting with n = 1.
2. Calculate the first four terms of the sequence bn = 2 + (−1)n , starting with n = 1.
3. Use Theorem 1 to determine the limit of the sequence bn =
sequence diverges.
3n + 1
or state that the
2n + 4
4. Use Theorem 1 to determine the limit of the sequence cn = 4(2n ) or state that the
sequence diverges.
361
en
or show that the sequence diverges (justifying
2n
each step using the appropriate Limit Laws or Theorems).
5. Determine the limit of the sequence yn =
√
n
or show that the sequence diverges
n+4
(justifying each step using the appropriate Limit Laws or Theorems).
6. Determine the limit of the sequence an = √
2
7. Determine the limit of the sequence bn = en −n or show that the sequence diverges
(justifying each step using the appropriate Limit Laws or Theorems).
3 − 4n
or show that the sequence diverges
2 + 7 · 4n
(justifying each step using the appropriate Limit Laws or Theorems).
8. Determine the limit of the sequence bn =
9. Show that an =
3n2
is strictly increasing. Find an upper bound.
n2 + 2
362
Solutions to Worksheet 10.1
1. Calculate the first four terms of the sequence bn = cos πn, starting with n = 1.
Setting n = 1, 2, 3, 4 into the formula for bn gives:
b1
b2
b3
b4
= cos π · 1 = cos π = −1
= cos π · 2 = cos 2π = 1
= cos π · 3 = cos 3π = −1
= cos π · 4 = cos 4π = 1
2. Calculate the first four terms of the sequence bn = 2 + (−1)n , starting with n = 1.
The first four terms of {bn } are obtained by setting n = 1, 2, 3, 4 in the formula for bn :
b1
b2
b3
b4
= 2 + (−1)1
= 2 + (−1)2
= 2 + (−1)3
= 2 + (−1)4
=2−1=1
=2+1=3
=2−1=1
=2+1=3
3. Use Theorem 1 to determine the limit of the sequence bn =
3n + 1
or state that the
2n + 4
sequence diverges.
Using Theorem 1 and the asymptotic behavior of rational functions we get:
3x + 1
3n + 1
3
= lim
=
lim
x→∞ 2x + 4
n→∞ 2n + 4
2
4. Use Theorem 1 to determine the limit of the sequence cn = 4(2n ) or state that the
sequence diverges.
By Theorem 1,
lim 4 · 2n = lim 4 · 2x = ∞.
n→∞
x→∞
Thus, the sequence 4 (2n ) diverges.
en
or show that the sequence diverges (justifying
2n
each step using the appropriate Limit Laws or Theorems).
e
en ! e "n
=
and
> 1. By the Limit of Geometric Sequences, proved in Example 6,
2n
2
2 ! "n
e
we conclude that lim
= ∞. Thus, the given sequence diverges to ∞.
n→∞ 2
√
n
6. Determine the limit of the sequence an = √
or show that the sequence diverges
n+4
(justifying each step using the appropriate Limit Laws or Theorems).
5. Determine the limit of the sequence yn =
363
We compute the limit using Theorem 1:
√
√
√
√x
1
1
n
x
x
=
=1
= lim
lim √
= lim √
= lim √x
4
4
x→∞ 1 + √
n→∞ n + 4
x→∞ x + 4
x→∞ √ + √
1+0
x
x
x
2
7. Determine the limit of the sequence bn = en −n or show that the sequence diverges
(justifying each step using the appropriate
%
&Limit Laws or Theorems).
# 2
$
1
We have lim x − x = lim x2 1 −
= ∞. Since the exponential function ex is
x→∞
x→∞
x
2
continuous, it follows from Theorem 4 that lim ex −x = ∞. Thus, by Theorem 1, we
n2 −n
conclude lim e
n→∞
x→∞
n2 −n
= ∞; that is, the sequence e
diverges.
3 − 4n
or show that the sequence diverges
2 + 7 · 4n
(justifying each step using the appropriate Limit Laws or Theorems).
Dividing the numerator and denominator by 4n yields:
8. Determine the limit of the sequence bn =
3 − 4n
=
an =
2 + 7 · 4n
n
3
− 44n
4n
n
2
+ 7·4
4n
4n
=
3
4n
2
4n
−1
+7
We now compute the limit using Limit Laws
for Sequences and the limit of the geometric
% &
n
1
1
= 0. We obtain:
sequence (see Example 6) lim n = lim
n→∞ 4
n→∞ 4
#
$
3
lim 43n − 1
3 lim 41n − lim 1
−
1
n→∞
n→∞
4n
#2
$ = n→∞ 1
lim an = lim 2
=
n→∞
n→∞ n + 7
−
lim 7
2 lim 4n
lim 4n + 7
4
n→∞
=
n→∞
n→∞
3·0−1
1
=−
2·0+7
7
3n2
is strictly increasing. Find an upper bound.
n2 + 2
3x2
We consider the function f (x) = 2
. Differentiating f yields:
x +2
6x (x2 + 2) − 3x2 · 2x
12x
f ′ (x) =
=
2
(x2 + 2)
(x2 + 2)2
9. Show that an =
f ′ (x) > 0 for x > 0, hence f is strictly increasing on this interval. It follows that
an = f (n) is also strictly increasing. Next, we show that M = 3 is an upper bound
for an by writing
3n2 + 6
3 (n2 + 2)
3n2
≤ 2
=
= 3.
an = 2
n +2
n +2
n2 + 2
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That is, an ≤ 3 for all n.