MATH 101, Section 212 (CSP)
Week 8: Marked Homework Solutions
2011 Mar 10
1. (a) [3] The limit as n → ∞
√ does not exist, the sequence is divergent. Divide both
numerator and denominator by n to get
√
√
2 n + √1n
2 n
2n + 1
√ =
≈
an =
for large positive n.
√1 − 3
1−3 n
−3
n
√
As n becomes larger and larger (“goes to ∞”), n increases without bound, √1n approaches
0, therefore an decreases without bound (“goes to −∞”) and there can be no finite number
L such that an is arbitrarily close to L for every sufficiently large n.
(b) [3] limn→∞ an = 1, the sequence is convergent. Divide both numerator and denominator
by n to get
n+1
1 + (−1)n
n + (−1)n+1
an =
=
for any positive integer n
n
1
(−1)n+1 = 1 if n is odd, (−1)n+1 = −1 if n is even. In either case
−1 ≤ (−1)n+1 ≤ 1
− n1 ≤
(−1)n+1
n
≤
1
n
for any positive integer n, therefore (since limn→∞ n1 = 0 and limn→∞ − n1
Squeeze Theorem
n+1
lim (−1)n = 0
= 0) by the
n→∞
and then by the Limit Laws we get limn→∞ an =
1+0
1
= 1.
(c) [3] The limit as does not exist, the sequence is divergent. Writing out the first few terms,
a1 = 1 + (−1)1 = 0,
a2 = 1 + (−1)2 = 2,
a3 = 1 + (−1)3 = 0,
a4 = 1 + (−1)4 = 2,
...
we realize that the sequence is just 0s and 2s,
{an } = {0, 2, 0, 2, 0, 2, 0, 2, . . . }.
There is no unique finite number L such that an is arbitrarily close to L for every sufficiently
large n.
(d) [3] limn→∞ an = 1, the sequence is convergent. Since the sine of any number is always
between −1 and 1, we have
−1 ≤ sin n ≤ 1
− n1 ≤
sin n
n
1
≤
1
n
for any positive integer n, therefore (since limn→∞ n1 = 0 and limn→∞ − n1
Squeeze Theorem (p. 679)
lim sinn n = 0.
= 0) by the
n→∞
(e) [3] The limit does not exist, the sequence is divergent. Similar to suggested homework
problem 11.1.45: since n → ∞, consider n > 106 = 1, 000, 000. Then
1,000,000
1,000,001 1,000,002
n!
n
2
n−1
1
an = 1,000,000
·
· 1,000,000
·
·
·
·
·
·
·
·
·
·
·
·
n =
1,000,000 1,000,000
1,000,000
1,000,000 1,000,000
1,000,000
The product of the first 1, 000, 000 terms is c = 1, 000, 000 ! · 10−6,000,000 , which is a positive
constant (it doesn’t matter how small it is, as long as it is positive). The next n − 1, 000, 001
terms are numbers which are all greater than or equal to 1, so their product is greater than
or equal to 1. Therefore
n
an ≥ c · 1 · 1,000,000
and the right hand side increases without bound (“goes to ∞”) for any positive constant c.
Therefore an also increases without bound (“goes to ∞”), so there can be no finite number
L such that an is arbitrarily close to L for every sufficiently large n.
2. [4] If every an is the same value, then an+1 = an for n = 1, 2, 3, . . . , and an = 4an (1 − an ).
After some algebra we then get
(4an − 3)an = 0,
thus an = 0 or an = 43 for n = 1, 2, 3, . . . and in particular, for n = 1. So {an } is a constant
sequence if and only if
a1 = 0 or a1 = 43 .
3. If a is positive and a < 2, then taking square roots we have
√
√
a < 2.
√
√
(a) [1] Now multiply both sides by the positive number a to get a < 2a.
√
√
(b) [1] On the other hand, multiplying both sides by 2 gives 2a < 2.
√
√
(c) [5] a1 = √ 2 is positive and a1 < 2. Therefore by (a) and (b) we have a1 < 2a1 < 2, i.e.
(since a2 = 2a1 )
a1 < a2 < 2.
.
√
Now √
a2 is positive and a2 < 2. Therefore by (a) and (b) we have a2 < 2a2 < 2, i.e. (since
a3 = 2a2 )
a2 < a3 < 2.
Now √
a3 is positive and a3 < 2. Therefore by (a) and (b) we have a3 <
a4 = 2a3 )
a3 < a4 < 2,
2
√
2a3 < 2, i.e. (since
“etc.” Intuitively, we can see that the same argument would work for every n (the precise
proof of “etc.” would use a technique called the “Principle of Mathetical Induction” on p.
77, as used in Example 11.1.13, p. 683, but you do not need to know this for MATH 101),
so we have
an < an+1 < 2 for every n = 1, 2, 3, . . . .
The inequality on the left says that {an } is increasing, and the inequality on the right says
that {an } is bounded above, by 2. Furthermore, every an is positive so the sequence is
bounded below, by 0. Then by the Monotonic Sequence Theorem, {an } is convergent.
(d) [3] By part (c), limn→∞ an = L exists. Therefore we also√have lim√n→∞ an+1 = L. On the
other hand, by Limit Laws we have limn→∞ an+1 = limn→∞ 2an = 2L, therefore
√
L = 2L.
√
√
√
Since L > 0 (why?) we can divide by L to get L = 2, therefore L = 2:
lim an = 2.
n→∞
q p
p √
√
3/4
(Alternatively, observe that a1 = 2 = 2 , a2 = 2 2 = 2 , a3 = 2 2 2 = 27/8 ,
n
“etc.” so, using the Principle of Induction again if we want a precise proof, an = 21−(1/2 )
and now use Limit Laws.)
4. (a) [5] Since both 21 < 1 and − 15 < 1, we have the two convergent geometric series
√
∞
X
1 n
2
n=0
(with a = 1 and r =
∞
X
n=0
1
2
− 51
1/2
= 1 + 12 +
1 2
2
+
1 3
2
+··· =
1
1−
1
2
=2
and
n
2
3
= 1 + − 15 + − 15 + − 15 + · · · =
1
=
1 − − 51
5
6
(with a = 1 and r = − 51 ). By Theorem 11.2.8 on p. 693, the given series is convergent with
∞ X
n=0
(b) [3] Since limn→∞
have
1
n
1
2n
+
(−1)n
5n
=
∞
X
n=0
1 n
2
+
∞
X
n=0
− 51
n
=2+
5
6
=
17
.
6
= 0 and cos x is continuous at x = 0, by Theorem 11.1.7, p. 680 we
lim cos n1 = cos 0 = 1,
n→∞
which is not 0, and therefore by the Test for Divergence, p. 692, the series
divergent.
3
P∞
n=1
cos n1 is
(c) [5] Use partial fractions to see that each partial sum is a telescoping sum. Put
A
B
C
D
6n + 3
= + 2+
+
2
+ 1)
n n
n + 1 (n + 1)2
n2 (n
and multiply by n2 (n + 1)2 to get
6n + 3 = An(n + 1)2 + B(n + 1)2 + Cn2 (n + 1) + Dn2 .
Substituting n = 0 gives B = 3, substituting n = −1 gives D = −3, taking the coefficient of
n3 on both sides gives 0 = A + C or C = −A, and taking the coefficient of n2 on both sides
gives 0 = 2A + B + C + D = 2A + 3 − A − 3 thus A = 0 and then C = 0. So we have
∞
∞ h
i
X
X
3
3
6n+3
.
=
−
n2 (n+1)2
n2
(n+1)2
n=1
n=1
Now writing out some partial sums we have
s1 =
s2 =
s3 =
3
12
and in general
sn =
3
12
−
3
22
+
3
22
−
−
3
12
3
22
3
32
−
+
+
3
22
3
22
3
32
3
12
−
3
,
22
3
22
3
32
−
3
32
3
32
+
−
−
3
42
+
=3−
−
+···+
3
42
3
n2
3
,
32
=3−
−
3
,
42
3
(n+1)2
=3−
3
(n+1)2
because every term except the first term and the last term cancels. Therefore
3
s = lim sn = lim 3 − (n+1)2 = 3 − 0 = 3
n→∞
n→∞
exists (as a unique finite number). The series is convergent, and its sum is
∞
X
6n+3
n2 (n+1)2
= 3.
n=1
5. [5] This is similar to Example 11.2.4, p. 690. Use the geometric series.
1.24123123123123 . . . = 1 +
=
124
100
=
124
100
2
101
4
102
+
+
+
123
105
+
∞
X
123
105
1−
1
103
=
124
100
+
=
124
100
+
=
123999
99900
=
41333
.
33300
+
123
105
123
105
123
+ 10
8 +
123
1
+ 105
103
1 n−1
3
10
n=1
123
105
123
99900
4
123
1011
123
+ 10
14 + . . .
1 2
1 3
+ 123
+...
103
105 103
P
P∞
P∞
n n
n
n−1
6. (a) [4] For any fixed x, this is a geometric series ∞
2
x
=
(2x)
=
n=1
n=1
n=1 (2x)(2x)
(with a = 2x and r = 2x), so the series is convergent if and only if |2x| < 1, i.e.
|x| <
and its sum is
∞
X
1
2
2x
, if |x| < 12 .
1 − 2x
n=1
P
n
2
3
(b) [4] For any fixed x, this is a geometric series ∞
n=0 sin x = 1+(sin x)+(sin x) +(sin x) +
. . . (with a = 1 and r = sin x), so the series is convergent if and only if | sin x| < 1, i.e.
x 6=
(2k+1)π
2
(2x)(2x)n−1 =
for all (positive, negative and zero) integers k
and its sum is
∞
X
n=0
sinn x =
1
,
1 − sin x
if x 6=
5
(2k+1)π
2
for all integers k.