Solutions to MATH 152 Fall 2008 Exam 3B
1.
A n→∞
lim (a2n − 3bn ) = ( lim an )2 − 3 lim bn = 22 + 3(−3) = −5.
n→∞
n→∞
2.
x
1
lim
= lim 1 = ∞
E Apply L'Hospital's Rule to x→∞
ln x x→∞
3.
E Complete
the square: (x
√
2
3.
r=
4.
x
− 2x + 1) + y + z = 2 + 1; (x − 1)2 + y 2 + z 2 = 3, so r2 = 3 and
B Using the Comparison Test.
essarily mean that
∞
X
2
2
(D) is NOT neessarily true beause lim an = 0 does not nen→∞
an is onvergent.
n=1
5.
n
n
= 1, the sequene (−1)n
A Sine n→∞
lim
alternates between −1 and 1, therefore the
n+1
n+1
terms of the series do not approah 0, whih means the series diverges by the Test for Divergene.
6.
bn
lim
= 1, whih means the series
D Let an = n1 . Then n→∞
an
or both diverge. Sine
Comparison Test.
∞
X
∞
X
bn and
n=1
an diverges (by Integral Test or P-Test),
n=1
∞
X
an either both onverge
n=1
∞
X
bn diverges by the Limit
n=1
7.
B
8.
B Sine the terms of both series approah zero, both ∞series onverge by ∞the Alternating Series
∞ X
1
1
1
, whih is a Telesoping Series:
=
−
(n + 3)(n + 4) n=1 n + 3 n + 4
n=1
1
1 1
1 1
1
1
1 1
+
+
+ ···+
= as N → ∞.
−
−
−
−
sN =
4 5
5 6
6 7
N +3 N +4
4
∞
X
1
. Both are P 4/3
3/4
n
n
n=1
n=1
series; in (IA), P > 1so the series onverges, and in (IIA), P < 1 so the series diverges. Therefore,
Test. To test absolute onvergene, we look at (IA)
X
1
and (IIA)
X
series (I) onverges absolutely, and series (II) onverges but not absolutely.
9.
A The series an be written as
∞ n−1
X
2
2
9
n=1
2
2
r = . The sum of the series is 9
3
1−
10.
E
2
3
The Malaurin series for cos x is
∞
∞
X
X
(−1)n (x2 )2n
(−1)n x4n
.
=
(2n)!
(2n)!
n=0
n=0
3
, whih is a geometri series with a =
2
and
9
2
= .
3
∞
X
(−1)n x2n
.
(2n)!
n=0
So the Malaurin series for cos(x2 ) is
11. (a) AB =< 2, 1, −2 >, BC =< 1, 0, 1 >. AB · BC = (2)(1) + (1)(0) + (−2)(1) = 0, so the sides
are perpendiular to eah other.
(b) |AB| =
√
3 2
.
2
p
√
√
1
22 + 12 + (−2)2 = 3, |BC| = 12 + 0 + 12 = 2, so the area is |AB||BC| =
2
1
∞
∞
X
X
xn
(−1)n
, so e−1 =
. Therefore, subtrating the rst term of the series gives us
n!
n!
n=0
n=0
∞
X
(−1)n
e−1 − 1 =
n!
n=1
12. ex =
13. f (−1) = 0; f ′ (x) = 8x3 + 1, so f ′ (−1) = −7; f ′′ (x) = 24x2 , so f ′′ (−1) = 24; f ′′′ (x) = 48x, so
f ′′′ (−1) = −48. The third degree Taylor Polynomial is f (−1) +
1)2 +
f ′′′ (−1)
(x + 1)3 = −7(x + 1) + 12(x + 1)2 − 8(x + 1)3 .
3!
f ′ (−1)
f ′′ (−1)
(x + 1) +
(x +
1!
2!
n+1
(x−1)
√
3n+1 √n+1 n
|x − 1|
<1
·√
14. (a) Applying the Ratio Test gives us absolute onvergene when lim (x−1)n = lim
n→∞ n→∞
3
n
+
1
√
n
3
n
and divergene when the limit is > 1. Sine the seond fration approahes 1, we have ab|x − 1|
< 1, |x − 1| < 3 whih makes the radius of onvergene
solute onvergene when
3
3.
(b) The series onverges when −3 < x − 1 < 3, −2 < x < 4. To nd the interval of onvergene,
∞
∞
X
X
(−1)n
(−3)n
√
√ , whih on=
n
3n n
n=1
n=1
∞
∞
X
X
1
3n
√
√ ,
=
verges by the Alternating Series test. When x = 4, the series beomes
n n
n
3
n=1
n=1
test the endpoints: When x = −2, the series beomes
whih diverges by the P-test or integral test. Therefore, the interval of onvergene is
−2 ≤ x < 4.
X
∞
(−1)n+1
∞ ∞
X
X
(−1)n+1
(−1)n+1 2n
S(x) dx =
x dx =
x dx = C +
x2n+1
15. (a)
(2n
+
1)!
(2n
+
1)!
(2n
+
1)(2n
+
1)!
n=1
n=1
n=1
2n+1
1/2
∞
∞
n+1
n+1
X
X
(−1)
1
(−1)
1/2
x2n+1 |0 =
S(x) dx =
(b)
.
(2n
+
1)(2n
+
1)!
(2n
+
1)(2n
+
1)!
2
0
n=1
n=1
2·4+1
9
1
1
1
1
=
.
() Sine the series is alternating |S−S3 | ≤ |a4 | =
(2 · 4 + 1)(2 · 4 + 1)! 2
9 · 9! 2
2n
2